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3,301
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23
| 2
|
In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$
|
Mark $DC=z$ $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ $FECD$ are the same. Mark the height to be $h$
Then, we have that $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$
From this, we get that $x=2z+y$
We also get that $\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)$
Simplifying, we get that $2x=z+3y$
Notice that we want $\tfrac{AB}{DC}=\tfrac{x}{z}$
Dividing the first equation by $z$ , we get that $\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})$
Dividing the second equation by $z$ , we get that $2(\tfrac{x}{z})=1+3(\tfrac{y}{z})$
Now, when we subtract the top equation from the bottom, we get that $\tfrac{x}{z}=\boxed{5}$
| 5
|
3,302
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23
| 3
|
In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$
|
Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid $x$ . The area of the smaller trapezoid is $A$ $h \cdot \frac {b_1 + b_2}{2}$ . The area of the larger trapezoid is $A$ $h \cdot \frac {b_1 + x}{2}$ . Since this problem asks for proportions, assume that $b_1 = 1$ and $b_2 = 2$
The smaller trapezoid has area $h$ while the larger trapezoid must have area $5h$ . We have the equation $\frac {x}{2} = 5$ $x$ = 10, and our answer is $\boxed{5}$
| 5
|
3,303
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24
| 1
|
Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$
|
Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero.
Then, $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$
Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2 \cdot 11$ , so $11|(a-b)(a+b)$ . However, the maximum value of $a-b$ is $9-1=8$ , so $11|a+b$ . The maximum value of $a+b$ is $9+8=17$ , so $a+b=11$
Then, we have $33^2(a-b) = m^2$ , so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$ . We know $a+b=11$ , and, for $a-b=1$ , adding equations to eliminate $b$ gives us $2a=12 \Longrightarrow a=6, b=5$ . Testing $a-b=4$ gives us $2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}$ , which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$ , and $x+y+m=\boxed{154}$
| 154
|
3,304
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24
| 2
|
Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$
|
The first steps are the same as Solution 1. Let $x = 10a+b, y = 10b+a$ , where we know that a and b are digits (whole numbers less than $10$ ).
Like Solution 1, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . This is where the solution diverges.
We know that the left side of the equation is a perfect square because $m$ is an integer. If we factor $99$ into its prime factors, we get $3^2\cdot 11$ . In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even. Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals $11$ while $(a-b)$ is a factor of nine (1 or 9). The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$ . The sum of the numbers is $\boxed{154}$
| 154
|
3,305
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24
| 3
|
Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$
|
Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$ , substituting and factoring, we get $99(a+b)(a-b) = m^2$
Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$
Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$ and $m = \sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$ , which we can also write as $(x-y)(x+y) = 11^2 \cdot 3^2$
A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$ , so $x+y+m$ $=$ $\boxed{154}$
| 154
|
3,306
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24
| 4
|
Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$
|
Continue the same as Solution $3$ until we get $33$ . Knowing that $33^2 = 1089$ , we have narrowed down our Pythagorean triples. We know that the $2$ other squares should be larger than $33^2$ , so we can start testing.
If we start testing the $40$ s, it is fruitless since the closest to $33^2$ would be $33 - 45 - 54$ which is not a Pythagorean triple. We can start by testing out the $50$ s, and it turns our that $33 - 56 - 65$ is a Pythagorean triple. Therefore, our answer is $33+56+65$ $\boxed{154}$
| 154
|
3,307
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25
| 1
|
subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$
|
The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ $25+100$ $26+99$ $27+98$ $\ldots$ $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{62}$
| 62
|
3,308
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25
| 2
|
subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$
|
"Cut" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $\boxed{62}$
| 62
|
3,309
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25
| 3
|
subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$
|
The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$ . So, we have to find two consecutive numbers, $n$ and $n+1$ , whose sum is $125$ . Setting up our equation, we have $n+(n+1) = 2n+1 = 125$ . When we solve for $n$ , we get $n =\boxed{62}$
| 62
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3,310
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25
| 4
|
subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$
|
We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get $B =\boxed{62}$
| 62
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3,311
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_1
| 1
|
You and five friends need to raise $1500$ dollars in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?
