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3,101
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_6
| 1
|
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 16\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 24$
|
The age of each person is a factor of $128 = 2^7$ . So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$ $32$ $8$ , or $2$ years old, respectively. Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$
| 18
|
3,102
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_7
| 1
|
By inserting parentheses, it is possible to give the expression \[2\times3 + 4\times5\] several values. How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$
|
The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions \begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + 4\times5) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*} are in fact all distinct. So the answer is $\boxed{4}$
| 4
|
3,103
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_8
| 1
|
In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$
|
Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$
| 17
|
3,104
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_11
| 1
|
How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ $2$ $3$ $3$ $5$ $5$ $5$
$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$
|
A seven-digit palindrome is a number of the form $\overline{abcdcba}$ . Clearly, $d$ must be $5$ , as we have an odd number of fives. We are then left with $\{a,b,c\} = \{2,3,5\}$ . There are $3!$ permutations of these three numbers, since each is reflected over the midpoint we only have to count the first there. Each of the $\boxed{6}$ will give us one palindrome.
| 6
|
3,105
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_11
| 2
|
How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ $2$ $3$ $3$ $5$ $5$ $5$
$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$
|
Say we have a 2 first. Then, we have a 2 pinned as the last digit, so we have to fill in the remaining digits with only 3's and 5's. We have 2 options for the second digit then, and the rest is fixed. This means that we have $2$ ways for this case.
Say we have a 3 first. By symmetry, this is the same as the 2 cases, so we have $2$ ways.
Say we have a 5 first. We then have a 5 in the middle. We can either have a 2 second or a 3 second. So we have $2$ ways.
This means that our answer is $2+2+2=\boxed{6}$
| 6
|
3,106
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_12
| 1
|
Distinct points $A$ $B$ $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
$\text{(A) } 3 \qquad \text{(B) } 4 \qquad \text{(C) } 5 \qquad \text{(D) } 6 \qquad \text{(E) } 7$
|
Consider the classical formula for triangle area: $\frac 12 \cdot b \cdot h$ .
Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.
Hence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are $1$ $2$ , and $3$ . Therefore there are only $\boxed{3}$ possible values for the area.
| 3
|
3,107
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_12
| 2
|
Distinct points $A$ $B$ $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
$\text{(A) } 3 \qquad \text{(B) } 4 \qquad \text{(C) } 5 \qquad \text{(D) } 6 \qquad \text{(E) } 7$
|
No matter what how we draw a triangle by selecting three non-linear points, its height will always remain the same. Therefore, we will only get different areas with different base-lengths. The possibilities are $1$ $2$ , and $3$ units for a total of $\boxed{3}$
| 3
|
3,108
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_16
| 1
|
Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$
|
[asy] unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C, dashed ); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); [/asy]
As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$ , hence $\angle OAC = \angle OCA = 30^\circ$ . Then $\angle AOC = 120^\circ$ , and from symmetry we have $\angle AOB = \angle COB = 60^\circ$ . Thus, this gives us $\angle ABO = \angle CBO = 30^\circ$
We know that $DO = AO$ , as $D$ lies on the circle. From $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$ , Hence $DO = \frac{BO}2$ , therefore $BD = BO - DO = \frac{BO}2$ , and $\frac{BD}{BO} = \boxed{12}$
| 12
|
3,109
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_16
| 2
|
Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$
|
[asy] unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); label("$S$",Sp,(S+SSW)); [/asy]
As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$ . Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.
We then have $\angle SCD = \angle SAD = 30^\circ$ , hence $D$ is the incenter of $\triangle ABC$ , and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$ , and as $SD = SO$ , we have $2\cdot SD = SD + SO = OD$ , therefore $BD=OD$ , and as before we conclude that $\frac{BD}{BO} = \boxed{12}$
| 12
|
3,110
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17
| 1
|
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$
|
For $c\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.
Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of the entire triangle is $\frac{3(3-c)}2$ , therefore the area of the shaded part is $\frac{7-3c}{2}$
The entire figure has area $5$ , hence we want the shaded part to have area $\frac 52$ . Solving for $c$ , we get $c=\boxed{23}$
| 23
|
3,111
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17
| 2
|
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$
|
The unit square is of area 1, so the five unit squares have area 5.
Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of area 3.5.
It can then be inferred: $(3-c) * 3 = 7$
$3-c=\frac{7}{3}$ , so $3-\frac{7}{3}=c$
$3-\frac{7}{3} = \frac{9-7}{3} = \frac{2}{3}$ $\boxed{23}$
| 23
|
3,112
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17
| 3
|
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$
|
The shaded space of the object can become a triangle by adding a unit square to its bottom. This triangle has a base length of $3-c$ and a height of $3$ . The area of this triangle region is now (using the formula $A=bh/2$ for a triangle) $(9-3c)/2$ . But, remember that we have to subtract the area of the extra unit square from this area to get the area of the shaded region.
If you add 3 unit squares to the top of the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3c)/2$ . After you subtract the area of the 3 extra unit squares, we have derived the area of the unshaded portion.
Since these areas are both equal, we set them equal to each other and solve for $c$ $(9-3c)/2 -1 = (9+3c)/2 -3$ $c$ is now solved to be $\frac{2}{3}$ $\boxed{23}$
| 23
|
3,113
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17
| 4
|
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$
|
We are looking for the area of the shaded region to be $\frac 52$ . We start by testing $(A) \frac 12$ . The area of the shaded region would be $\frac{(3-\frac 12)(3) }{2}-1=\frac {11}{4}$ when $c=\frac 12$ . This does not match our wanted answer.
We try $(B) \frac 35$ next. The area of the shaded region would be $\frac{(3-\frac 35)(3) }{2}-1=\frac {13}{5}$ when $c=\frac 35$ . This also does not match our desired answer.
We then try $(C) \frac 23$ . The area of the shaded region would be $\frac{(3-\frac 23)(3) }{2}-1=\frac {5}{2}$ when $c=\frac 23$ . Therefore, our answer is $\boxed{23}$
| 23
|
3,114
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_19
| 1
|
A particular $12$ -hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1$ , it mistakenly displays a $9$ . For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?
$\mathrm{(A)}\ \frac 12\qquad \mathrm{(B)}\ \frac 58\qquad \mathrm{(C)}\ \frac 34\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac {9}{10}$
|
The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$ . So the correct hour is displayed $\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$ , so the minutes that will not display correctly are $10, 11, 12, \dots, 19$ and $01, 21, 31, 41,$ and $51$ . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{12}$
| 12
|
3,115
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_21
| 1
|
What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$
|
The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore
is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed{4}$
| 4
|
3,116
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24
| 1
|
The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$
|
Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.
