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1,101
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25
| 3
|
There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$
|
Let's clear up the fractions: \[\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}\] \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] Notice that if we divide everything by $7$ then we would have: \[514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7\] Since $0 \le a_7<7$ and $a_7$ must be an integer, then we have $\frac{2}{7}=\frac{1}{7}a_7$ , so $a_7=2$
Similarly, if we divide everything by $6$ , then we would have: \[85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6\] Again, since $0 \le a_6<6$ and $a_6$ must be an integer, we have $\frac{4}{6}=\frac{1}{6}a_6$ , so $a_6=4$
The pattern repeats itself, so in the end we have $a_2=1$ $a_3=1$ $a_4=1$ $a_5=0$ $a_6=4$ $a_7=2$ . So $a_2+a_3+a_4+a_5+a_6+a_7=\boxed{9}$
| 9
|
1,102
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25
| 4
|
There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$
|
By multiplying both sides by $7$ we get
\[5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
since $0 \le a_2 < 2$ , if $a_2 = 0$ the rest of the right hand side will not add up to be $5$ , so $a_2 = 1$
\[\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_3 = 2$ $\frac76 a_3 = \frac73 > \frac32$ , so $a_3 = 1$
\[\frac13 = \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_4 = 2$ $\frac{7}{24} a_4 = \frac{7}{12} > \frac13$ , so $a_4 = 1$
\[\frac{1}{24} = \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_5 = 1$ $\frac{7}{120} a_5 = \frac{7}{120} > \frac{1}{24}$ , so $a_5 = 0$
\[\frac{1}{24} = \frac{7}{720} a_6 + \frac{a_7}{720}\]
$30 = 7 a_6 + a_7$ . Since $a_7 < 7$ $a_7 = 2$ and $a_6=4$
Therefore, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \boxed{9}$
| 9
|
1,103
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_26
| 1
|
Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$
|
We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\circ}$
Let the number of sides in our polygons be $3\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$
The integral angle of a regular $k$ -gon is $180 \frac{k-2}k$ . Therefore we are looking for integer solutions to:
\[360 = 180\left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Which can be simplified to:
\[2 = \left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Furthermore, we know that two of the polygons are congruent, thus WLOG $a=c$ . Our equation now becomes
\[2 = \left( 2\cdot\frac{a-2}a + \frac{b-2}b \right)\]
Multiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$
Using the standard technique for Diophantine equations , we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$
Remembering that $a,b\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$ $(2,4)$ $(4,2)$ , and $(8,1)$
These correspond to the following pairs $(a,b)$ $(5,10)$ $(6,6)$ $(8,4)$ , and $(12,3)$
The perimeters of the resulting polygon for these four cases are $14$ $12$ $14$ , and $\boxed{21}$
| 21
|
1,104
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_26
| 2
|
Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$
|
We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\boxed{21}$ sides.
| 21
|
1,105
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_28
| 1
|
Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$
|
Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in\{-1,1,2\}$
Also, order of the $x_i$ does not matter, so we are only interested in the counts of the variables of each type. Let $a$ of the $x_i$ be equal to $-1$ $b$ equal to $1$ , and $c$ equal to $2$
The conditions (ii) and (iii) simplify to: (ii) $2c + b - a = 19$ (iii) $4c + a + b = 99$ and we want to find the maximum and minimum of $2^3c + 1^3b + (-1)^3a = 8c + b - a$ over all non-negative solutions of the above two equations.
Subtracting twice (ii) from (iii) we get $b = 3a-61$ . By entering that into one of the two equations and simplifying we get $c=40-a$
Thus all the solutions of our system of equations have the form $(a,3a-61,40-a)$
As all three variables must be non-negative integers, we have $3a-61 \geq 0 \Rightarrow a \geq 21$ and $40-a\geq 0 \Rightarrow a \leq 40$
For $(a,b,c)$ of the form $(a,3a-61,40-a)$ the expression we are maximizing/minimizing simplifies to $320-8a+3a-61-a = 259 - 6a$ . Clearly, the maximum is achieved for $a=21$ and the minimum for $a=40$ . Their values are $M=133$ and $m=19$ , and therefore $\frac Mm = \boxed{7}$
| 7
|
1,106
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_28
| 2
|
Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$
|
Let $a =$ number of $2$ s, $b =$ number of $1$ s, $c =$ number of $-1$
\[4a+b+c=99\] \[2a+b-c=19\]
Multiplying the second equation by $2$ gives $4a+2b-2c=38$ . By subtracting this equation from the first equation we get $3c-b=61$ $3c=61+b$ . As we need to minimize the value of $c$ $c = 21$ $b = 2$ $a = 19$
$x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133$
Therefore, $\frac Mm = \frac{133}{19}=\boxed{7}$
| 7
|
1,107
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_2
| 1
|
Letters $A,B,C,$ and $D$ represent four different digits selected from $0,1,2,\ldots ,9.$ If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$
$\mathrm{(A) \ }13 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ }16 \qquad \mathrm{(E) \ } 17$
|
If we want $\frac{A+B}{C+D}$ to be as large as possible, we want to try to maximize the numerator $A+B$ and minimize the denominator $C+D$ . Picking $A=9$ and $B=8$ will maximize the numerator, and picking $C=0$ and $D=1$ will minimize the denominator.
Checking to make sure the fraction is an integer, $\frac{A+B}{C+D} = \frac{17}{1} = 17$ , and so the values are correct, and $A+B = 17$ , giving the answer $\boxed{17}$
| 17
|
1,108
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_3
| 1
|
If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which
then $\texttt{a+b+c =}$
$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$
|
Working from right to left, we see that $2 - b = 3$ . Clearly if $b$ is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from $a$ . Borrow $1$ from the digit $a$ , and you get $12 - b = 3$ , giving $b = 9$
Since $1$ was borrowed from $a$ , we have from the tens column $(a-1) - 8 = 7$ . Again for single digit integers this will not work. Again, borrow $1$ from $7$ , giving $10 + (a-1) - 8 = 7$ . Solving for $a$
$10 + a - 1 - 8 = 7$
$1 + a = 7$
$a = 6$
Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$ , giving $c = 2$
As a check, the problem is $762 - 489 = 273$ , which is a true sentence.
The desired quantity is $a + b + c = 6 + 9 + 2 = 17$ , and the answer is $\boxed{17}$
| 17
|
1,109
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_5
| 1
|
If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$
$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$
|
Divide both sides of the original equation by $2^{1995}$ , giving:
$2^3 - 2^2 - 2^1 + 1 = k$
$k = 3$ , and the answer is $\boxed{3}$
| 3
|
1,110
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_6
| 1
|
If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$
|
If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.
Since $45^2 = 2025$ , the factors should be as close to $44$ or $45$ as possible.
Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$
$37$ is relatively close to $44$ , and no numbers between $38$ and $44$ are factors of $1998$ . Thus, the two factors are $37$ and $2\cdot 3^3 = 54$ , and the difference is $54 - 37 = 17$ , and the answer is $\boxed{17}$
| 17
|
1,111
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_8
| 1
|
A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac 78$
|
The area of the trapezoid is $\frac{1}{3}$ , and the shorter base and height are both $\frac{1}{2}$ . Therefore, \[\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{56}\]
| 56
|
1,112
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_8
| 2
|
A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac 78$
|
Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}$ , we have \[b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})\] \[x+\frac{1}{2}=2((1-x)+\frac{1}{2})\] \[x=\frac{5}{6}\boxed{56}\]
| 56
|
1,113
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_9
| 1
|
A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by members of the audience?
$\mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36$
|
Assume that there are $100$ people in the audience.
$20$ people heard $60$ minutes of the talk, for a total of $20\cdot 60 = 1200$ minutes heard.
$10$ people heard $0$ minutes.
$\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\cdot 20 = 700$ minutes.
$35$ people heard $40$ minutes of the talk, for a total of $1400$ minutes.
Altogether, there were $1200 + 0 + 700 + 1400 = 3300$ minutes heard among $100$ people.
Thus, the average is $\frac{3300}{100} = 33$ minutes, and the answer is $\boxed{33}$
| 33
|
1,114
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_10
| 1
|
A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$
|
Let the length of the longer side be $x$ , and the length of the shorter side be $y$ . We are given that $2x+2y=14\implies x+y=7$ . However, note that $x+y$ is also the length of a side of the larger square. Thus the area of the larger square is $(x+y)^2=7^2=\boxed{49}$
| 49
|
1,115
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_10
| 2
|
A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$
|
Expand the small square so it basically equals the area of the large square. Two of the sides of the rectangles shrink to zero. The other two sides expand to equal the length of the large outer square, and have a length of $\frac{14}{2} = 7$ . Thus, the area of the larger square is $7^2=\boxed{49}$
| 49
|
1,116
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_11
| 1
|
Let $R$ be a rectangle. How many circles in the plane of $R$ have a diameter both of whose endpoints are vertices of $R$
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$
|
There are $6$ pairs of vertices of $R$ . However, both diagonals determine the same circle, therefore the answer is $\boxed{5}$
| 5
|
1,117
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_15
| 1
|
A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?
$\mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6$
|
$A_{\triangle} = \frac{s_t^2\sqrt{3}}{4}$
$A_{hex} = \frac{6s_h^2\sqrt{3}}{4}$ since a regular hexagon is just six equilateral triangles.
Setting the areas equal, we get:
$s_t^2 = 6s_h^2$
$\left(\frac{s_t}{s_h}\right)^2 = 6$
$\frac{s_t}{s_h} = \sqrt{6}$ , and the answer is $\boxed{6}$
| 6
|
1,118
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_19
| 1
|
How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$
|
The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$ . The length of the base is $10$ , thus the height must be $2$ . The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant.
| 4
|
1,119
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_20
| 1
|
Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers." Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers." Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?
