1968 AHSME Problems/Problem 19

Problem

Let $n$ be the number of ways $10$ dollars can be changed into dimes and quarters, with at least one of each coin being used. Then $n$ equals:

$\text{(A) } 40\quad \text{(B) } 38\quad \text{(C) } 21\quad \text{(D) } 20\quad \text{(E) } 19$

Solution

Because we have to use at least one dime and one quarter, this part of the selection is forced. Thus, the number of ways to partition $$10$ in the desired manner is equal to the number of ways to partition $$10-10\cent-25\cent=$9.65$ into quarters and dimes without restriction. However, $$9.65$ is not a multiple of $25\cent$ , so we must use dimes until we reach a multiple of $25\cent$ . We first reach such a multiple at $$9.25$ , and, because we have had no freedom up to this point, the number of ways to partition $$9.25$ into dimes and quarters will yield our desired answer. $$9.25$ can be partitioned into 37 quarters, and we can replace quarters 2 at a time with 5 dimes to partition it differently. We can replace anywhere from 0 to 18 pairs of quarters with dimes in this manner, giving us $\boxed{19}$ different partitions, which is answer choice $\fbox{E}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1968 AHSME Problems/Problem 19

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