Difference between revisions of "2004 AMC 10B Problems/Problem 13"

Latest revision as of 21:35, 30 September 2024

Problem

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?

$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11$

Solution 1

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 $\boxed{\mathrm{(B)\ }8}$ coins in the stack.

Using the above observation we can easily construct such a stack. A stack of $8$ dimes would have height $8\cdot 135=1080$ , thus we need to add $320$ . This can be done for example by replacing five dimes by nickels (for $+60\cdot 5 = +300$ ), and one dime by a penny (for $+20$ ).

Solution 2

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 $\boxed{(B)\,\, 8}.$

Solution 3

We notice that the thickness of 4 quarters is 7 mm. 7 is half of 14, so we multiply the 4 quarters by two and get $\boxed{(B)\,\, 8}$ .

Note

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 $\boxed{\mathrm{(B)\ }8}$ .

Similarly, we can simply take $8$ quarters to get $8\cdot 1.75=14$ .

@@ Line 37: / Line 37: @@
 Writing both sides in terms of mod 4, we have <math>-p-n-d-q \equiv 0  \pmod 4</math>.
-This means that the sum <math>p+n+d+q</math> is divisible by 4. Therefore, the answer must be <math>\boxed{(B)\,\, 8}.</math>\
+This means that the sum <math>p+n+d+q</math> is divisible by 4. Therefore, the answer must be <math>\boxed{(B)\,\, 8}.</math>
 ==Solution 3==
-We notice that the thickness of 4 quarters is 7 mm. 7 is half of 14, so we multiply the 4 quarters by two and get <math>\boxed{(B)\,\, 8}</math>\.
+We notice that the thickness of 4 quarters is 7 mm. 7 is half of 14, so we multiply the 4 quarters by two and get <math>\boxed{(B)\,\, 8}</math>.
 ==Note==

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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