$\textrm{(A)}\ 2 \qquad \textrm{(B)}\ 3 \qquad \textrm{(C)}\ 4 \qquad \textrm{(D)}\ 5 \qquad \textrm{(E)}\ \text{There is not enough information to determine the number.}$
|
Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$ $(1,3,9)$ $(1,4,8)$ $(1,5,7)$ $(2,3,8)$ $(2,4,7)$ $(2,5,6)$ , and $(3,4,6)$
If Casey saw the number $3$ , she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$
At this moment, if the last card contained a $10$ $9$ , or a $6$ , Tracy would know the other two numbers. (Note that Tracy is aware of the fact that $(3,4,6)$ was eliminated. If she saw the number $6$ , she would know that the other two are $2$ and $5$ .) This eliminates three more possibilities.
Thus before Stacy took her look, we are left with four possible cases: $(2,4,7)$ $(1,4,8)$ $(1,5,7)$ , and $(2,3,8)$ . As Stacy could not find out the exact combination, the middle number must be $\boxed{4}$
| 4
|
1,120
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_21
| 1
|
In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?
$\mathrm{(A) \ } \frac dh \qquad \mathrm{(B) \ } 0 \qquad \mathrm{(C) \ } \frac {d^2}h \qquad \mathrm{(D) \ } \frac {h^2}d \qquad \mathrm{(E) \ } \frac{d^2}{h-d}$
|
Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\frac sw = \frac h{h-d}$ . In the second race, Sunny's track length is $h+d$ . She will finish this track in $\frac{h+d}s$ . In this time, Windy will run the distance $w\cdot \frac{h+d}s = \frac{(h+d)(h-d)}h$ . This is less than $h$ , therefore Sunny is ahead. The exact distance between Windy and the finish is $h - \frac{(h+d)(h-d)}h = \frac{ h^2 - (h^2-d^2) }h = \boxed{2}$
| 2
|
1,121
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_23
| 1
|
The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$ , and for no other values of $k$ . Find $b-a$
$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$
|
Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$
The first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$ , which is a circle centered at $(6,3)$ with radius $7$
The second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$ , which is a circle centered at $(2,6)$ with radius $r=\sqrt{40+k}$
The distance between the two centers is $5$ , and therefore the two circles intersect if $2\leq r \leq 12$
From $\sqrt{40+k} \geq 2$ we get that $k\geq -36$ . From $\sqrt{40+k}\leq 12$ we get $k\leq 104$
Therefore $b-a = 104 - (-36) = \boxed{140}$
| 140
|
1,122
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_28
| 1
|
In triangle $ABC$ , angle $C$ is a right angle and $CB > CA$ . Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$ . If $AC/AD = 2/3$ , then $CD/BD = m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26$
|
Let $AC = 2$ $AD = 3$ $\cos \angle CAD = \frac23$
By the pythagorean theorem $CD = \sqrt{3^2-2^2} = \sqrt{5}$
$\sin \angle BDA = \sin (180^{\circ} - \angle BDA) = \sin \angle CDA = \cos \angle (90^{\circ} - CDA) = \cos \angle CAD = \frac23$
$\sin \angle BAD = \sqrt{ \frac{1-cos (2\angle BAD)}{2} } = \sqrt{ \frac{1-\cos \angle CAD}{2} } = \sqrt{ \frac{1-\frac23}{2} } = \frac{\sqrt{6}}{6}$
By the Law of Sine, $\frac{ \sin \angle BDA }{AB} = \frac{ \sin \angle BAD }{BD}$
$\frac{ \frac23 }{ \sqrt{2^2 + ( \sqrt{5} + BD)^2} } = \frac{ \frac{\sqrt{6}}{6} }{BD}$
$8BD^2 = 3(9+ 2BD \sqrt{5} + BD^2)$
$5BD^2 - 6 BD \sqrt{5} -27=0$
As $BD>0$ $BD = \frac{6 \sqrt{5} + \sqrt{ (6 \sqrt{5})^2 - 4 \cdot 5 (-27) } }{10} = \frac{9\sqrt{5}}{5}$
$\frac{CD}{BD} = \frac{\sqrt{5}}{\frac{9\sqrt{5}}{5}} = \frac59$
$5+9=\boxed{14}$
| 14
|
1,123
|
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_30
| 1
|
For each positive integer $n$ , let
Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 9$
|
We have $a_n = n(n+1)\dots (n+9)$
The value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$ , where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \min(x_n,y_n)$ . The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$
Clearly, the last non-zero digit is even iff $x_n - z_n > 0$ iff $x_n > y_n$
Thus we are looking for the smallest $n$ such that the power of $5$ that divides $a_n$ is at least equal to the power of $2$ that divides $a_n$
The number $a_n$ is a product of $10$ consecutive integers. Out of these, $5$ are divisible by $2$ . Out of those $5$ , at least $2$ are divisible by $4$ , and out of those $2$ , one is divisible by $8$ . Therefore $x_n\geq 5+2+1=8$ for all $n$
On the other hand, exactly $2$ of our ten integers are divisible by $5$ , and at most one of them can be divisible by a higher power of $5$ . As we need $y_n\geq x_n\geq 8$ , one of the integers from $n$ to $n+9$ must be divisible by $5^7 = 78125$ . Therefore $n\geq 78116$
We can now take numbers starting with $78116$ , and write each of them in the form $2^x 5^y r$ . We are looking for 10 consecutive rows where the sum of $y$ s is at least equal to the sum of $x$ s.
At this point we can stop, as we just found out that $a_{78117}$ is of the form $2^8 5^8 r_{78117}$ . Therefore the $k$ we seek is $k=78117$
Now all we need to do is to compute the last non-zero digit. As the powers of $2$ and $5$ that divide $a_{78117}$ are equal, the last non-zero digit is simply the product of the last digits of the ten $z$ s. This is $7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}$
| 9
|
1,124
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_1
| 1
|
If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$
|
From the units digit calculation, we see that the units digit of $a\times 3$ is $9$ . Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$ . As a double-check, that does work, since $23 \times 3 = 69$ , which is the first line of the multiplication.
The second line of the multiplication can be found by doing the multiplication $23\times b = 92$ . Dividing both sides by $23$ gives $b=4$
Thus, $a + b = 3 + 4 = 7$ , and the answer is $\boxed{7}$
| 7
|
1,125
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_2
| 1
|
The adjacent sides of the decagon shown meet at right angles. What is its perimeter?
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; dot(origin);dot((12,0));dot((12,1));dot((9,1));dot((9,7));dot((7,7));dot((7,10));dot((3,10));dot((3,8));dot((0,8)); draw(origin--(12,0)--(12,1)--(9,1)--(9,7)--(7,7)--(7,10)--(3,10)--(3,8)--(0,8)--cycle); label("$8$",midpoint(origin--(0,8)),W); label("$2$",midpoint((3,8)--(3,10)),W); label("$12$",midpoint(origin--(12,0)),S);[/asy]
$\mathrm{(A)\ } 22 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 34 \qquad \mathrm{(D) \ } 44 \qquad \mathrm{(E) \ }50$
|
The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$
The four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$
Thus, the perimeter is $2(12+10) = 44$ , which is option $\boxed{44}$
| 44
|
1,126
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_3
| 1
|
If $x$ $y$ , and $z$ are real numbers such that
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \ } 8 \qquad \mathrm{(D) \ } 12 \qquad \mathrm{(E) \ }50$
|
If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \ge 0$ with the equality iff $a=0$
In this case, $x-3=0$ $y-4=0$ , and $z-5=0$ . Adding the three equations and moving the constant to the right gives $x + y + z = 12$ , and the answer is $\boxed{12}$
| 12
|
1,127
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_4
| 1
|
If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$
|
Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$
We want a relationship between $a$ and $b$ . Dividing the second equation into the first will cancel the $c$ , so we try that and get:
$\frac{a}{b} = \frac{1.5}{1.25}$
$\frac{a}{b} = \frac{150}{125}$
$\frac{a}{b} = \frac{6}{5}$
$a = 1.2b$
In this case, $a$ is $1.2 - 1 = 0.2 = 20\%$ bigger than $b$ , and the answer is $\boxed{20}$
| 20
|
1,128
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_4
| 2
|
If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$
|
Arbitrarily assign a value to one of the variables. Since $c$ is the smallest variable, let $c = 100$
If $a$ is $50\%$ larger than $c$ , then $a = 150$
If $b$ is $25\%$ larger than $c$ , then $b = 125$
We see that $\frac{a}{b} = \frac{150}{125} = 1.2$ So, $a$ is $20\%$ bigger than $b$ , and the answer is $\boxed{20}$
| 20
|
1,129
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_5
| 1
|
A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles? [asy] defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,3)--foot((r,3),(0,1),(4,1))); draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]
$\mathrm{(A)\ } 35.2 \qquad \mathrm{(B) \ }76 \qquad \mathrm{(C) \ } 80 \qquad \mathrm{(D) \ } 84 \qquad \mathrm{(E) \ }86$
|
Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$
From the large rectangle, we see that the top has length $3w$ , the right has length $l + w$ , the bottom has length $2l$ , and the left has length $l + 2$
Since the perimeter of the large rectangle is $176$ , we know that $172 = 3w + l + w + 2l + l + w$ , or $172 = 5w + 4l$
From the top and bottom of the large rectangle, we know that $3w = 2l$ , or $l = 1.5w$
Plugging that into the first equation, we get $176 = 5w + 4(1.5)w$
$176 = 11w$
$w = 16$
$l = 1.5w = 24$
$P = 2l + 2w = 2(16 + 24) = 80$ , and the answer is $\boxed{80}$
| 80
|
1,130
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_7
| 1
|
The sum of seven integers is $-1$ . What is the maximum number of the seven integers that can be larger than $13$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
|
If the first six integers are $14$ , the last number can be $(-14\cdot 6) - 1 = -85$ . The sum of all seven integers will be $-1$
However, if all seven integers are over $13$ , the smallest possible sum is $14\cdot 7 = 98$
Thus, the answer is $6$ , which is option $\boxed{6}$
| 6
|
1,131
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_8
| 1
|
Mientka Publishing Company prices its bestseller Where's Walter? as follows:
$C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le n\le 48\\ 10n, &\text{if }49\le n\end{matrix}\right.$
where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $25$ books cost less than $24$ books. For how many values of $n$ is it cheaper to buy more than $n$ books than to buy exactly $n$ books?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$
|
Clearly, the areas of concern are where the piecewise function shifts value.