$\mathrm{(A) \ } 250\qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 1500 \qquad \mathrm{(D) \ } 7500 \qquad \mathrm{(E) \ } 9000$
|
There are $6$ people to split the $1500$ dollars among, so each person must raise $\frac{1500}6=250$ dollars. $\Rightarrow\boxed{250}$
| 250
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3,312
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_3
| 1
|
Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$
|
$20$ dollars is the same as $2000$ cents, and $1.45\%$ of $2000$ is $0.0145\times2000=29$ cents. $\Rightarrow\boxed{29}$
| 29
|
3,313
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_3
| 2
|
Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$
|
Since there can't be decimal values of cents, the answer must be $\Rightarrow\boxed{29}$
| 29
|
3,314
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_4
| 1
|
What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$
|
$|x-1|$ is the distance between $x$ and $1$ $|x-2|$ is the distance between $x$ and $2$
Therefore, the given equation says $x$ is equidistant from $1$ and $2$ , so $x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{32}$
| 32
|
3,315
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_4
| 2
|
What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$
|
We know that either $x-1=x-2$ or $x-1=-(x-2)$ . The first equation simplifies to $-1=-2$ , which is false, so $x-1=-(x-2)$ . Solving, we get $x=\frac{3}{2}\Rightarrow\boxed{32}$
| 32
|
3,316
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6
| 1
|
Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$
|
Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$
| 26
|
3,317
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6
| 2
|
Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$
|
Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$
| 26
|
3,318
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6
| 3
|
Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$
|
Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters. Adding them together gives the answer of $\boxed{26}$
| 26
|
3,319
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_7
| 1
|
A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?
$\mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134$
|
There are $5\times8=40$ oranges on the $1^{\text{st}}$ layer of the stack. The $2^{\text{nd}}$ layer that is added on top of the first will be a layer of $4\times7=28$ oranges. When the third layer is added on top of the $2^{\text{nd}}$ , it will be a layer of $3\times6=18$ oranges, etc.
Therefore, there are $5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100$ oranges in the stack. $\boxed{100}$
| 100
|
3,320
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8
| 1
|
A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$
|
We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.
After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\boxed{37}$ rounds until the game ends.
| 37
|
3,321
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8
| 2
|
A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$
|
Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.
First round: $15,14,13$ (given)
Second round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$
We see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$ th number in the sequence $1,4,7,10...$ which is $\boxed{37}$
| 37
|
3,322
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8
| 3
|
A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$
|
Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$ $A$ had $11$ tokens, and in round $17$ $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$ $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\boxed{37}$
| 37
|
3,323
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9
| 1
|
In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$
|
Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ $[\triangle CDB]$ . So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$ . Therefore, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{4}$
| 4
|
3,324
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9
| 2
|
In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$
|
Since $AE \perp AB$ and $BC \perp AB$ $AE \parallel BC$ . By alternate interior angles and $AA\sim$ , we find that $\triangle ADE \sim \triangle CDB$ , with side length ratio $\frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$ . Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{4}$
| 4
|
3,325
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9
| 3
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In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$
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Let $[X]$ represent the area of figure $X$ . Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{4}$
| 4
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3,326
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9
| 4
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In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$
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We want to figure out $Area(\triangle ADE) - Area(\triangle BDC)$ .
Notice that $\triangle ABC$ and $\triangle BAE$ "intersect" and form $\triangle ADB$
This means that $Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$ because $Area(\triangle ADB)$ cancels out, which can be seen easily in the diagram.
$Area(\triangle BAE) = 0.5 * 4 * 8 = 16$
$Area(\triangle ABC) = 0.5 * 4 * 16 = 12$
$Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{4}$
| 4
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3,327
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_11
| 1
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A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume , by what percent must the height be decreased?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60$
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When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$
| 36
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3,328
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_12
| 1
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Henry's Hamburger Haven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two,or three meat patties and any collection of condiments. How many different kinds of hamburgers can be ordered?
$\text{(A) \ } 24 \qquad \text{(B) \ } 256 \qquad \text{(C) \ } 768 \qquad \text{(D) \ } 40,320 \qquad \text{(E) \ } 120,960$
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For each condiment, a customer may either choose to order it or not. There are $8$ total condiments to choose from. Therefore, there are $2^8=256$ ways to order the condiments. There are also $3$ choices for the meat, making a total of $256\times3=768$ possible hamburgers. $\boxed{768}$
| 768
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3,329
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_13
| 1
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At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?
$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$
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If each man danced with $3$ women, then there will be a total of $3\times12=36$ pairs of men and women. However, each woman only danced with $2$ men, so there must have been $\frac{36}2 \Longrightarrow \boxed{18}$ women.
| 18
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3,330
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_13
| 2
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At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?
$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$
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Consider drawing out a diagram. Let a circle represent a man, and let a shaded circle represent a woman.
Then, we know that for every 2 men, there will be 3 woman using our diagram. Therefore, the ratio between the number of men and women is 2:3.
Hence, we know that:
$\frac{2}{3} = \frac{12}{x} \implies x = 18 \implies \boxed{18}.$
| 18
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3,331
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_14
| 1
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The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$
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Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{0}$ dimes.
| 0
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3,332
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_14
| 2
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The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$
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If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\boxed{0}.$
| 0
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3,333
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_15
| 1
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Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$
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Rewrite $\frac{(x+y)}x$ as $\frac{x}x+\frac{y}x=1+\frac{y}x$
We also know that $\frac{y}x<0$ because $x$ and $y$ are of opposite sign.
Therefore, $1+\frac{y}x$ is maximized when $|\frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.