[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted ); label("$X$",(0,0),S); label("$Y$",(R,0),SE); label("$Z$",rotate(20)*(R,0),ENE); draw( arc( (0,0), (1.5,0), rotate(20)*(1.5,0) ) ); label("$20^\circ$", rotate(10)*(1.75,0), E ); [/asy]
Each of the angles at $X$ is $\frac{180^\circ}9 = 20^\circ$ . From $\triangle XYZ$ , the degree measure of the smaller interior angle of the trapezoid is $\frac{180^\circ - 20^\circ}2 = 80^\circ$ , hence the degree measure of the larger interior angle is $180^\circ - 80^\circ = \boxed{100}$
| 100
|
3,117
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24
| 2
|
The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$
|
A decagon can be formed from the trapezoids and the base. The sum of the decagon's angles is $180(10-2)=1440^\circ$ . Letting the larger angle in each trapezoid be $x$ , the two angles formed by the line each measures $(180-x)^\circ$ . There are $8$ congruent angles left. Each of those angles measures $(360-2x)^\circ$ . Putting it all together: $8(360-2x)+2(180-x)=1440\implies x=\boxed{100}$
| 100
|
3,118
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24
| 3
|
The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$
|
If we reflect the arch across the line, we form an 18-gon. $\frac{180*(18-2)}{18} = 160^\circ$ so each interior angle of the 18-gon is $160^\circ$ . Let $x$ be the degree measure of the larger interior angle of a trapezoid. From the diagram, we see that $2x + 160 = 360$ , so $2x = 200$ and $x = 100$ , or $\boxed{100}$
| 100
|
3,119
|
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_6
| 1
|
A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?
$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7$
|
Since the three segments are all the same length, the triathlete's average speed is the harmonic mean of the three given rates. Therefore, the average speed is \[\frac{3}{\frac{1}{3}+\frac{1}{20}+\frac{1}{10}}=\frac{3}{\frac{29}{60}}=\frac{180}{29}\approx6\Rightarrow\boxed{6}\]
| 6
|
3,120
|
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_7
| 1
|
The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\] simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$
|
Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes
$\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$
$= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$
$= \boxed{9}$
| 9
|
3,121
|
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_10
| 1
|
Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$ . What is the area of $S_3$
$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4$
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Since the length ratio is $\frac{1}{\sqrt{2}}$ , then the area ratio is $\frac{1}{2}$ (since the area ratio between two similar 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \boxed{4}$
| 4
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3,122
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15
| 1
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Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$
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Set the time Ian traveled as $I$ , and set Han's speed as $H$ . Therefore, Jan's speed is $H+5.$
We get the following equation for how much Han is ahead of Ian: $H+5I = 70.$
The expression for how much Jan is ahead of Ian is: $2(H+5)+10I.$
This simplifies to: $2H+10+10I.$
However, this is just $2(H+5I)+10.$
Substitute, from the first equation, $H+5I$ as $70.$
Therefore, the answer is $140 + 10$ , which is $150$ , or $\boxed{150}$
| 150
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3,123
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15
| 2
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Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$
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We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ $H_t$ $J_s$ $J_t$ . The problem gives us 5 equations
\begin{align} H_s&=I_s+5 \\ H_t&=I_t+1 \\ J_s&=I_s+10 \\ J_t&=I_t+2 \\ H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}
Substituting equations $(1)$ and $(2)$ into $(5)$ gives:
\[(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)\]
We are asked the difference between Jan's and Ian's distances, or
\[J_s J_t-I_s I_t=x,\]
Where $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:
\[(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow\]
\[2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x\]
Substituting $(*)$ into this equation gives:
\[2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x\]
Therefore, the answer is $150$ miles or $\boxed{150}$
| 150
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3,124
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15
| 3
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Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$
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Let Ian drive $D$ miles, at a speed of $R$ , for some time $T$ (in hours). Hence, we have $D=RT$ . We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$ , for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$ . We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$ , where $x$ is how much further Jan traveled than Ian. We now have three equations: \[D= RT\] \[D + 70 = (R+5)(T+1) = RT + R + 5T + 5\] \[D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.\] Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: \[70 = 5T + R + 5 \Longrightarrow 5T + R = 65\] \[x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.\] Since $x = 150$ , our answer is $\boxed{150}$
| 150
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3,125
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15
| 4
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Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$
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Let Ian drive $d$ miles, $t$ hours, and at speed $s$
Ian's Equation: $d=s \cdot t.$
Han drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian.
Han's Equation: $d+70=(s+5) \cdot (t+1).$
Let Jan have driven $m$ miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian.
Jan's Equation: $m=(s+10) \cdot (t+2).$
Let's group the equations together:
$(1) \phantom{a} d=s \cdot t$
$(2) \phantom{a} d+70=(s+5) \cdot (t+1)$
$(3) \phantom{a} m=(s+10) \cdot (t+2)$
Let's see what we want to find: We want to find $n$ . The equation is $m=d+n$ where $n$ is the number of more miles traveled by Jan than Ian.
Onto the calculating part.
Expanding the second equation, we get
$d+70=st+5t+s+5.$
Note that $st=d$ by the first equation, so substituting we get
$d+70=d+5t+s+5.$
Simplifying gets us
$(4) \phantom{a} s+5t=65.$
(Note for the above process: You could have substituted $d$ with $st$ but that would lead you to the same result since $d-d=st-st=0.$
Let's look at the third equation. Expanding, we get
$m=st+10t+2s+20.$
Since we want $n$ we want the equation $m=d+n$ . We write the expanded third equation into this form since $st=d.$
$(5) \phantom{a} m=d+(10t+2s+20)$
Let's take a closer look at the $n$ section of the equation:
$(6) \phantom{a} n=10t+2s+20$
This looks very similar to equation 4, if you multiply equation 4 by $2$ you get
$(7) \phantom{a} 10t+2s=130$
Plugging equation 7 into equation 6 we have
$n=(10t+2s)+20 \Rightarrow n=130+20$
Calculating gets us
$(8) \phantom{a} n=150.$
Substituting equation 8 into equation 5 gets us
$m=d+n \Rightarrow m=d+150.$
The $n$ term is $150$ which is what we want to find so the answer is $\boxed{150}.$
| 150
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3,126
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_15
| 5
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Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$
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Since Han drove for $1$ hour and drove $70$ miles more than Ian during that hour, we know that Ian's speed is $65$ miles per hour since Han drove $5$ mph faster than him. Now Jan went $10$ mph faster than Ian for $2$ hours, so we can tell that she drove $75 \cdot 2$ miles more than Ian, therefore the answer is $\boxed{150}.$
| 150
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3,127
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23
| 1
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Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$
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First, choose the two letters to be repeated in each set. $\dbinom{5}{2}=10$ . Now we have three remaining elements that we wish to place into two separate subsets. There are $2^3 = 8$ ways to do so because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted (Take for example $S_{1} = \{a,b,c,d \}$ and $S_{2} = \{a,b,e \}$ ). Notice how $S_{1}$ and $S_{2}$ are interchangeable. Division by two will fix this problem. Thus we have:
$\dfrac{10 \times 8}{2} = 40 \implies \boxed{40}$
| 40
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3,128
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23
| 2
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Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$
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Another way of looking at this problem is to break it down into cases.