Since $C(25) = 11\cdot 25 = 275$ , we want to find the least value of $n$ for which $C(n) > 275$
If $n \le 24$ , then $C(n) = 12n$ , so for $C(n) > 275$ $12n > 275$ , which is equivalent to $n > 22.91$ . Thus, both $n=23$ and $n=24$ will be more expensive than $n=25$
Since $C(49) = 10\cdot 49 = 490$ , we want to find the least value of $n$ for which $C(n) > 490$
If $25 \le n \le 48$ , then $C(n) = 11n$ , so for $C(n) > 490$ , we have $11n > 490$ , leading to $n > 44.5$ . Thus, $n=45, 46, 47, 48$ will be more expensive than $n=49$
Thus, there are $2 + 4 = \boxed{6}$
| 6
|
1,132
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_11
| 1
|
In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$ $14$ $11$ , and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$ , what is the least number of points she could have scored in the tenth game?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$
|
The sum of the scores for games $6$ through $9$ is $68$ . The average in these four games is $\frac{68}{4} = 17$
The total points in all ten games is greater than $10\cdot 18 = 180$ . Thus, it must be at least $181$
There are at least $181 - 68 = 113$ points in the other six games: games $1-5$ and game $10$
Games $1-5$ must have an average of less than $17$ . Thus we cannot put more than $16 + 17 + 17 + 17 + 17 = 84$ points in those five games.
Thus, the tenth game must have at least $113 - 84 = 29$ points, and the answer is $\boxed{29}$
| 29
|
1,133
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_13
| 1
|
How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
|
Let $N = 10t + u$ , where $t$ is the tens digit and $u$ is the units digit.
The condition of the problem is that $10t + u + 10u + t$ is a perfect square.
Simplifying and factoring, we want $11(t+u)$ to be a perfect square.
Thus, $t+u$ must at least be a multiple of $11$ , and since $t$ and $u$ are digits, the only multiple of $11$ that works is $11$ itself.
Thus, $(t,u) = (2,9)$ is the first solution, and $(t,u) = (9,2)$ is the last solution. There are $8$ solutions in total, leading to answer $\boxed{8}$
| 8
|
1,134
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_14
| 1
|
The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$ . If the populations in the years $1994$ $1995$ , and $1997$ were $39$ $60$ , and $123$ , respectively, then the population in $1996$ was
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102$
|
Let $x$ be the population in $1996$ , and let $k$ be the constant of proportionality.
If $n=1994$ , then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$
Translating this sentence, $(x - 39) = k(60)$
Similarly, letting $n=1995$ gives the sentence $(123 - 60) = kx$
Since $kx = 63$ , we have $k = \frac{63}{x}$
Plugging this into the first equation, we have:
$(x - 39) = \frac{60\cdot 63}{x}$
$x - 39 = \frac{3780}{x}$
$x^2 - 39x - 3780 = 0$
$(x - 84)(x + 45) = 0$
Since $x>0$ , we must have $x=84$ , and the answer is $\boxed{84}$
| 84
|
1,135
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_15
| 1
|
Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$ , and $CE=12$ . The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$
|
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair F = midpoint(B--C); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(A--F); draw(rightanglemark(B,G,E,3));[/asy]
One median divides a triangle into $2$ equal areas, so all three medians will divide a triangle into $6$ equal areas.
The median $CE$ is divided into a $2:1$ ratio at centroid $G$ , so $GE = \frac{1}{3}\cdot CE = \frac{1}{3}\cdot 12 = 4$
Similarly, $BG = \frac{2}{3}\cdot 8 = \frac{16}{3}$
The area of the right triangle $\triangle BEG$ is $\frac{1}{2}\cdot\frac{16}{3}\cdot 4$
The area of the whole figure is $6\cdot \frac{1}{2}\cdot\frac{16}{3}\cdot 4 = 64$ , and the correct answer is $\boxed{64}$
| 64
|
1,136
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_15
| 2
|
Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$ , and $CE=12$ . The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$
|
Notice that if you were to draw in line ED, you would get an orthodiagonal quadrilateral with diagonals 8 and 12. The area is going to be equal to 48. Now we need to examine the triangle AED. If the area we are trying to find is denoted as A, we can tell that the area of AEC is A/2. The area of AED is going to be half of that since AD = DC so it would be A/4. This means that 48 is 3/4 of A, so naturally A is going to be 64. Giving $\boxed{64}$
| 64
|
1,137
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_16
| 1
|
The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):
\[\left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right]\]
Or you leave at least one row and one column unchanged:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right]\]
In the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums. (In the example given, the sum in row $x$ is the same as in column $x$ .)
In the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal. (In the second example given, row $3$ and column $3$ are untouched.)
Either way, $3$ changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right]\]
Letting the $*$ be a zero does indeed give $6$ different sums, so the answer is $4$ , which is option $\boxed{4}$
| 4
|
1,138
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_17
| 1
|
A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$ . The distance between the points of intersection is $0.5$ . Given that $k = a + \sqrt{b}$ , where $a$ and $b$ are integers, what is $a+b$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$
|
Since the line $x=k$ is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of $k$ for which the distance $|\log_5 x - \log_5 (x+4)| = \frac{1}{2}$
Since $\log_5 x$ is a strictly increasing function, we have:
$\log_5 (x + 4) - \log_5 x = \frac{1}{2}$
$\log_5 (\frac{x+4}{x}) = \frac{1}{2}$
$\frac{x+4}{x} = 5^\frac{1}{2}$
$x + 4 = x\sqrt{5}$
$x\sqrt{5} - x = 4$
$x = \frac{4}{\sqrt{5} - 1}$
$x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}$
$x = 1 + \sqrt{5}$
The desired quantity is $1 + 5 = 6$ , and the answer is $\boxed{6}$
| 6
|
1,139
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_18
| 1
|
A list of integers has mode $32$ and mean $22$ . The smallest number in the list is $10$ . The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10$ , the mean and median of the new list would be $24$ and $m+10$ , respectively. If were $m$ instead replaced by $m-8$ , the median of the new list would be $m-4$ . What is $m$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$
|
Let there be $n$ integers on the list. The list of $n$ integers has mean $22$ , so the sum of the integers is $22n$
Replacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$
The new mean of the list is $24$ , so the new sum of the list is also $24n$
Thus, we get $22n + 10 = 24n$ , leading to $n=5$ numbers on the list.
If there are $5$ numbers on the list with mode $32$ and smallest number $10$ , then the list is $\{10, x, m, 32, 32\}$
Since replacing $m$ with $m-8$ gives a new median of $m-4$ , and $m-4$ must be on the list of $5$ integers since $5$ is odd, $x = m-4$ , and the list is now $\{10, m-4, m, 32, 32\}$
The sum of the numbers on this list is $22n = 22\cdot 5 = 110$ , so we get:
$10 + m - 4 + m + 32 + 32 = 110$
$70 + 2m = 110$
$m = 20$ , giving answer $\boxed{20}$
| 20
|
1,140
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_24
| 1
|
A rising number, such as $34689$ , is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom{9}{5} = 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97^{\text{th}}$ number in the list does not contain the digit
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
|
The list starts with $12345$ . There are $\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$ , and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.
Thus, the $71^{\text{st}}$ number is $23456$ . There are $\binom{6}{3} = 20$ three-digit rising numbers that do not begin with a $1,2$ or $3$ , and thus $20$ five digit rising numbers that begin with a $23$
Thus, the $91^{\text{st}}$ number is $24567$ . Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{\text{th}}$ number, which does not contain the digit $\boxed{5}$
| 5
|
1,141
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_25
| 1
|
Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$ $\overrightarrow{BB^\prime}$ $\overrightarrow{CC^\prime}$ , and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$ . If $AA^{\prime} = 10$ $BB^{\prime}= 8$ $CC^\prime = 18$ , and $DD^\prime = 22$ and $M$ and $N$ are the midpoints of $A^{\prime} C^{\prime}$ and $B^{\prime}D^{\prime}$ , respectively, then $MN =$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
|
Let $ABCD$ be a unit square with $A(0,0,0)$ $B(0,1,0)$ $C(1,1,0)$ , and $D(1,0,0)$ . Assume that the rays go in the +z direction. In this case, $A^\prime(0,0,10)$ $B^\prime(0,1,8)$ $C^\prime(1,1,18)$ , and $D^\prime(1,0,22)$ . Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}, \frac{1}{2}, 14)$ and $N(\frac{1}{2}, \frac{1}{2}, 15)$ . The distance $MN$ is $15 - 14 = 1$ , and the answer is $\boxed{1}$
| 1
|
1,142
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_26
| 1
|
Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$
|
The product of two lengths with a common point brings to mind the Power of a Point Theorem
Since $PA = PB$ , we can make a circle with radius $PA$ that is centered on $P$ , and both $A$ and $B$ will be on that circle. Since $\angle APB = \widehat {AB} = 2 \angle ACB$ , we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the measure of inscribed angle $\angle ACB$ , which is true for all inscribed angles.
