# Difference between revisions of "1998 AHSME Problems/Problem 6"

## Problem

If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is

$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$

## Solution

If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.

Since $45^2 = 2025$, the factors should be as close to $44$ or $45$ as possible.

Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$.

$37$ is relatively close to $44$, and no numbers between $38$ and $44$ are factors of $1998$. Thus, the two factors are $37$ and $2\cdot 3^3 = 54$, and the difference is $54 - 37 = 17$, and the answer is $\boxed{C}$