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 $\boxed{C}$

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1998 AHSME Problems/Problem 6

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