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3.06,0.9); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); pair E = (0,4.5); dot(A);dot(B);dot(C);dot(P);dot(D);dot(E); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NW);label("$D$",D,NE + N);label("$P$",P,N);label("$E$", E, NW); draw(A--B--P--cycle); draw(A--C--B--cycle); draw(circle(P, 2.46)); draw(P--E);[/asy]
Since $PDB$ is a line, we have $PD + DB = PB$ , which gives $3 = DB + 2$ , or $DB = 1$
We now extend radius $PB = 3$ to diameter $EB = 6$ . Since $EDB$ is a line, we have $ED + DB = EB$ , which gives $ED + 1 = 6$ , or $ED = 5$
Finally, we apply the power of a point theorem to point $D$ . This states that $AD \cdot DC = DB \cdot DE$ . Since $DB = 1$ and $DE = 5$ , the desired product is $5$ , which is $\boxed{5}$
| 5
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1,143
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_26
| 2
|
Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$
|
Construct the angle bisector of $\angle APD,$ and let it intersect $AD$ at $E.$ From the angle bisector theorem, we have $AE=3a$ and $DE=2a$ for some $a.$ Then, note that $\angle EPD = \angle BCD = x,$ so $EPCB$ is cyclic. Then, $\frac{PD}{ED} = \frac{CD}{BD}$ or $\frac{2}{2x} = \frac{CD}{1}.$ Thus, $AD \cdot DC = 5x \cdot DC = 5,$ or $\boxed{5}.$
| 5
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1,144
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_27
| 1
|
Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$ . Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$
|
Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$
In this case, we know that $f(x+ 4) + f(x - 4) = f(x)$ . Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$ . Adding those two equations gives $f(x + 8) + f(x - 4) = 0$ after cancelling out common terms.
Again plugging in $x + 4$ in for $x$ in that last equation (in order to get $f(x)$ ), we find that $f(x) = -f(x + 12)$ . Now, plugging in $x+12$ for $x$ , we get $f(x + 12) = -f(x + 24)$ . This proves that $f(x) = f(x + 24)$ , so there is a period of $24$ , which gives answer $\boxed{24}$ . We now eliminate answers $A$ through $C$
| 24
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1,145
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_28
| 1
|
How many ordered triples of integers $(a,b,c)$ satisfy $|a+b|+c = 19$ and $ab+|c| = 97$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$
|
WLOG , let $a \ge 0$ , and let $a \ge b$ . We can say this because if we have one solution $(a,b) = (a_0, b_0)$ with $a_0 \ge 0$ and $a_0 > b_0$ , we really have the four solutions $(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)$ by the symmetry of the original problem.
Furthermore, we assert that these four solutions are distinct. We can say that $a > b$ , since if $a=b$ , we have $c = 19 - 2a$ for the first equation and either $c = 97 - a^2$ or $c = a^2 - 97$ for the second equation. Equating $19 - 2a = 97 - a^2$ gives no integer solution, while equating $19 - 2a = a^2 - 97$ also gives no integer solution.
Thus, we can now assume WLOG that $a \ge 0$ and $a > b$ , and each pair of $(a_0,b_0)$ that we get will generate four unique solutions: $(a_0, b_0), (b_0, a_0), (-a_0, -b_0), (-b_0, -a_0)$
We now divide the problem into $c \ge 0$ and $c < 0$
If $c \ge 0$ , we have $|a + b| + c = 19$ and $ab + c = 97$
Solving both equations for $c$ and equating them, we get that $ab - |a + b| = 78$ . Splitting these up, we find that either $ab - a - b = 78$ or $ab + a+ b = 78$ . Factoring both with SFFT gives $(a-1)(b-1) = 79$ or $(a +1)(b+1) = 79$ . We factor with the restrctions that $a \ge 0$ and $a > b$ . Since $79$ is prime, we have:
$a - 1 = 79$ and $b - 1 = 1$ , which leads to $(80,2)$
$a + 1 = 79$ and $b + 1 = 1$ , which leads to $(78, 0)$
Each of those solutions could generate $3$ more solutions, giving a total of $8$ potential solutions. However, in each the first set of four solutions, we have $|a + b| = 82$ , which from the original first equation $|a + b| + c = 19$ gives $c = 19 - 82$ , which contradicts our initial assumption that $c\ge 0$ . Similarly, for the second set of four solutions, we have $|a + b| = 78$ , which leads to $c = 19 - 78$ , also contradicting $c \ge 0$
If $c < 0$ , we have $|a + b| + c = 19$ and $ab - c = 97$ . We note that $19 - c$ must be positive whenever $c$ is negative, and thus $|a + b| = a + b$
Solving both equations for $c$ and using SFFT as above gives $(a+1)(b+1) = 117$ . Since $117 = 3^2\cdot 13$ , we factor with the restriction that $a\ge 0$ and $a > b$ . Thus, we can let $a+1 \in \{117, 39, 13\}$ , which means $a \in \{116, 38, 12\}$ . These give corresponding $b+1\in \{1, 3, 9\}$ , which leads to corresponding $b \in \{0, 2, 8\}$ . Combining the solutions, we have $(a,b) = (116, 0), (38, 2), (12, 8)$
Each of these three solutions permutes, negates, and permute-negates into $4$ solutions as described in the start of the solution, for a total of $12$ solutions.
Checking our solutions to ensure $c < 0$ , we find in the first set of four solutions, $|a + b| = 116$ , and thus $c = 19 - 116$ , which is indeed negative.
In the second set of four solutions, $|a + b| = 40$ , which leads to $c = 19 - 40$ , which is also negative.
Finally, in the third set of four solutoins, $|a + b| = 20$ , which leads to $c = 19 - 20$ , which is negative.
Thus, there are $12$ ordered triples, and the answer is $\boxed{12}$
| 12
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1,146
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_29
| 1
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Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$ . For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\\ \textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.}$
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Define a super-special number to be a number whose decimal expansion only consists of $0$ 's and $1$ 's. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\frac{1}{7}=0.142857142857\hdots$ . This can be done in $8$ numbers if we take \[0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots\] Now assume for sake of contradiction that we can do this with strictly less than $8$ super-special numbers (in particular, less than $10$ .) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a $1$ in that place. This means that in order to obtain the $8$ in $0.1428\hdots$ , there must be $8$ super-special numbers, so the answer is $\boxed{8}$
| 8
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1,147
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https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_30
| 1
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For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$
|
If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$ , because all positive binary numbers begin with a $1$ , and if you switch digits twice, you will have a $1$ at the end. Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive.
All of these odd numbers will have an even $D(n)$ $D(n) = 0$ will be given by the numbers $1, 11, 111, 1111, 11111, 111111$ , which is a total of $6$ numbers.
We skip $D(n) = 2$ for now, and move to $D(n) = 4$ , which is easier to count. The smallest $D(n) = 4$ happens when $n = 10101$ . To get another number such that $D(n) = 4$ , we may extend any of the five blocks of zeros or ones by one digit. This will form $110101, 100101, 101101, 101001, 101011$ , all of which are odd numbers that have $D(n) = 4$ . To find seven digit numbers that have $D(n) = 4$ , we can again extend any block by one, so long as it remains less than $1100001$ or under. There are five cases.
1) Extending $110101$ is impossible without going over $1100001$
2) Extending $100101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions work, giving $1000101, 1001101, 1001001, 1001011$
3) Extending $101101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions give $1001101, 1011101, 1011001, 1011011$ . However, $1001101$ already appeared in #2, giving only three new numbers.
4) Extending $101001$ at the first group is impossible. The other four extensions are $1001001, 1011001, 1010001, 1010011$ , but the first two are repeats. Thus, there are only two new numbers.
5) Extending $101011$ at the first group is impossible. The other four extensions give $1001011, 1011011, 1010011, 1010111$ , but only the last number is new.
Thus, there is $1$ five digit number, $5$ six digit numbers, and $4 + 3 + 2 + 1 = 10$ seven digit numbers under $1100001$ for which $D(n) = 4$ . That gives a total of $16$ numbers.
There smallest number for which $D(n) = 6$ is $1010101$ , which is under $98$ . Further extensions, as well as cases where $D(n) > 6$ , are not possible.
Thus, we know that there are $6$ odd numbers that have $D(n) = 0$ , and $16$ odd numbers that have $D(n) = 4$ , and $1$ number that has $D(n) = 6$ . The remaining odd numbers must have $D(n) = 2$ . This means there are $49 - 6 - 16 - 1 = 26$ numbers that have $D(n) = 2$ , which is option $\boxed{26}$
| 26
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1,148
|
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_30
| 2
|
For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$
|
For $D(n)$ to be $2$ $n$ must be in the form $1...10...01...1$ . Note that $n$ must be at least $3$ digits for $D(n)$ to be $2$
Case $1$ $n$ has $3$ digits
$n$ $101_2$ , there is only $1$ possible value for $n$ when $n$ has $3$ digits.
Case $2$ $n$ has $4$ digits
$n = 1$ _ _ $1_2$ , there could be $1$ zero or $2$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $2$ possible places. If there are $2$ zeros, there is only $1$ possible placement. Therefore, there are $2+1=3$ possible values for $n$ when $n$ has $4$ digits.
Case $3$ $n$ has $5$ digits
$n = 1$ _ _ _ $1_2$ , there could be $1$ $2$ or $3$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $3$ possible places. If there are $2$ zeros, there are $2$ possible placements. If there are $3$ zeros, there is only $1$ possible placement. Therefore, there are $3+2+1=6$ possible values for $n$ when $n$ has $5$ digits.
Case $4$ $n$ has $6$ digits
$n = 1$ _ _ _ _ $1_2$ , there could be $1$ $2$ $3$ or $4$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $4$ possible places. If there are $2$ zeros, there are $3$ possible placements. If there are $3$ zeros, there are $2$ possible placements. If there are $4$ zeros, there is only $1$ possible placement. Therefore, there are $4+3+2+1=10$ possible values for $n$ when $n$ has $6$ digits.
