1984 AIME Problem 2

Attempt Problem 2 of the 1984 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 1984 AIME solutions, or check the answer key.

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

The integer

n

is the smallest positive multiple of

15

such that every digit of

n

is either

8

0.

Compute

\frac{n}{15}.

Answer

Divisibility by

5

determines the last digit

Divisibility by

3

restricts the number of digits equal to

8

A multiple of

15

must end in

0

and have digit sum divisible by

3.

Because

8\equiv2\pmod3,

the number of

8

’s must be a positive multiple of

3.

The smallest possible number therefore has three

8

’s followed by

0,

namely

n=8880.

Thus

\frac{n}{15}=592.

Problem 2 in other years