First, our two subsets can have 2 and 5 elements. The 5-element subset (aka the set) will contain the 2-element subset. There are $\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets.
Second, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\cdot3=30$ ways to choose these sets.
This leads us to the answer:
$10+30=40 \implies \boxed{40}$
| 40
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3,129
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23
| 3
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Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$
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There are $\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We now must place the 3 remaining elements into the subsets. Using stars and bars, we can notate this as: $I I I X \rightarrow \dbinom{4}{3}=4$ . Thus, $4*10=\boxed{40}$
| 40
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3,130
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_24
| 1
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Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$
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$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$
So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{6}$
| 6
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3,131
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_24
| 2
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Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$
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I am going to share another approach to this problem.
A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$
Let us take this step by step. First, we consider $k^2.$
Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$
Before continuing, though, we must take note of the following:
\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}
Now, we continue with the calculation.
\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}
We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$
\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}
This implies that \begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}
Thus, \begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{6}
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3,132
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_1
| 1
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A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$
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The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\boxed{6}$
| 6
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3,133
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_2
| 1
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$4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$
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After reversing the numbers on the second and fourth rows, the block will look like this:
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$
The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{4}$
| 4
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_2
| 2
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$4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$
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Notice that at baseline the diagonals sum to the same number ( $52$ ). Therefore we need only compute the effect of the swap. The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{4}$
| 4
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_7
| 1
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An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$
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The area of the large triangle is $\frac{10^2\sqrt3}{4}$ , while the area of each small triangle is $\frac{1^2\sqrt3}{4}$ . Dividing these two quantities results in $100$ , therefore $\boxed{100}$ small triangles can fill the large one without overlap.
| 100
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3,136
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_7
| 2
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An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$
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[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]
The number of triangles is $1+3+\dots+19 = \boxed{100}$
| 100
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_8
| 1
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A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$
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The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$ . Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \geq 0$ which solves to $r\leq 8 \frac13$ $r$ must be an integer, so there are $\boxed{9}$ possible values of $r$ , and each gives us one solution.
| 9
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3,138
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_8
| 2
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A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$
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Let $x$ and $y$ be the number of roses and carnations bought. The equation should be $3x+2y = 50$ . Since $50$ is an even number, the product of $3x$ must be even and smaller than $50$ . You can try nonnegative even integers for $x$ and you will end up with the numbers $0$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ , and $16$ . There are $9$ numbers in total, so the answer is $\boxed{9}$
| 9
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3,139
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_8
| 3
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A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$
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Let $r$ represent the number of roses, and let $c$ represent the number of carnations. Then, we get the linear Diophantine equation, $3r+2c=50$ .
Using the Euclidean algorithm, we get the initial solutions to be $r_0=50$ and $c_0=-50$ , meaning the complete solution will be, $r=50+\frac{2}{\gcd(2,3)}$ $k=50+2k$ $c=-50-\frac{3}{\gcd(2,3)}k=-50-3k$
The solution range for which both $r$ and $c$ are positive is $17$ $\leq k$ $\leq$ $25$ . There are $\boxed{9}$ possible values for $k$
| 9
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3,140
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_12
| 1
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Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four
times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\mathrm{(A)}\ 2500\qquad\mathrm{(B)}\ 3000\qquad\mathrm{(C)}\ 3500\qquad\mathrm{(D)}\ 4000\qquad\mathrm{(E)}\ 4500$
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Every time the pedometer flips from $99999$ to
$00000$ Pete has walked $100000$ steps.
So, if the pedometer flipped $44$ times
Pete walked $44*100000+50000=4450000$ steps.
Dividing by $1800$ steps per mile gives $2472.\overline{2}$
This is closest to answer $\boxed{2500}$
| 500
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3,141
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13
| 1
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For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$
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Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the $2008^{\text{th}}$ term of the sequence is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{4015}$
| 15
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3,142
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13
| 2
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For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$
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Let $a_1, a_2, a_3, \cdots, a_n$ be the terms of the sequence. We know $\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n$ , so we must have $a_1 + a_2 + a_3 + \cdots + a_n = n^2$ . The sum of consecutive odd numbers down to $1$ is a perfect square, if you don't believe me, try drawing squares with the sum, so $a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1$ , so the answer is $a_{2008} = 2(2007) + 1 = \boxed{4015}$
| 15
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3,143
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13
| 3
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For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$
|
Let the mean be $\frac{(a)+(a+d)+(a+2d)+...+(a+(n-1)) \cdot d)}{n}$ $=\frac{n \cdot a}{n} + \frac{(1+2+3+...+(n-1)) \cdot d}{n}$ $=a + \frac{n \cdot (n-1) \cdot d}{2n}$ $=a+ \frac{(n-1) \cdot d}{2}$
Note that this is also equal to n
$a+ \frac{(n-1) \cdot d}{2}=n$ $\therefore 2a+ (n-1) \cdot d=2n$
1st term + nth term $=2n=2 \cdot 2008=4016$ Now note that, from previous solutions, the first term is 1, hence the 2008th term is $4016-1=\boxed{4015}$
| 15
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3,144
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13
| 4
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For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$
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From inspection, we see that the sum of the sequence is $n^2$ . We also notice that $n^2$ is the sum of the first $n$ odd integers. Because $4015$ is the only odd integer, $\boxed{4015}$ is the answer.
| 15
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3,145
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_15
| 1
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How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$
|
By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$
This means that $a^2=2b+1$
We know that $a,b>0$ and that $b<100$
We also know that $a^2$ is odd and thus $a$ is odd, since the right side of the equation is odd. $2b$ is even. $2b+1$ is odd.
So $a=1,3,5,7,9,11,13$ , but if $a=1$ , then $b=0$ . Thus $a\neq1.$
$a=3,5,7,9,11,13$
The answer is $\boxed{6}$
| 6
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3,146
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_18
| 1
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Bricklayer Brenda takes $9$ hours to build a chimney alone, and bricklayer Brandon takes $10$ hours to build it alone. When they work together, they talk a lot, and their combined output decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?
$\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900$
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Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.