This occurs at $(-4,2)$ , so $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{12}$
| 12
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3,334
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_15
| 2
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Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$
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If the answer choice is valid, then it must satisfy $\frac{(x+y)}x$ . We use answer choices from greatest to least since the question asks for the greatest value.
Answer choice $\text{(E)}$ . We see that if $\frac{(x+y)}x = 1$ then
$x+y=x$ and $y=0$ . However, $0$ is not in the domain of $y$ , so $\text{(E)}$ is incorrect.
Answer choice $\text{(D)}$ , however, we can find a value that satisfies $\frac{x+y}{x}=\frac{1}{2}$ which simplifies to $x+2y=0$ , such as $(-4,2)$
Therefore, $\boxed{12}$ is the greatest.
| 12
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3,335
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_15
| 3
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Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$
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As $-4\leq x\leq-2$ , we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or $x+y$ must also be as small as possible.
So we pick our smallest value for $y$ , which is $2$
Now if we if set our value of $x$ to its lowest, our expression becomes $\frac{(-2+2)}{-2} = \frac{0}{2}$ As we decrease our $x$ value, we see that our numerator decrease from $0$ and our denominator decrease from $-2$ at the same rate. So decreasing our $x$ value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number.
So we also pick the smallest value for $x$ , which is $-4$ . We know have $\frac{(-4+2)}{-4} = \frac{-2}{-4} = \frac{1}{2}$
Therefore, $\boxed{12}$ is our greatest possible value.
| 12
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3,336
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_16
| 1
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The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?
2004 AMC 10A problem 16.png
$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 19\qquad \mathrm{(E) \ } 20$
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Since there are five types of squares: $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,$ and $5 \times 5.$ We must find how many of each square contain the black shaded square in the center.
If we list them, we get that
Thus, the answer is $1+4+9+4+1=19\Rightarrow\boxed{19}$
| 19
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3,337
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_16
| 2
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The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?
2004 AMC 10A problem 16.png
$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 19\qquad \mathrm{(E) \ } 20$
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We use complementary counting. There are only $2\times2$ and $1\times1$ squares that do not contain the black square. Counting, there are $12$ $2\times2$ squares, and $25-1 = 24$ $1\times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible $(25$ $1\times1$ squares $16$ $2\times2$ squares $9$ $3\times3$ squares $4$ $4\times4$ squares and $1$ $5\times5$ square), therefore there are $55-36 = 19$ squares that contain the black square, which is $\boxed{19}$
| 19
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3,338
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_17
| 1
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Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$
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Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350\Longrightarrow\boxed{350}$
| 350
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3,339
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_17
| 2
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Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$
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The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{350}$
| 350
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3,340
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18
| 1
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sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{1}$
| 1
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3,341
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18
| 2
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sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ . Thus, we get the following equations:
$9r=11+d\Rightarrow d=9r-11$
$9r^2=29+2d$
Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ . If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{1}$
| 1
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3,342
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18
| 3
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sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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Let the three numbers be, in increasing order, $9,y,z$
Hence, we have that $9-y=y-z\implies 9+z=2y$
Also, from the second part of information given, we get that
$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2$
Plugging back in...
$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$
Simplifying, we get that $z^2-10z-551=0$
Applying the quadratic formula, we get that $z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}$
Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$
However, the problem asks for the minimum value of the third term in a geometric progression.
Hence, the answer is $-19+20=\boxed{1}$
| 1
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3,343
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18
| 4
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sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1}$
| 1
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3,344
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18
| 5
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sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\boxed{1}$
| 1
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3,345
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18
| 6
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sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\boxed{1}$
| 1
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3,346
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_19
| 1
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A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?
