1998 CEMC Gauss (Grade 7) Problems/Problem 25

Problem

Two natural numbers, $p$ and $q,$ do not end in zero. The product of any pair, $p$ and $q,$ is a power of 10 (that is, 10, 100, 1000, 10 000 , ...). If $p > q$ , the last digit of $p-q$ cannot be

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$

Solution

If the product $pq$ is a power of $10,$ then $p$ must be in the form $5^n$ and $q$ must be in the form $2^n.$

Start looking at small values of $n$ and subtract:

$p=5, \quad q=2, \quad \rightarrow \quad p-q \equiv 3$ $p=25, \quad q=4, \quad \rightarrow \quad p-q \equiv 1$ $p=125, \quad q=8, \quad \rightarrow \quad p-q \equiv 7$ $p=625, \quad q=16, \quad \rightarrow \quad p-q \equiv 9$

This pattern continues in groups of $4$ , and the only number not included is $\boxed{\textbf{(C)}\ 5}.$

-edited by coolmath34

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1998 CEMC Gauss (Grade 7) Problems/Problem 25

Problem

Solution