Using $w = rt$ , we get $x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)$ . Solving for $x$ , we get $\boxed{900}$
| 900
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3,147
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_21
| 1
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Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$
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For the first man, there are $10$ possible seats. For each subsequent man, there are $4$ $3$ $2$ , or $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$
| 480
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3,148
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_21
| 2
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Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$
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Label the seats ABCDEFGHIJ, where A is the top seat. The first man has $10$ possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are $4! = 24$ ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be $10\cdot 2\cdot 24\cdot x = 480x$ possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of $480$ . So, our answer is $\boxed{480}$
| 480
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3,149
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_22
| 1
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Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?
$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$
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There are $\frac{6!}{3!\cdot2!\cdot1!}=60$ total orderings.
Suppose we order the red and white beads first. If these two colors are ordered first, we must make sure that no neighboring beads are the same color, or only one pair of neighboring beads are the same color. There are five possible orderings:
$R\ W R\ W R$
$R\ R\ W R\ W$
$W R\ R\ W R$
$R\ W R\ R\ W$
$W R\ W R\ R$
For the first case, there are $6$ possible places we can put the blue bead. For the other 4 cases, there is only one place we can put the blue bead, which is between the pair of red beads. The desired probability is $\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{16}$
| 16
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3,150
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_23
| 1
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A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$ . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the whole floor. How many possibilities are there for the ordered pair $(a,b)$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
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Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$ . With this information we can make the equation:
\begin{eqnarray*} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4a - 4b + 8 \\ ab - 4a - 4b + 8 &=& 0 \end{eqnarray*} Applying Simon's Favorite Factoring Trick , we get
\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}
Since $b > a$ , then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$ , or $(a-4) = 2$ and $(b-4) = 4$ . This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\boxed{2}$
| 2
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3,151
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24
| 1
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Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$
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To start off, draw a diagram like in solution two and label the points. Create lines $\overline{AC}$ and $\overline{BD}$ . We can call their intersection point $Y$ . Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$ , angle $BAC$ equals $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle $CYB$ equals $120$ degrees.
Extend $\overline{CD}$ and $\overline{AB}$ and let their intersection be $E$ . Since angle $BEC$ plus angle $CYB$ equals $180$ degrees, quadrilateral $YCEB$ is a cyclic quadrilateral.
Next, draw a line from point $Y$ to point $E$ . Since angle $YBC$ and angle $YEC$ point to the same arc, angle $YEC$ is equal to $5$ degrees. Since $EYD$ is an isosceles triangle (based on angle properties) and $YAE$ is also an isosceles triangle, we can find that $YAD$ is also an isosceles triangle. Thus, each of the other angles is $\frac{180-120}{2}=30$ degrees. Finally, we have angle $BAD$ equals $30+55=\boxed{85}$ degrees.
| 85
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3,152
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24
| 2
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Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$
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First, connect the diagonal $DB$ , then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$ . Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$ , the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$ . Because angle $ABC$ is $70^\circ$ , we get angle $ABD$ is $65^\circ$ . Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$ , we see that angle $BDE$ is also $65^\circ$ , and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$
Now, because we drew $ED = DC$ , triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.
Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$ . Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$ . Then, $CEB$ is $(x-60)^\circ$ . Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$ . Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$ . Because angle $CBE$ equals angle $CEB$ , we get \[x-60=110-x\] , solving into $x=\boxed{85}$
| 85
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3,153
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24
| 3
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Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$
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[asy] unitsize(3 cm); pair A, B, C, D; A = (0,0); B = dir(85); C = B + dir(-25); D = C + dir(-35); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B))); draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C))); draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D))); dot("$A$", A, SW); dot("$B$", B, N); dot("$C$", C, NE); dot("$D$", D, SE); label("$I$", 6/7*C); [/asy]
Let the unknown $\angle BAD$ be $x$
First, we draw diagonal $BD$ and $AC$ $I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$
Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$ . Expanding on this, we can fill in a couple more angles. $\angle ABD = 70^\circ - 5^\circ = 65^\circ$ $\angle ACD = 170^\circ - 55^\circ = 115^\circ$ $\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ$ $\angle BIC = \angle AID = 180^\circ - 60^\circ = 120^\circ$
We can rewrite $\angle CAD$ and $\angle BDA$ in terms of $x$ $\angle CAD = x - 55^\circ$ and $\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x$
Let us relabel $AB = BC = CD = a$ and $AD = b$
By Rule of Sines on $\triangle ACD$ and $\triangle ABD$ respectively, $\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}$ , and $\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}$
In a more convenient form, $\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
and $\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}$
$\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
Now, by identity $\sin(\theta) = \sin(180^\circ-\theta)$ $\sin(65^\circ) = \sin(115^\circ)$
Therefore, $\sin(115^\circ-x) = \sin(x-55^\circ).$ This equation is only satisfied by option $\boxed{85}$
| 85
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3,154
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25
| 1
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Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$
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Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$ .
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$ .
We have $D(0)=200$
The truck always moves for $20$ seconds, then stands still for $30$ . During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$ , hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$ .
During the remaining $30$ seconds $D(t)$ decreases by $150$
From this observation it is obvious that after four full cycles, i.e. at $t=200$ , we will have $D(t)=0$ for the first time.
During the fifth cycle, $D(t)$ will first grow from $0$ to $100$ , then fall from $100$ to $-50$ . Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$ , then fall from $50$ to $-100$ . Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.
During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$ , then fall from $0$ to $-150$ . Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5}$ meetings.
| 5
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3,155
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25
| 2
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Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$
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The truck takes $20$ seconds to go from one pail to the next and then stops for $30$ seconds at the new pail. Thus it sets off from a pail every 50 sec. Let $t$ denote the time elapsed and write $t=50k + \Delta$ , where $\Delta \in [0,50)$ . In this time Michael has traveled $5t = 250k+5\Delta$ feet. What about the truck? In the first $50k$ seconds the truck covers $k$ pails, i.e. $200k$ feet so it moves $200+200k$ feet from Michael's starting point. Then we have two cases:
(a) if $\Delta < 20$ , then the truck travels an additional $10\Delta$ feet. For them to intersect we must have $200+200k + 10\Delta = 250k + 5\Delta$ . Solving, we get $\Delta = 10k - 40$ . Since $\Delta$ must lie in the interval $[0,20)$ we get $k \in \{4, 5\}$
(b) if $\Delta \in [20, 50)$ , then the truck travels an additional $200$ feet. For them to intersect we must have $200+200k + 200 = 250k + 5\Delta$ . Solving, we get $\Delta = 10(8-k)$ . Since $\Delta$ must lie in the interval $[20,50)$ we get $k \in \{4, 5, 6\}$
Thus Michael intersects with the truck $5$ times, which is option $\boxed{5}$
| 5
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25
| 3
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Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$
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We make a chart by seconds in increments of ten.