[asy] size(250);defaultpen(linewidth(0.8)); draw(ellipse(origin, 3, 1)); fill((3,0)--(3,2)--(-3,2)--(-3,0)--cycle, white); draw((3,0)--(3,16)^^(-3,0)--(-3,16)); draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); filldraw(ellipse((0, 16), 3, 1), white, black); draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((-3,2)--(0,-1)^^(-3,1)--(-1,-0.89)); draw((0,-1)--(0,15), dashed); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4)); draw((-7,0)--(-5,0)^^(-7,16)--(-5,16)); draw((3,-3)--(-3,-3), Arrows(6)); draw((-6,0)--(-6,16), Arrows(6)); draw((-2,9)--(-1,9), Arrows(3)); label("$3$", (-1.375,9.05), dir(260), fontsize(7)); label("$A$", (0,15), N); label("$B$", (0,-1), NE); label("$30$", (0, -3), S); label("$80$", (-6, 8), W);[/asy]
$\mathrm{(A) \ } 120 \qquad \mathrm{(B) \ } 180 \qquad \mathrm{(C) \ } 240 \qquad \mathrm{(D) \ } 360 \qquad \mathrm{(E) \ } 480$
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The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base $3$ and height $80$ . Thus, we get $3\times80=240\Rightarrow\boxed{240}$
| 240
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3,347
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_20
| 1
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Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$
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Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=x\sqrt{2}$ . If we go angle chasing, we find out that $\angle AEB=75^{\circ}$ , thus $\angle ABE=15^{\circ}$ $\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ . Thus $\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ , or $AE=\frac{x(\sqrt{3}-1)}{2}$ . Thus $AB=\frac{x(\sqrt{3}+1)}{2}$ , and $[ABE]=\frac{x^2}{4}$ , and $[DEF]=\frac{x^2}{2}$ . Thus the ratio of the areas is $\boxed{2}$
| 2
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3,348
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_20
| 2
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Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$
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WLOG, let the side length of $ABCD$ be 1. Let $DE = x$ . It suffices that $AE = 1 - x$ . Then triangles $ABE$ and $CBF$ are congruent by HL, so $CF = AE$ and $DE = DF$ . We find that $BE = EF = x \sqrt{2}$ , and so, by the Pythagorean Theorem, we have $(1 - x)^2 + 1 = 2x^2.$ This yields $x^2 + 2x = 2$ , so $x^2 = 2 - 2x$ . Thus, the desired ratio of areas is \[\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{2}.\]
| 2
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3,349
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_20
| 3
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Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$
|
Assume $AB=1$ . Then, $FC$ is $x$ and $ED$ is $1-x$ . We see that using $HL$ $FCB$ is congruent to EAB. Using Pythagoras of triangles $FCB$ and $FDE$ we get $2{(1-x)}^2=x^2+1$ . Expanding, we get $2x^2-4x+2=x^2+1$ . Simplifying gives $x^2-4x+1=0$ solving using completing the square (or other methods) gives 2 answers: $2-\sqrt{3}$ and $2+\sqrt{3}$ . Because $x < 1$ $x=2-\sqrt{3}$ . Using the areas, the answer is $\boxed{2}$
| 2
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3,350
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_20
| 4
|
Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$
|
First, since $\bigtriangleup BEF$ is equilateral and $ABCD$ is a square, by the Hypothenuse Leg Theorem, $\bigtriangleup ABE$ is congruent to $\bigtriangleup CBF$ . Then, assume length $AB = BC = x$ and length $DE = DF = y$ , then $AE = FC = x - y$ $\bigtriangleup BEF$ is equilateral, so $EF = EB$ and $EB^2 = EF^2$ , it is given that $ABCD$ is a square and $\bigtriangleup DEF$ and $\bigtriangleup ABE$ are right triangles. Then we use the Pythagorean theorem to prove that $AB^2 + AE^2 = EB^2$ and since we know that $EB^2 = EF^2$ and $EF^2 = DE^2 + DF^2$ , which means $AB^2 + AE^2 = DE^2 + DF^2$ . Now we plug in the variables and the equation becomes $x^2 + (x+y)^2 = 2y^2$ , expand and simplify and you get $2x^2 - 2xy = y^2$ . We want the ratio of area of $\bigtriangleup DEF$ to $\bigtriangleup ABE$ . Expressed in our variables, the ratio of the area is $\frac{y^2}{x^2 - xy}$ and we know $2x^2 - 2xy = y^2$ , so the ratio must be $2$ . So, the answer is $\boxed{2}$
| 2
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3,351
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_21
| 1
|
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)
$\mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4}$
|
Let the area of the shaded region be $S$ , the area of the unshaded region be $U$ , and the acute angle that is formed by the two lines be $\theta$ . We can set up two equations between $S$ and $U$
$S+U=9\pi$
$S=\dfrac{8}{13}U$
Thus $\dfrac{21}{13}U=9\pi$ , and $U=\dfrac{39\pi}{7}$ , and thus $S=\dfrac{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}$
Now we can make a formula for the area of the shaded region in terms of $\theta$
$\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}$
Thus $3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{7}$
| 7
|
3,352
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_21
| 2
|
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)
$\mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4}$
|
As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of $\theta$
$\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi$
So, the shaded region is $3\theta+3\pi$ . This means that the unshaded region is $9\pi-(3\theta+3\pi)$
Also, the shaded region is $\frac{8}{13}$ of the unshaded region. Hence, we can now make an equation and solve for $\theta$
$3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta$
Simplifying, we get $63\theta=9\pi\implies \theta=\boxed{7}$
| 7
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3,353
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_1
| 1
|
Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?
$\mathrm{(A) \ } 297 \qquad \mathrm{(B) \ } 330\qquad \mathrm{(C) \ } 363\qquad \mathrm{(D) \ } 396\qquad \mathrm{(E) \ } 726$
|
There are $22-12+1=11$ rows of $33$ seats, giving $11\times 33=\boxed{363}$ seats.
| 363
|
3,354
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_2
| 1
|
How many two-digit positive integers have at least one $7$ as a digit?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30$
|
Ten numbers $(70,71,\dots,79)$ have $7$ as the tens digit. Nine numbers $(17,27,\dots,97)$ have it as the ones digit. Number $77$ is in both sets.
