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Problem 723

AMC 10/12, early questions

Combinatorics Difficulty 3.4 Find the answer Japan Junior Mathematical Olympiad · Japan

There are four spade cards with numbers $1$ , $2$ , $3$ , $4$ , six heart cards with numbers $1$ , $2$ , $3$ , $\ldots$ , $6$ and eight diamond cards with numbers $1$ , $2$ , $3$ , $\ldots$ , $8$ . Suppose you choose three cards, one from each group. How many possible choices are there if the total of the numbers on the chosen cards must be a multiple of $7$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

28 ways

Let $a$ , $b$ , $c$ be the number on the spade, heart and diamond card chosen, respectively. If $a + c$ is not a multiple of $7$ , let $k$ be the remainder obtained when $a + c$ is divided by $7$ . Then $k$ satisfies $1 \le k \le 6$ , and $a + b + c$ becomes a multiple of $7$ when and only when $b = 7 - k$ . On the other hand if $a + c$ is a multiple of $7$ , then $a + b + c$ cannot be a multiple of $7$ , since $1 \le b \le 6$ . Thus, the number of possible choices satisfying the requirement of the problem is the total number of choices for $(a, c)$ , which equals $4 \times 8 = 32$ , minus the number of cases for which $a + c$ is a multiple of $7$ . The latter possibility occurs in 4 ways, namely, $(a, c) = (1, 6), (2, 5), (3, 4), (4, 3)$ . Therefore, the answer we seek is $32 - 4 = 28$ ways.

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