$\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline 0&0&200\\ 10&50&300\\ 20&100&400\\ 30&150&400\\ 40&200&400\\ 50&250&400\\ 60&300&500\\ 70&350&600\\ 80&400&600\\ 90&450&600\\ 100&500&600\\ 110&550&700\\ 120&600&800\\ 130&650&800\\ 140&700&800\\ 150&750&800\\ 160&800&900\\ 170&850&1000\\ 180&900&1000\\ 190&950&1000\\ 200&1000&1000\\ 210&1050&1100\\ 220&1100&1200\\ 230&1150&1200\\ 240&1200&1200\\ 250&1250&1200\\ 260&1300&1300\\ 270&1350&1400\\ 280&1400&1400\\ 290&1450&1400\\ 300&1500&1400\\ 310&1550&1500\\ 320&1600&1600\\ 330&1650&1600\\ 340&1700&1600\\ 350&1750&1600\\ 360&1800&1700\\ 370&1850&1800\\ 380&1900&1800\\ 390&1950&1800\\ 400&2000&1800\\ \end{tabular}$
Notice at 200, 240, 260, 280, and 320 seconds, Michael and the garbage truck meet. It is clear that they met at these times, and will meet no more. Thus the answer is $\boxed{5}$
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https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25
| 4
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Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$
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This solution might be time consuming, but it is pretty rigorous. Also, throughout the solution refer to the graph in solution 1 to understand this one more. Lets first start off by defining the position function for Michael. We let $M(t) = 5t$ , where $t$ is the amount of seconds passed. Now, lets define the position function of the truck as two independent functions. It is obvious, graphically, that the position function of the truck is a piecewise function alternating between linear lines and constant lines. Lets focus on the linear pieces of the truck's position function. Let the $kth$ linear part of the truck's position function be denoted as $L_k(t)$ . Then through algebra, it is found that $L_k(t) = 10t + 500 - 300k$ ${50k - 50 <= t <= 50k - 30}$ . Now, lets move on to the constant pieces, which is a lot easier in terms of algebra. Let the $ith$ constant part of the truck's position function be denoted as $C_i(t)$ . Then, again, through algebra we obtain $C_i(t) = 200 * (i + 1)$ ${50i - 30 <= t <= 50i}$ . Now, let me stress that $i$ and $k$ are disjoint, which is why I used different variable names. We are interested in where $C_i(t)$ and $L_k(t)$ intersect $M(t)$ or $5t$ $L_k(t)$ intersects $M$ at $(60k - 100, 300k - 500)$ . However, the only $k$ values that actually work are $5 <= k <= 7$ because of the domain restrictions on $L_k(t)$ . Similarly, we also see that for $C_i(t)$ it intersects at $(40i + 40, 200i + 200)$ . The only $i$ values that work is $4 <= i <= 7$ . However, some pair $(i, k)$ might yield the same exact intersection point. Checking this through simple algebra we see that $(4,5)$ and $(7,7)$ do indeed yield the same intersection point. Thus, our answer is $(7 - 5 + 1) + (7 - 4 + 1) - 2 = 7 - 2 = 5$ or $\boxed{5}$
| 5
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3,158
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_10
| 1
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The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
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Let $x$ be the number of the children and the mom. The father, who is $48$ , plus the sum of the ages of the kids and mom divided by the number of kids and mom plus $1$ (for the dad) = $20$ . This is because the average age of the entire family is $20.$ This statement, written as an equation, is: \[\frac{48+16x}{x+1}=20\] \[48+16x=20x+20\] \[4x=28\] \[x=7\]
$7$ people - $1$ mom = $6$ children.
Therefore, the answer is $\boxed{6}$
| 6
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_10
| 2
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The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
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Let $m$ be the Mom's age.
Let the number of children be $x$ and their average age be $y$ . Their age totaled up is simply $xy$
We have the following two equations:
$\frac{m+48+xy}{2+x}=20$ , where $m+48+xy$ is the family's total age and $2+x$ is the total number of people in the family.
$\frac{m+48+xy}{2+x}=20$
$m+48+xy=40+20x$
The next equation is $\frac{m+xy}{1+x}=16$ , where $m+xy$ is the total age of the Mom and the children, and $1+x$ is the number of children along with the Mom.
$\frac{m+xy}{1+x}=16$
$m+xy=16+16x$
We know the value for $m+xy$ , so we substitute the value back in the first equation.
$m+48+xy=40+20x$
$(16+16x)+48=40+20x$
$x=6$
Earlier, we set $x$ to be the number of children. Therefore, there are $\boxed{6}$ children.
| 6
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3,160
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_13
| 1
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Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$
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Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \frac{x}{7}$ . We get $x-2r=\frac{x}{7}$ , so $\frac{6}{7}x = 2r$ , and $\frac{x}{r} = \frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\frac{7}{3}-1 = \frac{4}{3}$ , so the answer is $\boxed{34}$
| 34
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3,161
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_13
| 2
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Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$
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[asy] draw((0,0)--(7,0)); dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0)); label("$H$",(0,0),S); label("$Y$",(3,0),S); label("$P$",(6,0),S); label("$S$",(7,0),S); [/asy] Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{34}$
| 34
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3,162
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16
| 1
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Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$
|
The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity . The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$ , since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed{58}$
| 58
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3,163
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16
| 2
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Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$
|
If we don't know our parity rules, we can check and see that $ad-bc$ is only even when $ad$ and $bc$ are of the same parity (as stated above). From here, we have two cases.
Case 1: $odd-odd$ (which must be $o \cdot o-o \cdot o$ ). The probability for this to occur is $\left(\frac 12\right)^4 = \frac 1{16}$ , because each integer has a $\frac 12$ chance of being odd.
Case 2: $even-even$ (which occurs in 4 cases: $(e \cdot e-e \cdot e$ ), ( $o \cdot e-o \cdot e$ ) (alternating of any kind), and ( $e \cdot e-o \cdot e$ ) with its reverse, ( $o \cdot e-e \cdot e$ ).
Our first subcase of case 2 has a chance of $\frac 1{16}$ (same reasoning as above).
Our second subcase of case 2 has a $\frac 14$ chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
Our third subcase of case 2 has a $\frac 18$ chance, because the 1st, 2nd, and either 3rd or 4th flip matter.
Our fourth subcase of case 2 has a $\frac 18$ chance, because it's the same, just reversed.