Thus the result is $10+9-1=18 \Rightarrow$ $\boxed{18}$
| 18
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3,355
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_2
| 2
|
How many two-digit positive integers have at least one $7$ as a digit?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30$
|
We use complementary counting. The complement of having at least one $7$ as a digit is having no $7$ s as a digit.
We have $9$ digits to choose from for the first digit and $10$ digits for the second. This gives a total of $9 \times 10 = 90$ two-digit numbers.
But since we cannot have $7$ as a digit, we have $8$ first digits and $9$ second digits to choose from.
Thus there are $8 \times 9 = 72$ two-digit numbers without a $7$ as a digit.
$90$ (The total number of two-digit numbers) $- 72$ (The number of two-digit numbers without a $7$ $= 18 \Rightarrow$ $\boxed{18}$
| 18
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3,356
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_3
| 1
|
At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $48$ free throws. How many free throws did she make at the first practice?
$\mathrm{(A) \ } 3 \qquad \mathrm{(B) \ } 6 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 15$
|
At the fourth practice she made $48/2=24$ throws, at the third one it was $24/2=12$ , then we get $12/2=6$ throws for the second practice, and finally $6/2=3\Rightarrow\boxed{3}$ throws at the first one.
| 3
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3,357
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_4
| 1
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A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$
$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$
|
The product of all six numbers is $6!=720$ . The products of numbers that can be visible are $720/1$ $720/2$ , ..., $720/6$ .
The answer to this problem is their greatest common divisor -- which is $720/L$ , where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$ .
Clearly $L=60$ and the answer is $720/60=\boxed{12}$
| 12
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3,358
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_4
| 2
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A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$
$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$
|
The product P can be one of the following six numbers excluding the number that is hidden under, so we have:
\begin{align*}
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\
1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\
1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\
1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\
1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5
\end{align*}
The largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \cdot 3$
Hence $P=3\cdot2^2=\boxed{12}$
| 12
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3,359
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_5
| 1
|
In the expression $c\cdot a^b-d$ , the values of $a$ $b$ $c$ , and $d$ are $0$ $1$ $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$
|
If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ If $b=0$ , the expression evaluates to $c-d\leq 2$ Case $d=0$ remains.
In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$ . Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$ , where $c\cdot a^b=1\cdot 3^2=\boxed{9}$
| 9
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3,360
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_7
| 1
|
On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9$
|
Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $\boxed{5}$
| 5
|
3,361
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_7
| 2
|
On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9$
|
Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $\frac{60}{3}=20$ times. Thus $d=7\cdot20=140$ , and the sum of the digits is $\boxed{5}$
| 5
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3,362
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_8
| 1
|
Minneapolis-St. Paul International Airport is $8$ miles southwest of downtown St. Paul and $10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?
$\mathrm{(A)\ }13\qquad\mathrm{(B)\ }14\qquad\mathrm{(C)\ }15\qquad\mathrm{(D)\ }16\qquad\mathrm{(E)\ }17$
|
The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs $8$ miles and $10$ miles long. The hypotenuse length is $\sqrt{8^2 + 10^2}\approx12.8$ , and thus the answer is $\boxed{13}$
| 13
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3,363
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_10
| 1
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A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }8\qquad\mathrm{(C)\ }9\qquad\mathrm{(D)\ }10\qquad\mathrm{(E)\ }11$
|
The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\boxed{10}$
| 10
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3,364
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_12
| 1
|
An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ $d=YZ$ , and $e=XY$ . What is the area of the annulus?
[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]
$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$
|
The area of the large circle is $\pi b^2$ , the area of the small one is $\pi c^2$ , hence the shaded area is $\pi(b^2-c^2)$
From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{2}$
| 2
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3,365
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_12
| 2
|
An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ $d=YZ$ , and $e=XY$ . What is the area of the annulus?
[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]
$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$
|
Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\boxed{2}$
| 2
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3,366
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_13
| 1
|
In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11$
|
All numbers in this solution will be in hundredths of a millimeter.
The thinnest coin is the dime, with thickness $135$ . A stack of $n$ dimes has height $135n$
The other three coin types have thicknesses $135+20$ $135+40$ , and $135+60$ . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set $\{135n, 135n+20, 135n+40, \dots, 195n\}$
If we take an odd $n$ , then all the possible heights will be odd, and thus none of them will be $1400$ . Hence $n$ is even.
If $n<8$ the stack will be too low and if $n>10$ it will be too high. Thus we are left with cases $n=8$ and $n=10$
If $n=10$ the possible stack heights are $1350,1370,1390,\dots$ , with the remaining ones exceeding $1400$
Therefore there are $\boxed{8}$ coins in the stack.
| 8
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3,367
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_13
| 2
|
In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11$
|
Let $p,n,d$ , and $q$ be the number of pennies, nickels, dimes, and quarters used in the stack.
