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 $11?$

$\frac{2}{5}$

$\frac{1}{5}$

$\frac{1}{6}$

$\frac{1}{11}$

$\frac{1}{15}$

Answer: B

Concepts:digits divisibility basic counting basic probability

Difficulty rating: 2300

Small Hint:

Start from $99999$ ; a digit sum of $43$ means a total deficit of $2$

Big Hint:

The possibilities are one $7$ or two $8$ s; apply the divisibility-by- $11$ alternating-sum test

Solution:

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.

1970 AMC 12 Problem 31

Answer: B

Difficulty rating: 2300

Small Hint:

Big Hint:

Solution:

Problem 31 in Other Years