Case $4$ $n$ has $7$ digits
Given by the problem, $n \le 1100001_2$ . We could first count the number of $n$ that is in the form of $10$ _ _ _ _ $1_2$ , where the digits between the second and last digit are in the form of $0...01...1$
There could be $0$ $1$ $2$ $3$ or $4$ zeros. Therefore, there are $5$ possible values for $n$ when $n=10$ _ _ _ _ $1_2$
There is also the case were $n = 1100001_2$ . Therefore, there are $5+1=6$ possible values for $n$ when $n$ has $7$ digits.
Therefore, the answer is $1+3+6+10+6=\boxed{26}$
| 26
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1,149
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_1
| 1
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The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline &\texttt{2 4 5 6}\end{tabular}$
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$
|
Doing the addition as is, we get $641 + 852 + 973 = 2466$ . This number is $10$ larger than the desired sum of $2456$ . Therefore, we must make one of the three numbers $10$ smaller.
We may either change $641 \rightarrow 631$ $852 \rightarrow 842$ , or $973 \rightarrow 963$ . Either change results in a valid sum. The largest digit that could be changed is thus the $7$ in the number $973$ , and the answer is $\boxed{7}$
| 7
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1,150
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_2
| 1
|
Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$
|
If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \cdot 10 = 30$ dollars. He got $6$ dollars more than this.
He gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\frac{6}{2} = 3$ days, and the answer is $\boxed{3}$
| 3
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1,151
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_2
| 2
|
Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$
|
If Walter had done his chores for $10$ days exceptionally well, he would have earned $5 \cdot 10 = 50$ dollars. He got $50 - 36 = 14$ dollars less than this.
He gets $2$ dollars docked from his pay if he doesn't do his chores well. Therefore, he didn't do his chores well on $\frac{14}{2} = 7$ days. The other $10 - 7 = 3$ days, he did them exceptionally well. Therefore, the answer is $\boxed{3}$
| 3
|
1,152
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_2
| 3
|
Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$
|
Let $b$ be the number of days Walter does his chores but doesn't do them well, and let $w$ be the number of days he does his chores exceptionally well.
$b + w = 10$ since there are $10$ days Walter does chores.
$3b + 5w = 36$ since $3b$ is the amount he earns from doing his chores not well, and $5w$ is the amount he earnes from doing his chores exceptionally well, and those two sum to $36$ dollars.
Multiply the first equation by $3$ to get:
$3b + 3w = 30$
$3b + 5w = 36$
Subtract the first equation from the second equation to get:
$5w - 3w = 36 - 30$
$2w = 6$
$w = 3$
Thus, he does his chores well $3$ days, and the answer is $\boxed{3}$
| 3
|
1,153
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_3
| 1
|
$\frac{(3!)!}{3!}=$
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$
|
The numerator is $(3!)! = 6!$
The denominator is $3! = 6$
Using the property that $6! = 6 \cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$ , which is answer $\boxed{120}$
| 120
|
1,154
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_4
| 1
|
Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$ . The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$
|
First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$
We have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greater than $9$ , the median will be the highest it can possibly be. Thus, the maximum median is the fifth piece of data in the list, which is $8$ , giving an answer of $\boxed{8}$
| 8
|
1,155
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_7
| 1
|
A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$ , which of the following could be the age of the youngest child?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$
|
The bill for the three children is $9.45 - 4.95 = 4.50$ . Since the charge is $0.45$ per year for the children, they must have $\frac{4.50}{0.45} = 10$ years among the three of them.
The twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when the first twin was born and the second twin was born). If we let the twins be $5$ years old, that leaves $10 - 2\cdot 5 = 0$ years leftover for the youngest child. But if the twins are each $4$ years old, then the youngest child could be $10 - 2\cdot 4 = 2$ years old, which is choice $\boxed{2}$
| 2
|
1,156
|
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_10
| 1
|
How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$
|
There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments. However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{28}$
| 28
|
1,157
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_10
| 2
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How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$
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Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
A cube has $12$ edges: Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.
A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$
A cube has $4$ spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{28}$
| 28
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1,158
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_12
| 1
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A function $f$ from the integers to the integers is defined as follows:
\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]
Suppose $k$ is odd and $f(f(f(k))) = 27$ . What is the sum of the digits of $k$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$
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Going out the final step, if you have $f(f(f(k))) = 27$ , you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$
If you doubled either of these, $k$ would not be odd. So you must subtract $3$
If you subtract $3$ from $51$ , you would compute $f(48)$ , which would halve it, and not add the $3$ back.
If you subtract $3$ from $108$ , you would compute $f(105)$ , which would add the $3$ back.
Thus, $f(f(f(105))) = f(f(108)) = f(54) = 27$ , and $105$ is odd. The desired sum of the digits is $6$ , and the answer is $\boxed{6}$
| 6
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1,159
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_14
| 1
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Let $E(n)$ denote the sum of the even digits of $n$ . For example, $E(5681) = 6+8 = 14$ . Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$
|
The problem is asking for the sum of all the even digits in the numbers $1$ to $100$ . We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits. This gives the list:
$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$
There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.
Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$ , and the correct answer is $\boxed{400}$
| 400
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1,160
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_23
| 1
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The sum of the lengths of the twelve edges of a rectangular box is $140$ , and
the distance from one corner of the box to the farthest corner is $21$ . The total surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$
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Let $x, y$ , and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so: \[4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.\] The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$ . Thus, we have $\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y^2 + z^2 = 21^2$
Our target expression is the surface area of the box:
\[S = 2xy + 2xz + 2yz.\]
Since $S$ is a symmetric polynomial of degree $2$ , we try squaring the first equation to get:
\[35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.\]
Substituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$ , so $S = (35 + 21)(35 - 21) = 56\cdot 14 = 784$ , which is option $\boxed{784}$
| 784
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1,161
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24
| 1
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The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$
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The sum of the first $1$ numbers is $1$
The sum of the next $2$ numbers is $2 + 1$
The sum of the next $3$ numbers is $2 + 2 + 1$
In general, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$ ", where the word "next" follows the pattern established above.
Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$ , we find:
$f(50) = 1275$
$f(49) = 1225$
Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$ s to hit $1234$
We want to find:
$\sum_{n=1}^{49} 1 + 2(n-1)$
$\sum_{n=1}^{49} 2n - 1$
$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$
$2 \sum_{n=1}^{49} n - 49$
$2\cdot \frac{49\cdot 50}{2} - 49$
$49^2$
$2401$
Thus, the sum of the first $1225$ terms is $2401$ . We have to add $9$ more $2$ s to get to the $1234^{th}$ term, which gives us $2419$ , or option $\boxed{2419}$
| 419
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1,162
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24
| 2
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The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$
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The $k$ th appearance of 1 is at position $1 + 2 + \dots + k = \frac{k(k + 1)}{2}$ . Then there are $k$ 1's and $\frac{k(k + 1)}{2} - k = \frac{k(k - 1)}{2}$ 2's among the first $\frac{k(k + 1)}{2}$ numbers, so the sum of these $\frac{k(k + 1)}{2}$ terms is $k + k(k - 1) = k^2$
When $k = 49$ $\frac{k(k + 1)}{2} = 1225$ , and when $k = 50$ $\frac{k(k + 1)}{2} = 1275$
The sum of the first 1225 terms is $49^2 = 2401$ . The numbers in positions 1226 through 1234 are all 2's, so their sum is $(1234 - 1226 + 1) \cdot 2 = 18$ . Therefore, the sum of the first 1234 terms is $2401 + 18 = \boxed{2419}$
| 419
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1,163
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25
| 1
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Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
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Complete the square to get \[(x-7)^2 + (y-3)^2 = 64.\] Applying Cauchy-Schwarz directly, \[64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.\] \[40 \ge 3x+4y-33\] \[3x+4y \le 73.\] Thus our answer is $\boxed{73}$
| 73
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1,164
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25
| 2
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Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
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The first equation is a circle , so we find its center and radius by completing the square $x^2 - 14x + y^2 - 6y = 6$ , so \[(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.\]
So we have a circle centered at $(7,3)$ with radius $8$ , and we want to find the max of $3x + 4y$
The set of lines $3x + 4y = A$ are all parallel , with slope $-\frac{3}{4}$ . Increasing $A$ shifts the lines up and/or to the right.
We want to shift this line up high enough that it's tangent to the circle, but not so high that it misses the circle altogether. This means $3x + 4y = A$ will be tangent to the circle.
Imagine that this line hits the circle at point $(a,b)$ . The slope of the radius connecting the center of the circle, $(7,3)$ , to tangent point $(a,b)$ will be $\frac{4}{3}$ , since the radius is perpendicular to the tangent line.
