Difference between revisions of "1984 AIME Problems/Problem 2"

Revision as of 21:31, 10 February 2009

Problem

The integer $\displaystyle n$ is the smallest positive multiple of $\displaystyle 15$ such that every digit of $\displaystyle n$ is either $\displaystyle 8$ or $\displaystyle 0$ . Compute $\frac{n}{15}$ .

Solution

Any multiple of 15 is a multiple of 5 and a multiple of 3.

Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.

The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$ . For this number to be divisible by 3, $a$ must be divisible by 3. Thus $n$ must have at least three copies of the digit 8.

The smallest number which meets these two requirements is 8880. Thus $\frac{8880}{15} = 592$ is our answer.

@@ Line 5: / Line 5: @@
 Any multiple of 15 is a multiple of 5 and a multiple of 3.
-Any multiple of 5 ends in 0 or 5; since <math>n</math> only contains the digits 0 and 8, the [[units digit]] of <math>n</math> must be 0.
+Any multiple of 5 ends in 0 or 5; since <math>n</math> only contains the digits 0 and 8, the units [[digit]] of <math>n</math> must be 0.
 The sum of the digits of any multiple of 3 must be [[divisible]] by 3.  If <math>n</math> has <math>a</math> digits equal to 8, the sum of the digits of <math>n</math> is <math>8a</math>.  For this number to be divisible by 3, <math>a</math> must be divisible by 3.  Thus <math>n</math> must have at least three copies of the digit 8.

1984 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1984 AIME Problems/Problem 2"

Revision as of 21:31, 10 February 2009

Problem

Solution

See also