1970 AMC 12 Problem 31

Question

If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to

43,

what is the probability that this number will be divisible by

11?

\frac{2}{5}

\frac{1}{5}

\frac{1}{6}

\frac{1}{11}

\frac{1}{15}

Start from

99999

; a digit sum of

43

means a total deficit of

2

The possibilities are one

7

or two

8

s; apply the divisibility-by-

11

alternating-sum test

The maximum five-digit digit sum is

45,

so a sum of

43

has total deficit

2

from

99999.

Either one digit is

7

and the others are

9

, giving

5

numbers, or two digits are

8

and the others are

9

, giving

\binom52=10

numbers. Thus there are

15

numbers total.

Using the divisibility-by-

11

test, exactly

97999,\qquad 99979,\qquad 98989

are divisible by

11.

The probability is therefore

\frac{3}{15}=\frac{1}{5}.

Therefore, the correct answer is B.

Problem 31 in other years