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2004 AMC 10B Problems/Problem 13

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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

All numbers in this solution will be in hundreds 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.

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).

See also

2004 AMC 10B (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
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