1989 AIME Problem 3

Attempt Problem 3 of the 1989 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AIME solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Suppose

n

is a positive integer and

d

is a single digit in base

10.

Find

n

\frac{n}{810}=0.d25d25d25\ldots.

Answer

Convert the repeating three-digit block into a fraction with denominator

999

Use the integrality of

n

to determine which digit makes

4d+1

divisible by

37

The repeating decimal is

\frac{100d+25}{999}.

Thus

\begin{aligned}n&=\frac{810(100d+25)}{999}\\&=\frac{750(4d+1)}{37}.\end{aligned}

Since

0\leq d\leq9,

the only way

4d+1

can be divisible by

37

4d+1=37,

d=9

and

n=750.

Problem 3 in other years