From the conditions above, we get the following equation:
\[155p+195n+135d+175q=1400.\]
Then we divide each side by five to get
\[31p+39n+27d+35q=280.\]
Writing both sides in terms of mod 4, we have $-p-n-d-q \equiv 0 \pmod 4$
This means that the sum $p+n+d+q$ is divisible by 4. Therefore, the answer must be $\boxed{8}.$
| 8
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3,368
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_13
| 3
|
In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11$
|
We can easily add up $1.55\text{\ mm}$ and $1.95\text{\ mm}$ to get $3.50\text{\ mm}$ . We multiply that by $4$ to get $14\text{\ mm}$ . Since this works and it requires 8 coins, the answer is clearly $\boxed{8}$
| 8
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3,369
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_17
| 1
|
The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$
|
If Jack's current age is $\overline{ab}=10a+b$ , then Bill's current age is $\overline{ba}=10b+a$
In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$
We are given that $10a+b+5=2(10b+a+5)$
Thus $8a=19b+5 \Rightarrow a=\dfrac{19b+5}{8}$
For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\frac{19b+5}8$ is not an integer, and for $b\geq 4$ $a$ is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=\boxed{18}$
| 18
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3,370
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_17
| 2
|
The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$
|
Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.
The age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$
Thus Bill's age is in the set $\{13,22,31,40,49,58,67,76,85,94\}$
As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options $\{13,49,58,67\}$
Checking each of them, we see that only $13$ works, and gives the solution $31-13=\boxed{18}$
| 18
|
3,371
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_19
| 1
|
In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$
|
We already know that $a_1=2001$ $a_2=2002$ $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ $a_6=2003+2000-2005=1998$ $a_7=2000+2005-1998=2007$ , and so on.
We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$
| 0
|
3,372
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_19
| 2
|
In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$
|
Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$
Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots$
From the values given in the problem statement we see that $a_3=a_1+2$
From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$
From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$
Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}$
| 0
|
3,373
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https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_21
| 1
|
Let $1$ $4$ $\ldots$ and $9$ $16$ $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$
$\mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007$
|
The two sets of terms are $A=\{ 3k+1 : 0\leq k < 2004 \}$ and $B=\{ 7l+9 : 0\leq l<2004\}$
Now $S=A\cup B$ . We can compute $|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|$ . We will now find $|A\cap B|$
Consider the numbers in $B$ . We want to find out how many of them lie in $A$ . In other words, we need to find out the number of valid values of $l$ for which $7l+9\in A$
The fact " $7l+9\in A$ " can be rewritten as " $1\leq 7l+9 \leq 3\cdot 2003 + 1$ , and $7l+9\equiv 1\pmod 3$ ".
The first condition gives $0\leq l\leq 857$ , the second one gives $l\equiv 1\pmod 3$
Thus the good values of $l$ are $\{1,4,7,\dots,856\}$ , and their count is $858/3 = 286$
Therefore $|A\cap B|=286$ , and thus $|S|=4008-|A\cap B|=\boxed{3722}$
| 722
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3,374
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_21
| 2
|
Let $1$ $4$ $\ldots$ and $9$ $16$ $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$
$\mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007$
|
We can start by finding the first non-distinct term from both sequences. We find that that number is $16$ . Now, to find every
other non-distinct terms, we can just keep adding $21$ . We know that the last terms of both sequences are $1+3\cdot 2003$ and
$9+7\cdot 2003$ . Clearly, $1+3\cdot 2003$ is smaller and that is the last possible common term of both sequences. Now, we can
create the inequality $16+21k \leq 1+3\cdot 2003$ . Using the inequality, we find that there are $286$ common terms. There are 4008
terms in total. $4008-286=\boxed{3722}$
| 722
|
3,375
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_1
| 1
|
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$
|
The first $2003$ even counting numbers are $2,4,6,...,4006$
The first $2003$ odd counting numbers are $1,3,5,...,4005$
Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$
$= 1+1+1+...+1 = \boxed{2003}$
| 3
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3,376
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_1
| 2
|
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$
|
Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{2003}$
| 3
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3,377
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_1
| 3
|
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$
|
The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).
Sum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).
Knowing this, plug $2003$ for $n$
$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow$ $\boxed{2003}$
| 3
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3,378
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_1
| 4
|
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$
|
In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$ . And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$ . When compared to the sum of odd numbers, we see that each of the $n$ th term in the series of even numbers differ by $1$ . For example, take series $O$ and $E_1$ . The first terms are $1$ and $0$ . Their difference is $|1-0|=1$ . Similarly, take take series $O$ and $E_2$ . The first terms are $1$ and $2$ . Their difference is $|1-2|=1$ . Since there are $2003$ terms in each set, the answer $\boxed{2003}$
| 3
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3,379
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_1
| 5
|
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$
|
We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \cdot 2003$ , and the answer is $\boxed{2003}$
| 3
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3,380
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_2
| 1
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Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $2366, how many members are in the League?