So we have a point, $(7,3)$ , and a slope of $\frac{4}{3}$ that represents the slope of the radius to the tangent point. Let's start at the point $(7,3)$ . If we go $4k$ units up and $3k$ units right from $(7,3)$ , we would arrive at a point that's $5k$ units away. But in reality we want $5k = 8$ to reach the tangent point, since the radius of the circle is $8$
Thus, $k = \frac{8}{5}$ , and we want to travel $4\cdot \frac{8}{5}$ up and $3\cdot \frac{8}{5}$ over from the point $(7,3)$ to reach our maximum. This means the maximum value of $3x + 4y$ occurs at $\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)$ , which is $\left(\frac{59}{5}, \frac{47}{5}\right).$
Plug in those values for $x$ and $y$ , and you get the maximum value of $3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}$
| 73
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1,165
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25
| 3
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Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
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Let the tangent point be $P$ , and the tangent line's x-intercept be $Q$ . Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle $OPK$ is 3-4-5, $OP=8$ , so $OK = \frac{5}{3}*8 = \frac{40}{3}$ . Note that the horizontal distance from $O$ to the origin is $7$ , and the horizontal distance from K to Q is 4, ( $\frac{4}{3}$ of its y coordinate), so the x-intercept is $7+4+OK = 73/3$ . The value of $3x+4y$ is 73 at point $Q$ . Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of $P$ $\boxed{73}$
| 73
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1,166
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25
| 4
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Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
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Let $z = 3x + 4y$ . Solving for $y$ , we get $y = (z - 3x)/4$ . Substituting into the given equation, we get \[x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,\] which simplifies to \[25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.\]
This quadratic equation has real roots in $x$ if and only if its discriminant is nonnegative, so \[(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,\] which simplifies to \[-64z^2 + 4224z + 32704 \ge 0,\] which can be factored as \[-64(z + 7)(z - 73) \ge 0.\] The largest value of $z$ that satisfies this inequality is $\boxed{73}$
| 73
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1,167
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25
| 5
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Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
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First, we move all the non-constant terms of the constraint to one side and assign it to the function $g(x,y)$ \[g(x,y)=x^2+y^2-14x-6y.\] Since we are trying to maximize $f(x,y)=3x+4y$ , we need to solve for $x$ and $y$ in the system \[\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}\] We have that \begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*} and \begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}. To solve the original system, we can solve for $x$ and $y$ in terms of $\lambda$ using our equations from the gradients, then substitute them into the first equation. We have that $x=\frac{3}{2}\lambda+7$ and $y=2\lambda+3$ . Substituting into the first equation, we have that \begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*} Using the solutions of $x$ and $y$ in terms of $\lambda$ that we found earlier, we have that \[3x+4y=\frac{25}{2}\lambda+33.\] Because we are trying to maximize this function, we will use the positive solution for $\lambda$ . Therefore, after substituting, we have that the largest value of $g(x,y)$ that satisfies $f(x,y)=6$ is $\boxed{73}$
| 73
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1,168
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_26
| 1
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An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
$\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$
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Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$
The number of ways to select four red marbles out of the set of marbles without replacement is:
\[\binom{r}{4} = \frac{r!}{24\cdot (r -4)!}\]
The number of ways to select one white and three red marbles is:
\[\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}\]
The number of ways to select one white, one blue, and two red marbles is:
\[\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}\]
The number of ways to select one marble of each colors is:
\[\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr\] Setting the first and second statements equal, we find:
\[\frac{r!}{24\cdot (r -4)!} = \frac{w\cdot r!}{6\cdot (r - 3)!}\]
\[r - 3 = 4w\]
Setting the first and third statements equal, we find:
\[\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}\]
\[(r-3)(r-2) = 12wb\]
Setting the last two statements equal, we find:
\[\frac{wb\cdot r!}{2(r-2)!} = wbgr\]
\[r - 1 = 2g\]
These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.
From the first equation, we know that $r$ must be $3$ more than a multiple of $4$ , or that $r \equiv 3 \mod 4$
Putting the first equation into the second equation, we find $r-2 = 3b$ . Therefore, $r \equiv 2 \mod 3$ . Using the Chinese Remainder Theorem , we find that $r \equiv 11 \mod 12$
The third equation gives no new restrictions on $r$ ; it is already odd by the first equation.
Thus, the minimal positive value of $r$ is $11$ . This requires $g=\frac{r - 1}{2} = 5$ by the third equation, and $w = \frac{r-3}{4} = 2$ by the first equation. Finally, the second equation gives $b = \frac{(r-3)(r-2)}{12w} = 3$
The minimal total number of marbles is $11 + 5 + 2 + 3 = \boxed{21}$
| 21
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1,169
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27
| 1
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Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$
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The two equations of the balls are
\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]
\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]
Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$ , and the second ball goes from $1 \pm 4.5$ . The only integer value that $z$ can be is $z=5$
Plugging that in to both equations, we get:
\[x^2 + y^2 \le \frac{23}{4}\]
\[x^2 + y^2 \le \frac{17}{4}\]
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
From here, we do casework, noting that $|x|, |y| \le 3$
For $x=\pm 2$ , we must have $y=0$ . This gives $2$ points.
For $x = \pm 1$ , we can have $y\in \{-1, 0, 1\}$ . This gives $2\cdot 3 = 6$ points.
For $x = 0$ , we can have $y \in \{-2, -1, 0, 1, 2\}$ . This gives $5$ points.
Thus, there are $\boxed{13}$
| 13
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1,170
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27
| 2
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Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$
|
Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis while the y-axis remains the same).
The spheres now become circles with centers at $(1,0)$ and $(\frac{21}{2},0)$ . They have radii $\frac{9}{2}$ and $6$ , respectively.
Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\frac{21}{2},0)$
The point on circle $A$ closest to the center of circle $B$ is $(\frac{11}{2},0)$ . The point on circle B closest to the center of circle $A$ is $(\frac{11}{2},0)$
Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent "dome" shapes on either end. Because the tops of the "domes" are at $(0,0,\frac{9}{2})$ and $(0,0,\frac{11}{2})$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\frac{9}{2}$ and $\frac{11}{2}$ ). Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle. The radius of the circle will be the distance from the z-axis.
Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles. Let's call that altitude $h$
We know all three side lengths of this triangle: $\frac{9}{2}$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$ ). We can now find the area of the triangle using Heron's formula:
\[s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20\]
\[A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}=\sqrt{110}\]
Using the area of the triangle, we can find that altitude $h$ from the x-axis:
\[\sqrt{110}=\frac{h*\frac{19}{2}}{2}\]
\[h=\frac{4*\sqrt{110}}{19}\]
Remember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.
Going back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.
The radius $h$ of the circle in a 2-dimensional plane is $\frac{4\sqrt{110}}{19}$ , a little greater than $2$ . We know that $h>2$ because $4\sqrt{110}>4*10$ , and $\frac{40}{19}>2$
Finally, looking at a circle with radius slightly larger than $2$ , we see that there are $\boxed{13}$
| 13
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1,171
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27
| 3
|
Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$
|
Note that the spheres are on the same $x$ and $y$ axis. Therefore, we can draw the spheres so that only the $x$ and $z$ axis are featured.
$A = (1, 0, 5)$ $\sqrt{1^2+4^2} = \sqrt{17}<\frac92$ $A$ is inside the smaller sphere. $\sqrt{1^2+(\frac{11}{2})^2} = \frac{\sqrt{126}}{2}<6$ $A$ is inside the larger sphere. Point $A$ is inside the intersection.
$B = (2, 0, 5)$ $\sqrt{2^2+4^2} = \sqrt{20}<\frac92$ $B$ is inside the smaller sphere. $\sqrt{2^2+(\frac{11}{2})^2} = \frac{\sqrt{137}}{2}<6$ $B$ is inside the larger sphere. Point $B$ is inside the intersection.
$C = (3, 0, 5)$ $\sqrt{3^2+4^2} = \sqrt{25}>\frac92$ $C$ is outside the smaller sphere. $\sqrt{3^2+(\frac{11}{2})^2} = \frac{\sqrt{157}}{2}>6$ $C$ is outside the larger sphere. Point $C$ is outside the intersection.
The intersection of $2$ spheres is a circle, by drawing the circle flat we can see that there are $4$ more points within the intersection.
We can see that points with the same $x$ -coordinates with $A$ are still inside the circle. $B$ is very close to the circle, the points with the same $x$ -coordinates with $B$ are not necessarily inside the circle.
For example, $D = (2,1,5)$ $\sqrt{2^2+1^2+4^2}=\sqrt{21}>\frac92$ $D$ is outside the smaller sphere. $\sqrt{2^2 + 1^2 + (\frac{11}{2})^2} = \frac{\sqrt{141}}{2}<6$ $D$ is inside the larger sphere. Point $D$ is outside the intersection.
Therefore, the answer is $9+4 = \boxed{13}$
| 13
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1,172
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https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29
| 1
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If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$
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Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.
$30 \rightarrow p^{29}$
$2 \cdot 15 \rightarrow pq^{14}$
$3\cdot 10 \rightarrow p^2q^9$
$5\cdot 6 \rightarrow p^4q^5$
$2\cdot 3\cdot 5 \rightarrow pq^2r^4$
The variables $p, q, r$ are different prime factors, and one of them must be $3$ . We now try to count the factors of $2n$ , to see which prime factorization is correct and has $28$ factors.
In the first case, $p=3$ is the only possibility. This gives $2n = 2\cdot p^{28}$ , which has $2\cdot {29}$ factors, which is way too many.
In the second case, $p=3$ gives $2n = 2q^{14}$ . If $q=2$ , then there are $16$ factors, while if $q\neq 2$ , there are $2\cdot 15 = 30$ factors.
In the second case, $q=3$ gives $2n = 2p3^{13}$ . If $p=2$ , then there are $3\cdot 13$ factors, while if $p\neq 2$ , there are $2\cdot 2 \cdot 13$ factors.
In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$ . If $q=2$ , then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$ , there are $2\cdot 2\cdot 10$ factors.
In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$ . If $p=2$ , then there are $4\cdot 9$ factors, while if $p \neq 2$ , there are $2\cdot 3\cdot 9$ factors.
In the fourth case, $p=3$ gives $2n = 2\cdot 3^3\cdot q^5$ . If $q=2$ , then there are $7\cdot 4= 28$ factors. This is the factorization we want.
Thus, $3n = 3^4 \cdot 2^5$ , which has $5\cdot 6 = 30$ factors, and $2n = 3^3 \cdot 2^6$ , which has $4\cdot 7 = 28$ factors.
