Question

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by $11,$ given that the sum of its digits is $61?$

$\dfrac{3}{14}$

$\dfrac{3}{11}$

$\dfrac{2}{7}$

$\dfrac{4}{11}$

$\dfrac{3}{7}$

Solution:

A digit sum of $61 = 63 - 2$ means all seven digits are $9$ except for a total deficit of $2,$ which gives $\binom{2 + 6}{6} = 28$ numbers. For divisibility by $11$ we need $O - E \equiv 0 \pmod{11},$ where $O$ sums the $4$ odd-position digits and $E$ the $3$ even ones. Write the deficits as $d_O + d_E = 2.$ Then $O - E = 9 - d_O + d_E = 11 - 2 d_O,$ a multiple of $11$ only when $d_O = 0.$ So all of the deficit $2$ falls on the $3$ even positions, giving $\binom{2 + 2}{2} = 6$ ways. The probability is $\tfrac{6}{28} = \tfrac{3}{14}.$ Therefore, the answer is A.

2025 AMC 10B Problem 22

Solution:

Problem 22 in Other Years