$\mathrm{(A) \ } 77\qquad \mathrm{(B) \ } 91\qquad \mathrm{(C) \ } 143\qquad \mathrm{(D) \ } 182\qquad \mathrm{(E) \ } 286$
|
Since T-shirts cost $5$ dollars more than a pair of socks, T-shirts cost $5+4=9$ dollars.
Since each member needs $2$ pairs of socks and $2$ T-shirts, the total cost for $1$ member is $2(4+9)=26$ dollars.
Since $2366$ dollars was the cost for the club, and $26$ was the cost per member, the number of members in the League is $2366\div 26=\boxed{91}$
| 91
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3,381
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_3
| 1
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A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?
$\mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\%$
|
The volume of the original box is $15\cdot10\cdot8=1200.$
The volume of each cube that is removed is $3\cdot3\cdot3=27.$
Since there are $8$ corners on the box, $8$ cubes are removed.
So the total volume removed is $8\cdot27=216$
Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{18}.$
| 18
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3,382
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_4
| 1
|
It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$
|
Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.
Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.
Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{3}$
| 3
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3,383
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_4
| 2
|
It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$
|
The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or $\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}$ (for speeds $x$ and $y$ ). Mary's speed going to school is $2\,\text{km/hr}$ , and her speed coming back is $6\,\text{km/hr}$ . Plugging the numbers in, we get that the average speed is $\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{3}$
| 3
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3,384
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_5
| 1
|
Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
|
Using factoring:
$2x^{2}+3x-5=0$
$(2x+5)(x-1)=0$
$x = -\frac{5}{2}$ or $x=1$
So $d$ and $e$ are $-\frac{5}{2}$ and $1$
Therefore the answer is $\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{0}$
| 0
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3,385
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_5
| 2
|
Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
|
We can use the sum and product of a quadratic (a.k.a Vieta):
$(d-1)(e-1)=de-(d+e)+1 \implies\text{product}-\text{sum}+1 \implies \dfrac{c}{a}-\left(-\dfrac{b}{a}\right)+1 \implies \dfrac{b+c}{a}+1= \dfrac{5}{-5}+1=\boxed{0}$
| 0
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3,386
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_5
| 3
|
Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
|
By inspection, we quickly note that $x=1$ is a solution to the equation, therefore the answer is
$(d-1)(e-1)=(1-1)(e-1)=\boxed{0}$
| 0
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3,387
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_5
| 4
|
Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
|
The form $(d-1)(e-1)$ resembles the factored form for the quadratic, namely $(x-d)(x-e)$ based on the information given. Note putting in 1 for $x$ in the that quadratic immediately yields the desired expression. Thus, $2(1)^2+3(1)-5 = \boxed{0}$
| 0
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3,388
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_6
| 1
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Define $x \heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$ . Which of the following statements is not true?
$\mathrm{(A) \ } x \heartsuit y = y \heartsuit x$ for all $x$ and $y$
$\mathrm{(B) \ } 2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$
$\mathrm{(C) \ } x \heartsuit 0 = x$ for all $x$
$\mathrm{(D) \ } x \heartsuit x = 0$ for all $x$
$\mathrm{(E) \ } x \heartsuit y > 0$ if $x \neq y$
|
We start by looking at the answers. Examining statement C, we notice:
$x \heartsuit 0 = |x-0| = |x|$
$|x| \neq x$ when $x<0$ , but statement C says that it does for all $x$
Therefore the statement that is not true is $\boxed{0}$
| 0
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3,389
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_7
| 1
|
How many non- congruent triangles with perimeter $7$ have integer side lengths?
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$
|
By the triangle inequality , no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$
Since all sides must be integers, the largest possible length of a side is $3$ . Therefore, all such triangles must have all sides of length $1$ $2$ , or $3$ . Since $2+2+2=6<7$ , at least one side must have a length of $3$ . Thus, the remaining two sides have a combined length of $7-3=4$ . So, the remaining sides must be either $3$ and $1$ or $2$ and $2$ . Therefore, the number of triangles is $\boxed{2}$
| 2
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3,390
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_10
| 1
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The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
|
2003amc10a10solution.gif
Let the squares be labeled $A$ $B$ $C$ , and $D$
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$ , and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.
Therefore, squares $1$ $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.
Squares $4$ $5$ $6$ $7$ $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.