In this case, $6n = 3^4\cdot 2^6$ , which has $5\cdot 7 = 35$ factors, and the answer is $\boxed{35}$
| 35
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| 2
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If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$
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Because $2n$ has $28$ factors and $3n$ has $30$ factors, we should rewrite the number $n = 2^{e_1}3^{e_2}... p_n^{e_n}$ As the formula for the number of divisors for such a number gives: $(e_1+1)(e_2+1)... (e_n+1)$ We plug in the variations we need to make for the cases $2n$ and $3n$ $2n$ has $(e_1+2)(e_2+1)(e_3+1)... (e_n+1) = 28$ $3n$ has $(e_1+1)(e_2+2)(e_3+1)...(e_n+1) = 30$
If we take the top and divide by the bottom, we get the following equation: $\frac{(e_1+2)(e_2+1)}{(e_1+1)(e_2+2)} = \frac{14}{15}$ . Letting $e_1=x$ and $e_2 = y$ for convenience and expanding this out gives us: $xy-13x+16y+2=0$
We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into: $(x+16)(y-13) +2 + 208 = 0$ or $(x+16)(y-13) = - 210$
As we want to be dealing with rather reasonable numbers for $x$ and $y$ , we try to make the $x+16$ term the slightly larger term and the $y-13$ term the slightly smaller term. This effect is achieved when $x+16 = 21$ and $y-13 = -10$ . Therefore, $x = 5, y = 3$ . We get that this already satisfies the requirements for the number we are looking for, and we take $(5+2)(3+2) = \boxed{35}$
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If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$
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Let $\, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \,$ be the prime factorization of $\, n$ . Then the number of positive divisors of $\, n \,$ is $\, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \;$ . In view of the given information, we have \[28 = (e_1 + 2)(e_2 + 1)P\] and \[30 = (e_1 + 1)(e_2 + 2)P,\] where $\, P = (e_3 + 1)(e_4 + 1) \cdots \;$ . Subtracting the first equation from the second, we obtain $\, 2 = (e_1 - e_2)P, \,$ so either $\, e_1 - e_2 = 1 \,$ and $\, P = 2, \,$ or $\, e_1 - e_2 = 2 \,$ and $\, P = 1$ . The first case yields $\, 14 = (e_1 + 2)e_1 \,$ and $\, (e_1 + 1)^2 = 15$ ; since $\, e_1 \,$ is a nonnegative integer, this is impossible. In the second case, $\, e_2 = e_1 - 2 \,$ and $\, 30 = (e_1 + 1)e_1, \,$ from which we find $\, e_1 = 5 \,$ and $\, e_2 = 3$ . Thus $\, n = 2^5 3^3, \,$ so $\, 6n = 2^6 3^4 \,$ has $\, (6+1)(4+1) = \boxed{35} \,$ positive divisors.
| 35
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| 1
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A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$
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In hexagon $ABCDEF$ , let $AB=BC=CD=3$ and let $DE=EF=FA=5$ . Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$ . Similarly, $\angle CBE =\angle CFE=60^{\circ}$ . Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$ $Q$ that of $\overline{BE}$ and $\overline{AD}$ , and $R$ that of $\overline{CF}$ and $\overline{AD}$ . Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. [asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); real angleUnit = 15; draw(Circle(origin,1)); pair D = dir(22.5); pair C = dir(3*angleUnit + degrees(D)); pair B = dir(3*angleUnit + degrees(C)); pair A = dir(3*angleUnit + degrees(B)); pair F = dir(5*angleUnit + degrees(A)); pair E = dir(5*angleUnit + degrees(F)); draw(A--B--C--D--E--F--cycle); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F); draw(A--D^^B--E^^C--F); label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW); [/asy]
Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$ . Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, \[\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.\] It follows that \[\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.\] Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}. \blacksquare$
| 409
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| 2
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A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$
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All angle measures are in degrees.
Let the first trapezoid be $ABCD$ , where $AB=BC=CD=3$ . Then the second trapezoid is $AFED$ , where $AF=FE=ED=5$ . We look for $AD$
Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$ , if we drew $AC$ , we would see $\angle BCA=\angle BAC$ . Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ $\widehat{AB}$ means arc AB). Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$
Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$ . Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$ . Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$ $\angle CDE=120$ as well.
Now I focus on triangle $FAB$ . By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$ , so $BF=7$ . Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$ , we can now use the Law of Sines to get: \[\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.\]
Now I focus on triangle $AFD$ $\angle AFD=3\phi$ and $\angle ADF=\theta$ , and we are given that $AF=5$ , so \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.\] We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$ , but we need to find $\sin{3\phi}$ . Using various identities, we see \begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\ &=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\ &=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\ &=\sin{\phi}(3-4\sin^2{\phi})\\ &=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343} \end{align*} Returning to finding $AD$ , we remember \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.\] Plugging in and solving, we see $AD=\frac{360}{49}$ . Thus, the answer is $360 + 49 = 409$ , which is answer choice $\boxed{409}$
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A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$
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Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$ . Major arc $\overarc{AB}$ $=240^{\circ}$ $\angle ACB = 120^{\circ}$
By the Law of Cosine, $AB = \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$
$\frac{\angle AOB}{2} = 60^{\circ}$ , therefore, $r = \frac{\frac{AB}{2}}{\sin 60^{\circ}} = \frac{\frac{7}{2}}{ \frac{\sqrt{3}}{2} } = \frac{7\sqrt{3}}{3}$
$\sin \frac{\theta}{2} = \frac{\frac32}{r} = \frac{3}{2r} = \frac{3}{2 \cdot \frac{7\sqrt{3}}{3}} = \frac{3 \sqrt{3}}{14}$
Let $x$ be the length of the chord, $\sin \frac{3 \theta}{2} = \frac{\frac{x}{2}}{r}$
By the triple angle formula, $\sin \frac{3 \theta}{2} = 3 \cdot \sin \frac{\theta}{2} - 4 \cdot \sin(\frac{ \theta}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$
$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$
| 409
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| 4
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A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$
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Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$
$\sin \frac{\alpha}{2} = \frac{\frac32}{r}$ $\sin \frac{\alpha}{2} = \frac{3}{2r}$
$\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}$ $\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{2r}$
$\frac{\sin \frac{\alpha}{2}}{\sin (60^{\circ} - \frac{\alpha}{2}) } = \frac{\frac{3}{2r}}{\frac{5}{2r}} = \frac35$ $5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})$
$5 \cdot \sin \frac{\alpha}{2} = 3 ( \sin 60^{\circ} \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \cos 60^{\circ}) = 3 ( \frac{\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2} - \frac12 \cdot \sin \frac{\alpha}{2})$
$13 \cdot \sin \frac{\alpha}{2} = \frac{3\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2}$
Let $\sin \frac{\alpha}{2} = a$ $\cos \frac{\alpha}{2} = \sqrt{1-a^2}$ $13a = \frac{3\sqrt{3}}{2} \cdot \sqrt{1-a^2}$
$169a^2 = 27-27a^2$ $196a^2=27$ $\sin \frac{\alpha}{2} = a = \sqrt{\frac{27}{196}} = \frac{3 \sqrt{3}}{14}$
Let $x$ be the length of the chord, $\sin \frac{3 \alpha}{2} = \frac{\frac{x}{2}}{r}$
By the triple angle formula, $\sin \frac{3 \alpha}{2} = 3 \cdot \sin \frac{\alpha}{2} - 4 \cdot \sin(\frac{ \alpha}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$
$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$
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| 5
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A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$
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Note that major arc $\overarc{AE}$ is two thirds of the circumference, therefore, $\angle AFE = 120^{\circ}$
By the Law of Cosine, $AE= \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$
By the Ptolemy's theorem of quadrilateral $ABDE$ $AD \cdot BE = AB \cdot DE + BD \cdot AE$ $AD = BE$ $AD^2= 3 \cdot 5 + 7^2 = 64$ $AD = 8$
By the Ptolemy's theorem of quadrilateral $ABCD$ $AC \cdot BD = BC \cdot AD + AB \cdot CD$ $7AC = 3 \cdot 8 + 3 \cdot 5 = 39$ $AC = \frac{39}{7}$
By the Ptolemy's theorem of quadrilateral $ABCF$ $AC \cdot BF = AB \cdot CF + BC \cdot AF$ $AC = BF$ $(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3$ $CF = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$
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https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_13
| 1
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The addition below is incorrect. The display can be made correct by changing one digit $d$ , wherever it occurs, to another digit $e$ . Find the sum of $d$ and $e$
$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }$
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If we change $0$ , the units column would be incorrect.
If we change $1$ , then the leading $1$ in the sum would be incorrect.
However, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.
Changing the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$ . This does not match up to the bottom.
Changing the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$ . This is the number on the bottom if the $2$ s were changed to $6$ s.
Thus $d=2$ and $e=6$ . so $d+e= 8 \boxed{8}$
| 8
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https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_14
| 1
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If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$
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Substituting $x = -3$ , we get \[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\] But $f(-3) = 2$ , so $81a - 9b + 2 = 2$ , which means $81a - 9b = 0$ . Then \[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\]
| 8
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https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_22
| 1
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pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$ , although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$
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Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple . We know that $31$ and either $25,\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$ , to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \boxed{745}$
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| 1
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A list of five positive integers has mean $12$ and range $18$ . The mode and median are both $8$ . How many different values are possible for the second largest element of the list?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$
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Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$ , at least two of the five numbers must be $8$ . The last number we denote as $b$
Then their average is $\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26$ . Clearly $a \le 8$ . Also we have $b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a$ . Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$ ; listing shows that all such values work. The answer is $\boxed{6}$
| 6
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https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_29
| 1
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For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$
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$2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.
We can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$ , we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$ , we count the exact same case when we put those prime factors which were in $b$ into $c$
Thus, our total number of cases is $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{40}.$
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| 2
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For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$
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The prime factorization of $2310$ is $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11.$ Therefore, we have the equation \[abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\] where $a, b, c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a, b, c,$ which gives $3^5 = 243$ solutions for $(a, b, c).$ However, three of these solutions have two $1$ s and one $2310,$ which contradicts the fact that $a, b, c$ must be distinct. Because each prime factor appears only once, all other solutions have $a, b, c$ distinct. Correcting for this, we get $243 - 3 = 240$ ordered triples $(a, b, c)$ where $a, b, c$ are all distinct.
Finally, since order does not matter, we must divide by $3!,$ the number of ways to order $a, b, c.$ This gives the final answer, \[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\]
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A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$
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Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.
Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$ . The midpoint of this diagonal is at $\left(\frac 32,\frac 32,\frac 32\right)$ . The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac 92$
The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$ . Therefore the cube is intersected by this plane if and only if $x+y+z\in\{2,3,4\}$
There are six cubes such that $x+y+z=2$ : permutations of $(1,1,0)$ and $(2,0,0)$ Symmetrically, there are six cubes such that $x+y+z=4$ Finally, there are seven cubes such that $x+y+z=3$ : permutations of $(2,1,0)$ and the central cube $(1,1,1)$
That gives a total of $\boxed{19}$ intersected cubes.
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| 2
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A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$
|
Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\sqrt{3}$ , the altitude of the top vertex of the newly placed cube is $3\sqrt{3}$ . The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\frac{3\sqrt{3}}{2}$ , so that the height of the plane is $\frac{3\sqrt{3}}{2}$
By symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.
We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$ . The base of the pyramid is an equilateral triangle with side length of $\sqrt{3^2+3^2}=3\sqrt{2}$ . The height of the triangle is $\frac{ \sqrt{3} }{2} \cdot 3\sqrt{2}$ . The distance of the center of the triangle to the vertex is $\frac23 \cdot \frac{ \sqrt{3} }{2} \cdot 3\sqrt{2} = \sqrt {6}$ . Therefore, $h = \sqrt{3^2-(\sqrt {6})^2} = \sqrt{9-6}=\sqrt{3}$ . The altitude of the pyramid at the bottom of the cube is also $h$ . The altitude in the middle is $3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}$
The altitude of the vertex at the top is $3\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $2\sqrt{3}$ . The altitude of the third highest $3$ vertices are all $\sqrt{3}$ . The altitude of the bottom-most vertex is $0$
By scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $\frac{2\sqrt{3}}{3}$ . The altitude of the third highest $3$ vertices are all $\frac{\sqrt{3}}{3}$ . The altitude of the bottom-most vertex is $0$
The length of the space diagonal of a unit cube is $\sqrt{3}$ . The highest vertex of the bottom-most unit cube has an altitude of $\sqrt{3}$ . As $\sqrt{3} < \frac{3\sqrt{3}}{2}$ , therefore, the plane will not pass through the unit cube at the bottom.
For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{\sqrt{3}}{3} + \sqrt{3} = \frac{4\sqrt{3}}{3}$ . As $\frac{4\sqrt{3}}{3} < \frac{3\sqrt{3}}{2}$ , therefore, the plane will not pass through the next $3$ unit cubes.
For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}$ . As $\frac{5\sqrt{3}}{3} > \frac{3\sqrt{3}}{2}$ , therefore, the plane will pass through the next $3$ unit cubes.
So at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.
Thus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\boxed{19}$ unit cubes.
| 19
|
1,188
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_2
| 1
|
A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle? [asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,2.5)); label("14", (6.5,6)); label("35", (6.5,2.5)); [/asy]
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 25$
|
[asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path BG=shift(0,-0.5)*(B--G); path BF=shift(0.5,0)*(B--F); path FC=shift(0.5,0)*(F--C); path DH=shift(0,0.5)*(D--H); draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(BG,L=Label("$7$",position=MidPoint,align=(0,-1)),arrow=Arrows(),bar=Bars,red); draw(BF,L=Label("$5$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(FC,L=Label("$2$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(DH,L=Label("$3$",position=MidPoint,align=(0,1)),arrow=Arrows(),bar=Bars,red); label("$6$", (1.5,6)); label("$15$", (1.5,2.5),blue); label("$14$", (6.5,6)); label("$35$", (6.5,2.5)); [/asy]
We can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\times 5=\boxed{15}$
| 15
|
1,189
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_3
| 1
|
How many of the following are equal to $x^x+x^x$ for all $x>0$
$\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textbf{IV:}\ (2x)^{2x}$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
|
We look at each statement individually.
$\textbf{I:}\ 2x^x$ . We note that $x^x+x^x=x^x(1+1)=2x^x$ . So statement $\textbf{I}$ is true.
$\textbf{II:}\ x^{2x}$ . We find a counter example which is $x=1$ $2\neq 1$ . So statement $\textbf{II}$ is false.
$\textbf{III:}\ (2x)^x$ . We see that this statement is equal to $2^xx^x$ $x=2$ is a counter example. $8\neq 16$ . So statement $\textbf{III}$ is false.
$\textbf{IV:}\ (2x)^{2x}$ . We see that $x=1$ is again a counter example. $2\neq 4$ . So statement $\textbf{IV}$ is false.
Therefore, our answer is $\boxed{1}$
| 1
|
1,190
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_4
| 1
|
In the $xy$ -plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$
|
We see that the center of this circle is at $\left(\frac{-5+25}{2},0\right)=(10,0)$ . The radius is $\frac{30}{2}=15$ . So the equation of this circle is \[(x-10)^2+y^2=225.\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\boxed{10}$
| 10
|
1,191
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_5
| 1
|
Pat intended to multiply a number by $6$ but instead divided by $6$ . Pat then meant to add $14$ but instead subtracted $14$ . After these mistakes, the result was $16$ . If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$
|
We reverse the operations that he did and then use the correct operations. His end result is $16$ . Before that, he subtracted $14$ which means that his number after the first operation was $30$ . He divided by $6$ so his number was $180$
Now, we multiply $180$ by $6$ to get $1080$ . Finally, $1080+14=1094$ . Since $1094>1000$ , our answer is $\boxed{1000}$
| 0
|
1,192
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_9
| 1
|
If $\angle A$ is four times $\angle B$ , and the complement of $\angle B$ is four times the complement of $\angle A$ , then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$
|
Let $\angle A=x$ and $\angle B=y$ . From the first condition, we have $x=4y$ . From the second condition, we have \[90-y=4(90-x).\] Substituting $x=4y$ into the previous equation and solving yields \begin{align*}90-y=4(90-4y)&\implies 90-y=360-16y\\&\implies 15y=270\\&\implies y=\boxed{18}
| 18
|
1,193
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_12
| 1
|
If $i^2=-1$ , then $(i-i^{-1})^{-1}=$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}$
|
We simplify step by step as follows: \begin{align*}(i-i^{-1})^{-1}&=\frac{1}{i-i^{-1}}\\&=\frac{1}{i-\frac{1}{i}}\\&=\frac{1}{\left(\frac{i^2-1}{i}\right)}\\&=\frac{i}{i^2-1}\\&=\boxed{2}
| 2
|
1,194
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_18
| 1
|
Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$ . If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$ [asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$
|
We solve for $\angle A$ as follows: \[4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.\] That means that minor arc $\widehat{BC}$ has measure $40^\circ$ . We can fit a maximum of $\frac{360}{40}=\boxed{9}$ of these arcs in the circle.
| 9
|
1,195
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_19
| 1
|
Label one disk " $1$ ", two disks " $2$ ", three disks " $3$ $, ...,$ fifty disks " $50$ ". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501$
|
We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle.
We can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks.
From disks numbered from 10 to 50, we can pick up at most 9 disks to prevent picking up 10. There are 50-10+1 = 41 different numbers from 10 to 50. We pick up 9 from each number, therefore, we multiply $41 \cdot 9 = 369$ . In total, the maximum number we can pick up without picking up 10 of the same kind is $369+45=414$ . We need one more disk to guarantee a complete set of 10. Therefore, the answer is $\boxed{415}$
| 415
|
1,196
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_24
| 1
|
A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$ . The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$
|
The minimum range occurs in the set $\{7,7,12,12,12\}$ , so the answer is $\boxed{5}$
| 5
|
1,197
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_25
| 1
|
If $x$ and $y$ are non-zero real numbers such that \[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\] then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$
|
We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$
Solving for $y$ in the top equation gives us $3-x$ . Plugging this in gives us: $x^3-x^2+3x=0$ . Since we're told $x$ is not zero, we can divide by $x$ , giving us: $x^2-x+3=0$
The discriminant of this is $(-1)^2-4(1)(3)=-11$ , which means the equation has no real solutions.
We conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$ . Negating the top equation gives us $x-y=-3$ . We seek $x-y$ , so the answer is $\boxed{3}$
| 3
|
1,198
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_26
| 1
|
A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$ .) If $m=10$ , what is the value of $n$ [asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 26$
|
To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\frac{(n-2)*180}{n}$ , the measure of the decagon's interior angle is $\frac{8*180}{10} = 144$ degrees.
The regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\boxed{5}$
| 5
|
1,199
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_28
| 1
|
In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
|
The line with $x$ -intercept $a$ and $y$ -intercept $b$ is given by the equation $\frac{x}{a} + \frac{y}{b} = 1$ . We are told $(4,3)$ is on the line so
\[\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12\]
Since $a$ and $b$ are integers, this equation holds only if $(a-4)$ is a factor of $12$ . The factors are $1, 2, 3, 4, 6, 12$ which means $a$ must be one of $5, 6, 7, 8, 10, 16$ . The only members of this list which are prime are $a=5$ and $a=7$ , so the number of solutions is $\boxed{2}$
| 2
|
1,200
|
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_28
| 2
|
In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
|
Let $C = (4,3)$ $DF=a$ , and $AD=b$ . As stated in the problem, the $x$ -intercept $DF=a$ is a positive prime number, and the $y$ -intercept $AD=b$ is a positive integer.
Through similar triangles, $\frac{AB}{BC}=\frac{CE}{EF}$ $\frac{b-3}{4}=\frac{3}{a-4}$ $(a-4)(b-3)=12$
The only cases where $a$ is prime are: \[\begin{cases} a-4=1 & a=5 \\ b-3=12 & b=15 \end{cases}\]
\[and\]
\[\begin{cases} a-4=3 & a=7 \\ b-3=4 & b=5 \end{cases}\]
So the number of solutions are $\boxed{2}$
| 2
|
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