id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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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 $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\text... | 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, the... | 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 $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\text... | 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_... | 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 large... | 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... | 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 large... | 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 = 2... | 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 $\frac Mm =$
$\ma... | 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 co... | 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 $\frac Mm =$
$\ma... | 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 + \cd... | 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}... | 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) -... | 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... | 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 ... | 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 ... | 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... | 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 memb... | 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 tal... | 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 s... | 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 ... | 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 fini... | 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 $h$ , t... | 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 ... | 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) \ ... | 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... | 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... | 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 multipli... | 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... | 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 c... | 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,... | 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... | 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 $25$ b... | 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 m... | 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 poin... | 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... | 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 mult... | 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{(... | 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) = ... | 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(... | [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... | 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(... | 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 ... | 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\... | 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 ... | 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... | 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})... | 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 ... | 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... | 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)}... | 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... | 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 $DD^\prime = 22$ an... | 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},... | 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 = (... | 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 ci... | 5 |
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 = (... | 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 \... | 5 |
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 thos... | 24 |
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 solut... | 12 |
1,146 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_29 | 1 | 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\q... | 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, 0.010... | 8 |
1,147 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_30 | 1 | 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 $D(n) = 2$
$\t... | 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 ... | 26 |
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 $D(n) = 2$
$\t... | 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... | 26 |
1,149 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_1 | 1 | 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. T... | 7 |
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\... | 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 $\bo... | 3 |
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\... | 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 |
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\... | 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 ea... | 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 $9$ ,... | 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?
$... | 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... | 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 t... | 28 |
1,157 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_10 | 2 | How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$ | 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... | 28 |
1,158 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_12 | 1 | 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... | 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 ... | 6 |
1,159 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_14 | 1 | 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 ... | 400 |
1,160 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_23 | 1 | 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
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$ | 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 = 2... | 784 |
1,161 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24 | 1 | 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\t... | 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 ... | 419 |
1,162 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24 | 2 | 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\t... | 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 $k = 49$ $\frac{k(k + 1)}{2} = 1225$... | 419 |
1,163 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | 1 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | 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 |
1,164 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | 2 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | 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 +... | 73 |
1,165 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | 3 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | 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 t... | 73 |
1,166 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | 4 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | 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 ... | 73 |
1,167 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | 5 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | 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 ... | 73 |
1,168 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_26 | 1 | 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
(... | 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 o... | 21 |
1,169 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27 | 1 | 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... | 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... | 13 |
1,170 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27 | 2 | 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... | 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 $6$ , r... | 13 |
1,171 | 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... | 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 ... | 13 |
1,172 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | 1 | 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?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | 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 \... | 35 |
1,173 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | 2 | 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?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | 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)... ... | 35 |
1,174 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | 3 | 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?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | 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) \c... | 35 |
1,175 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | 1 | 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 $m$ a... | 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... | 409 |
1,176 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | 2 | 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 $m$ a... | 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 BA... | 409 |
1,177 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | 3 | 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 $m$ a... | 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^{... | 409 |
1,178 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | 4 | 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 $m$ a... | 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{\al... | 409 |
1,179 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | 5 | 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 $m$ a... | 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 ... | 409 |
1,180 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_13 | 1 | 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 ... | 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 $2$ ... | 8 |
1,181 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_14 | 1 | If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$ | 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 |
1,182 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_22 | 1 | 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) \ ... | 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-er... | 745 |
1,183 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_25 | 1 | 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?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$ | 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 \Longrig... | 6 |
1,184 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_29 | 1 | For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$ | $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . 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, ... | 40 |
1,185 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_29 | 2 | For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$ | 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 on... | 40 |
1,186 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_30 | 1 | 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{... | 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 diagon... | 19 |
1,187 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_30 | 2 | 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{... | 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 i... | 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,... | [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=... | 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^x... | 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... | 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 $10... | 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),... | 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 labe... | 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 pi... | 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... | 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)... | 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 regula... | 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 ... | 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: \[\begi... | 2 |
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