We sum these, and get our answer of $\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed{58}.$
| 58
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3,164
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_17
| 1
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Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 5700$
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$3 \cdot 5^2m$ must be a perfect cube, so each power of a prime in the factorization for $3 \cdot 5^2m$ must be divisible by $3$ . Thus the minimum value of $m$ is $3^2 \cdot 5 = 45$ , which makes $n = \sqrt[3]{3^3 \cdot 5^3} = 15$ . The minimum possible value for the sum of $m$ and $n$ is $\boxed{60}.$
| 60
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3,165
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_17
| 2
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Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 5700$
|
First, we need to prime factorize $75$ $75$ $5^2 \cdot 3$ . We need $75m$ to be in the form $x^3y^3$ . Therefore, the smallest $m$ is $5 \cdot 3^2$ $m$ = 45, and since $5^3 \cdot 3^3 = 15^3$ , our answer is $45 + 15$ $\boxed{60}$
| 60
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3,166
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20
| 1
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Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$
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Note that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = (a^{k} + a^{-k})^2,$ from which \[a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.\] We apply this result twice to get the answer: \begin{align*} a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \\ &= [(a + a^{-1})^2 - 2]^2 - 2 \\ &= \boxed{194} ~Azjps (Fundamental Logic)
| 194
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3,167
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20
| 2
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Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$
|
Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$
Squaring both sides of $a^2+a^{-2}=14$ gives $a^4+a^{-4}+2=196,$ from which $a^4+a^{-4}=\boxed{194}.$
| 194
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3,168
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20
| 4
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Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$
|
Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$
Applying the Binomial Theorem, we raise both sides of $a+a^{-1}=4$ to the fourth power: \begin{align*} \binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \\ a^4+4a^2+6+4a^{-2}+a^{-4}&=256 \\ \left(a^4+a^{-4}\right)+4\left(a^2+a^{-2}\right)&=250 \\ \left(a^4+a^{-4}\right)+4(14)&=250 \\ a^4+a^{-4}&=\boxed{194} ~MRENTHUSIASM
| 194
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3,169
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20
| 5
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Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$
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We multiply both sides of $4=a+a^{-1}$ by $a,$ then rearrange: \[a^2-4a+1=0.\]
We apply the Quadratic Formula to get $a=2\pm\sqrt3.$
By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of $a$ produce the same value of $a^4+a^{-4}:$ \begin{align*} a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \left(2+\sqrt{3}\right)^{-4} \\ &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&(*) \\ &=\boxed{194}$
| 194
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3,170
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20
| 7
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Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$
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Note that \[a^{4} + a^{-4} = (a^{2} + a^{-2})^{2} - 2.\] We guess that $a^{2} + a^{-2}$ is an integer, so the answer must be $2$ less than a perfect square. The only possibility is $\boxed{194}.$
| 194
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3,171
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21
| 1
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sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$
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We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.
Let $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length $s\sqrt{3}$ where $s$ is a side of the cube, the ratio of a side of the inner square to a side of the outer square is $\frac{1}{\sqrt{3}}$ (since the side of the outer square = the diagonal of the inner square). So we have $\frac{S}{24} = \left(\frac{1}{\sqrt{3}}\right)^2$ . Thus $S = 8\Rightarrow \mathrm{\boxed{8}$
| 8
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3,172
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21
| 2
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sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$
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Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives \[l^2 + l^2 + l^2 = 2^2 = 4.\] Hence each face has surface area \[l^2 = \frac{4}{3} \ \text{square meters}.\] So the surface area of the inscribed cube is $6\cdot \frac43 = \boxed{8}$ square meters.
| 8
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3,173
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21
| 3
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sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$
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First of all, it is pretty easy to see the length of each edge of the bigger cube is $2$ so the radius of the sphere is $1$ . We know that when a cube is inscribed in a sphere, assuming the edge length of the square is $x$ the radius can be presented as $\frac {\sqrt3x}{2} = 1$ , we can solve that $x=\frac {2\sqrt3}{3}$ and now we only need to apply the basic formula to find the surface and we got our final answer as $\frac {2\sqrt3}{3}*\frac {2\sqrt3}{3}*6=8$ and the final answer is $\boxed{8}$
| 8
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3,174
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_22
| 1
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A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$
$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$
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A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$ , so $S$ must be divisible by $37\ \mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{37}$
| 37
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3,175
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_23
| 1
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How many ordered pairs $(m,n)$ of positive integers , with $m \ge n$ , have the property that their squares differ by $96$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$
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Find all of the factor pairs of $96$ $(1,96),(2,48),(3,32),(4,24),(6,16),(8,12).$ You can eliminate $(1,96)$ and ( $3,32)$ because you cannot have two numbers add to be an even number and have an odd difference at the same time without them being a decimal. You only have $4$ pairs left, so the answer is $\boxed{4}$
| 4
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3,176
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25
| 1
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For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$
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It is well-known that $n \equiv S(n)\equiv S(S(n)) \pmod{9}.$ Substituting, we have that \[n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.\] Since $n \leq 2007,$ we must have that $\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of $3$ because $n$ is a multiple of $3.$
Then, we compute the corresponding values of $S(S(n)).$
Finally, we may compute the corresponding values of $n$ using the fact that $n=2007-S(n)-S(S(n)).$
Notice how all conditions are designed to be satisfied except whether $S(n)$ is accurate with respect to $n.$ So, the only thing that remains is to check this. We may eliminate, for example, when $n=2007$ we have $S(n)=9$ while the table states that it is $0.$ Proceeding similarly, we obtain the following table.
It follows that there are $\boxed{4}$ possible values for $n.$ ~samrocksnature
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3,177
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25
| 2
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For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$
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Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$
Proof of claim:
We examine the positive integers mod $9$ . Here are the cases.
Case 1. $n \equiv 1 \pmod 9$ . Now, we examine $S(n)$ modulo $9$ .
Case 1.1. The tens digit of $n$ is different from the tens digit of the largest multiple of $9$ under $n$ . (In other words, this means we will carry when adding from the perfect multiple of $9$ under $n$ .)
Observe that when we carry, i.e. Add $1$ onto $1989$ to obtain $1990$ , the units digit decreases by $9$ while the tens digit increases by $1$ . This means that the sum of the digits decreases by $8$ in total, and we have $-8 \equiv 1 \pmod 9$ , so the "mod 9" of the sum increases by $1$ . This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by $1$
Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by $1$ which means that the sum is also equivalent to $1 \pmod 9$
This means that $S(n) \equiv 1 \pmod 9$ and similarly letting the next $n=S(n)$ $S(S(n)) \equiv 1 \pmod 9$ . Summing these, we have $n+S(n)+S(S(n)) \equiv 3 \pmod 9$ . Clearly, no integers of this form will satisfy the condition because $2007$ is a perfect multiple of $9$
Case 2. $n \equiv 2 \pmod 9$
In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to $2$ mod $9$ . Then we can apply similar arguments to get $S(n) \equiv 2 \pmod 9$ and $S(S(n)) \equiv 2 \pmod 9$ , so adding gives $n+S(n)+S(S(n)) \equiv 6 \pmod 9$
It is trivial to see that for $n \equiv k \pmod 9$ , for $0 \leq k \leq 8$ , we must have $n+S(n)+S(S(n)) \equiv 3k \pmod 9$ . Only when $k=0, 3, 6$ is $3k$ a multiple of $9$ , which means that $n$ must be a multiple of $3$
Now, we find the integers. Again, consider two cases: Integers that are direct multiples of $9$ and integers that are multiples of $3$ but not $9$
Case 1. $n$ is a multiple of $9$ . An integer of the form $\overline{20ab}$ will not work since the least such integer is $2007$ which already exceeds our bounds. Thus, we need only consider the integers of the form $\overline{19ab}$ . The valid sums of the digits of $n$ are $18$ and $27$ in this case.
