Competition problem — Combinatorics

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Combinatorics Difficulty 5.1 AIME, harder Find the answer

8. Given $1 \leqslant x, y, z \leqslant 6$ .

The number of cases where the product of the positive integers $x, y, z$ is divisible by 10 is
$\qquad$ kinds.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

8. 72 .
(1) The number of ways to choose $x, y, z$ is $6^{3}$ ;
(2) The number of ways to choose $x, y, z$ without taking $2, 4, 6$ is $3^{3}$ ; (3) The number of ways to choose $x, y, z$ without taking 5 is $5^{3}$ ;
(4) The number of ways to choose $x, y, z$ without taking $2, 4, 5, 6$ is $2^{3}$ . Therefore, the number of ways for the product of $x, y, z$ to be divisible by 10 is $6^{3}-3^{3}-5^{3}+2^{3}=72$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.