Thus the answer is $\boxed{6}$
| 6
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3,391
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_10
| 2
|
The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
|
Another way to think of it is that a cube missing one face has $5$ of its $6$ faces. Since the shape has $4$ faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. $1$ $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\boxed{6}$
| 6
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3,392
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_11
| 1
|
The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$
|
$AMC10+AMC12=123422$
$AMC00+AMC00=123400$
$AMC+AMC=1234$
$2\cdot AMC=1234$
$AMC=\frac{1234}{2}=617$
Since $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$
Therefore, $A+M+C = 6+1+7 = \boxed{14}$
| 14
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3,393
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_11
| 2
|
The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$
|
We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422$ . Solving for $x$ , we get $x=61710$ . Therefore, the sum $A+M+C= \boxed{14}$
| 14
|
3,394
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_11
| 3
|
The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$
|
Consider the place values of the digits of $AMC10$ and $AMC12$
When we add $AMC10$ and $AMC12$ $C + C$ must result in a units digit of $4$ , meaning $C$ is either $2$ or $7$ . Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$ , and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$ , so $M$ is either $1$ or $6$ . Again, we must look at the digit before $M$ , or $A$ $A$ is even, so $M$ must be less than $5$ , or else the ten would be carried over. Ergo, $M$ is $1$ . Nothing is carried over, so we have $A + A = 12$ , and $A = 6$ . Therefore, the sum of $A$ $M$ , and $C$ is $6 + 1 + 7 = \boxed{14}$
| 14
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3,395
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_13
| 1
|
The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?
$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$
|
Let the numbers be $x$ $y$ , and $z$ in that order. The given tells us that
\begin{eqnarray*}y&=&7z\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\ x+y+z&=&32z+7z+z=40z=20\\ z&=&\frac{20}{40}=\frac{1}{2}\\ y&=&7z=7\cdot\frac{1}{2}=\frac{7}{2}\\ x&=&32z=32\cdot\frac{1}{2}=16 \end{eqnarray*}
Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \boxed{28}$
| 28
|
3,396
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_13
| 2
|
The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?
$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$
|
Alternatively, we can set up the system in equation form:
\begin{eqnarray*}1x+1y+1z&=&20\\ 1x-4y-4z&=&0\\ 0x+1y-7z&=&0\\ \end{eqnarray*}
Or, in matrix form $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} =\begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$
To solve this matrix equation, we can rearrange it thus:
$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} ^{-1} \begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$
Solving this matrix equation by using inverse matrices and matrix multiplication yields
$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{7}{2} \\ 16 \\ \end{bmatrix}$
Which means that $x = \frac{1}{2}$ $y = \frac{7}{2}$ , and $z = 16$ . Therefore, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28 \Rightarrow \boxed{28}$
| 28
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3,397
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_14
| 2
|
Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$ $e$ , and $10d+e$ , where $d$ and $e$ are single digits. What is the sum of the digits of $n$
$\mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24$
|
Since $d$ is a single digit prime number, the set of possible values of $d$ is $\{2,3,5,7\}$
Since $e$ is a single digit prime number and is the units digit of the prime number $10d+e$ , the set of possible values of $e$ is $\{3,7\}$
Using these values for $d$ and $e$ , the set of possible values of $10d+e$ is $\{23,27,33,37,53,57,73,77\}$
Out of this set, the prime values are $\{23,37,53,73\}$
Therefore the possible values of $n$ are:
$2\cdot3\cdot23=138$
$3\cdot7\cdot37=777$
$5\cdot3\cdot53=795$
$7\cdot3\cdot73=1533$
The largest possible value of $n$ is $1533$
So, the sum of the digits of $n$ is $1+5+3+3=12 \Rightarrow \boxed{12}$
| 12
|
3,398
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_16
| 1
|
What is the units digit of $13^{2003}$
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9$
|
$13^{2003}\equiv 3^{2003}\pmod{10}$
Since $3^4=81\equiv1\pmod{10}$
$3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}$
Therefore, the units digit is $7 \Rightarrow\boxed{7}$
| 7
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3,399
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_16
| 2
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What is the units digit of $13^{2003}$
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9$
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Since we are looking for the units digit of $13^{2003}$ , we only have to focus on the units digit of the base (13) as none of the other digits of the base affect the units digit of the resulting value.
By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern: \[3^1=3\] \[3^2=9\] \[3^3=27\] \[3^4=81,\] giving us the rotation $3-9-7-1.$
As this cycle resets every time the index increases by 4, we know that this cycle ends on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the third term of our pattern: $7$
Therefore, the units digit of our expression is $7 \Rightarrow\boxed{7}$
| 7
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3,400
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_18
| 1
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What is the sum of the reciprocals of the roots of the equation $\frac{2003}{2004}x+1+\frac{1}{x}=0$
$\mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003}$
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Multiplying both sides by $x$
$\frac{2003}{2004}x^{2}+1x+1=0$
Let the roots be $a$ and $b$
The problem is asking for $\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}$
By Vieta's formulas
$a+b=(-1)^{1}\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$
$ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}$
So the answer is $\frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{1}$
| 1
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