Case 1.1. The sum of the digits is $18$ . This means that $S(n)=18, S(S(n))=9$ , so $n=2007-18-9=1980$ . Clearly this number satisfies our constraints.
Case 1.2. The sum of the digits is $27$ . This means that $S(n)=27, S(S(n))=9$ , ,so $n=2007-27-9=1971$ . Since the sum of the digits of $1971$ is not $27$ , this does not work.
This means that there is $1$ integer in this case.
Case 2. $n$ is a multiple of $3$ , not $9$ .
.
Case 2.1. Integers of the form $\overline{20ab}$ . Then $S(n)=3$ or $S(n)=6$ ; it is trivial to see that $S(n)=6$ exceeds our bounds, so $S(n)=3$ and $n=2007-6=2001$
Case 2.2. Integers of the form $\overline{19ab}$ . Then $S(n)=12, 15, 21, 24$ and we consider each case separately.
Case 2.2.1. Integers with $S(n)=12$ . That means $n=2007-12-3=1992$ which clearly does not work.
Case 2.2.2. Integers with $S(n)=15$ . That means $n=2007-15-6=1986$ which also does not work
Case 2.2.3. Integers with $S(n)=21$ . That means $n=2007-21-3=1983$ which is valid.
Case 2.2.4. Integers with $S(n)=24$ . That means $n=2007-24-6=1977$ which is also valid.
We have considered every case, so there are $\boxed{4}$ integers that satisfy the given condition.
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25
| 3
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For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$
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Let the number of digits of $n$ be $m$ . If $m = 5$ $n$ will already be greater than $2007$ . Notice that $S(n)$ is always at most $9m$ . Then if $m = 3$ $n$ will be at most $999$ $S(n)$ will be at most $27$ , and $S(S(n))$ will be even smaller than $27$ . Clearly we cannot reach a sum of $2007$ , unless $m = 4$ (i.e. $n$ has $4$ digits).
Then, let $n$ be a four digit number in the form $1000a + 100b + 10c + d$ . Then $S(n) = a + b + c + d$
$S(S(n))$ is the sum of the digits of $a + b + c + d$ . We can represent $S(S(n))$ as the sum of the tens digit and the ones digit of $S(n)$ . The tens digit in the form of a decimal is
$\frac{a + b + c + d}{10}$
To remove the decimal portion, we can simply take the floor of the expression,
$\lfloor\frac{a + b + c + d}{10}\rfloor$
Now that we have expressed the tens digit, we can express the ones digit as $S(n) -10$ times the above expression, or
$a + b + c + d - 10\lfloor\frac{a + b + c + d}{10}\rfloor$
Adding the two expressions yields the value of $S(S(n))$
$= a + b + c + d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$
Combining this expression to the ones for $n$ and $S(n)$ yields
$1002a + 102b + 12c + 3d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$
Setting this equal to $2007$ and rearranging a bit yields
$12c + 3d = 2007 - 1002a - 102b + 9\lfloor\frac{a + b + c + d}{10}\rfloor$
$\Rightarrow$ $4c + d = 669 - 334a - 34b + 3\lfloor\frac{a + b + c + d}{10}\rfloor$
(The reason for this slightly weird arrangement will soon become evident)
Now we examine the possible values of $a$ . If $a \ge 3$ $n$ is already too large. $a$ must also be greater than $0$ , or $n$ would be a $3$ -digit number. Therefore, $a = 1 \, \text{or} \, 2$ . Now we examine by case.
If $a = 2$ , then $b$ and $c$ must both be $0$ (otherwise $n$ would already be greater than $2007$ ). Substituting these values into the equation yields
$d = 1 + 3\lfloor\frac{2 + d}{10}\rfloor$
$\Rightarrow$ $d=1$
Sure enough, $2001 + (2+1) + 3=2007$
Now we move onto the case where $a = 1$ . Then our initial equation simplifies to
$4c + d = 335 - 34b + 3\lfloor\frac{1 + b + c + d}{10}\rfloor$
Since $c$ and $d$ can each be at most $9$ , we substitute that value to find the lower bound of $b$ . Doing so yields
$34b \ge 290 + 3\lfloor\frac{19 + b}{10}\rfloor$
The floor expression is at least $3\lfloor\frac{19}{10}\rfloor=3$ , so the right-hand side is at least $293$ . Solving for $b$ , we see that $b \ge 9$ $\Rightarrow$ $b=9$ . Again, we substitute for $b$ and the equation becomes
$4c + d = 29 + 3\lfloor\frac{10 + c + d}{10}\rfloor$
$\Rightarrow$ $4c + d = 32 + 3\lfloor\frac{c + d}{10}\rfloor$
Just like we did for $b$ , we can find the lower bound of $c$ by assuming $d = 9$ and solving:
$4c + 9 \ge 29 + 3\lfloor\frac{c + 9}{10}\rfloor$
$\Rightarrow$ $4c \ge 20 + 3\lfloor\frac{c + 9}{10}\rfloor$
The right hand side is $20$ for $c=0$ and $23$ for $c \ge 1$ . Solving for c yields $c \ge 6$ . Looking back at the previous equation, the floor expression is $0$ for $c+d \le 9$ and $3$ for $c+d \ge 10$ . Thus, the right-hand side is $32$ for $c+d \le 9$ and $35$ for $c+d \ge 10$ . We can solve these two scenarios as systems of equations/inequalities:
$4c+d = 32$
$c+d \le 9$
and
$4c+d=35$
$c+d \ge 10$
Solving yields three pairs $(c, d):$ $(8, 0)$ $(8, 3)$ ; and $(7, 7)$ . Checking the numbers $1980$ $1983$ , and $1977$ ; we find that all three work. Therefore there are a total of $4$ possibilities for $n$ $\Rightarrow$ $\boxed{4}$
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3,179
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_1
| 1
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Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
$\mathrm{(A)}\ 678 \qquad \mathrm{(B)}\ 768 \qquad \mathrm{(C)}\ 786 \qquad \mathrm{(D)}\ 867 \qquad \mathrm{(E)}\ 876$
|
There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \[2\cdot(12\cdot8+10\cdot8)-60=2\cdot176-60=292\] We have three bedrooms, so she must paint $292\cdot3=\boxed{876}$ square feet of walls.
| 876
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3,180
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_2
| 1
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Define the operation $\star$ by $a \star b = (a+b)b.$ What is $(3 \star 5) - (5 \star 3)?$
$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16$
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Substitute and simplify. \[(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{16}\]
| 16
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3,181
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https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_2
| 2
|
Define the operation $\star$ by $a \star b = (a+b)b.$ What is $(3 \star 5) - (5 \star 3)?$
$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16$
|
Note that $(a \star b) - (b \star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2$ . We can substitute $a=3$ and $b=5$ to get $5^2 - 3^2 = \boxed{16}$
| 16
|
3,182
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_3
| 1
|
A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$
|
The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{24}$
| 24
|
3,183
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_3
| 2
|
A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$
|
Alternatively, we can use the harmonic mean to get $\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = \boxed{24}$
| 24
|
3,184
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_4
| 1
|
The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$
$\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60$
|
Because all the central angles of a circle add up to $360^\circ,$
\begin{align*} \angle BOC + \angle AOB + \angle AOC &= 360\\ 120 + 140 + \angle AOC &= 360\\ \angle AOC &= 100. \end{align*}
Therefore, the measure of $\text{arc}AC$ is also $100^\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\angle ABC = \boxed{50}$
| 50
|
3,185
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_6
| 1
|
The $2007 \text{ AMC }10$ will be scored by awarding $6$ points for each correct response, $0$ points for each incorrect response, and $1.5$ points for each problem left unanswered. After looking over the $25$ problems, Sarah has decided to attempt the first $22$ and leave only the last $3$ unanswered. How many of the first $22$ problems must she solve correctly in order to score at least $100$ points?
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$
|
Sarah is leaving $3$ questions unanswered, guaranteeing her $3 \times 1.5 = 4.5$ points. She will either get $6$ points or $0$ points for the rest of the questions. Let $x$ be the number of questions Sarah answers correctly. \begin{align*} 6x+4.5 &\ge 100\\ 6x &\ge 95.5\\ x &\ge 15.92 \end{align*} The number of questions she answers correctly has to be a whole number, so round up to get $\boxed{16}$
| 16
|
3,186
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_7
| 1
|
All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150$
|
$AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition: \[\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{150}\]
| 150
|
3,187
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_8
| 1
|
CHIKEN NUGGIEs
|
Case $1$ : The numbers are separated by $1$
We this case with $a=0, b=1,$ and $c=2$ . Following this logic, the last set we can get is $a=7, b=8,$ and $c=9$ . We have $8$ sets of numbers in this case.
Case $2$ : The numbers are separated by $2$
This case starts with $a=0, b=2,$ and $c=2$ . It ends with $a=5, b=7,$ and $c=9$ . There are $6$ sets of numbers in this case.
Case $3$ : The numbers start with $a=0, b=3,$ and $c=6$ . It ends with $a=3, b=6,$ and $c=9$ . This case has $4$ sets of numbers.
It's pretty clear that there's a pattern: $8$ sets, $6$ sets, $4$ sets. The amount of sets per case decreases by $2$ , so it's obvious Case $4$ has $2$ sets. The total amount of possible five-digit numbers is $8+6+4+2=\boxed{20}$
| 20
|
3,188
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_12
| 1
|
Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$
|
Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{5} Note that actual values were not found.
| 5
|
3,189
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_14
| 1
|
Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$
|
If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\%$ of the group are girls, \begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20&=3p\\ p&=20\end{align*} The number of girls initially in the group is $0.4p=0.4(20)=\boxed{8}$
| 8
|
3,190
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_14
| 2
|
Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$
|
Let $x$ be the number of people initially in the group and $g$ the number of girls. $\frac{2}{5}x = g$ , so $x = \frac{5}{2}g$ . Also, the problem states $\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \boxed{8}$
| 8
|
3,191
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_15
| 1
|
The angles of quadrilateral $ABCD$ satisfy $\angle A=2 \angle B=3 \angle C=4 \angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?
$\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180$
|
The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{173}\]
| 173
|
3,192
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_16
| 1
|
A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$
|
We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{93}$
| 93
|
3,193
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_16
| 2
|
A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$
|
Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$ . Simplifying this equation yields $j = \boxed{93}$
| 93
|
3,194
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_20
| 1
|
A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$
|
There are $25$ ways to choose the first square. The four remaining squares in its row and column and the square you chose exclude nine squares from being chosen next time.
There are $16$ remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.
Finally, the last square has $9$ remaining choices.
The number of ways to choose $3$ squares is $25 \cdot 16 \cdot 9,$ but the order in which you chose the squares does not matter as the blocks are indistinguishable, so we divide by $3!$
\[\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{600}\]
| 600
|
3,195
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_20
| 2
|
A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$
|
Once we choose our three squares, we will have occupied three separate columns $(A, B, C)$ and three separate rows. There are ${5 \choose 3} \times {5 \choose 3}$ ways to choose these rows and columns.
There are $3$ ways to assign the square in column $A$ to a row, $2$ ways to assign the square in column $B$ to one of the remaining two rows, and poor square in column $C$ doesn't get to choose. $:($
In total, we have \[{5 \choose 3} \times {5 \choose 3} \times 3!\] which totals out to $\boxed{600}$
| 600
|
3,196
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_20
| 3
|
A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$
|
We know that there are $\binom{25}{3}=2300$ ways to choose three blocks. However, the restriction clearly limits the number of ways we can choose our blocks. Hence, only $\text{(A)}$ $\text{(B)}$ , or $\text{(C)}$ could be the correct answer. Clearly, there are more than $125$ ways, thus yielding $\boxed{600}$ ways.
| 600
|
3,197
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_24
| 1
|
Let $n$ denote the smallest positive integer that is divisible by both $4$ and $9,$ and whose base- $10$ representation consists of only $4$ 's and $9$ 's, with at least one of each. What are the last four digits of $n?$
$\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944$
|
For a number to be divisible by $4,$ the last two digits have to be divisible by $4.$ That means the last two digits of this integer must be $4.$
For a number to be divisible by $9,$ the sum of all the digits must be divisible by $9.$ The only way to make this happen is with nine $4$ 's. However, we also need one $9.$
The smallest integer that meets all these conditions is $4444444944$ . The last four digits are $\boxed{4944}$
| 944
|
3,198
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_1
| 1
|
Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
$\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35$
|
The $5$ sandwiches cost $5\cdot 3=15$ dollars. The $8$ sodas cost $8\cdot 2=16$ dollars. In total, the purchase costs $15+16=\boxed{31}$ dollars.
| 31
|
3,199
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_3
| 1
|
The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$
|
Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{18}.$
| 18
|
3,200
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_3
| 2
|
The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$
|
We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct.
| 18
|
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