Maths Olympiad Prep

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Combinatorics Difficulty 3.4 AMC 10/12 Find the answer Japan

There are four spade cards with numbers 11, 22, 33, 44, six heart cards with numbers 11, 22, 33, \ldots, 66 and eight diamond cards with numbers 11, 22, 33, \ldots, 88. 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 77?

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

Solution

28 ways

Let aa, bb, cc be the number on the spade, heart and diamond card chosen, respectively. If a+ca + c is not a multiple of 77, let kk be the remainder obtained when a+ca + c is divided by 77. Then kk satisfies 1k61 \le k \le 6, and a+b+ca + b + c becomes a multiple of 77 when and only when b=7kb = 7 - k. On the other hand if a+ca + c is a multiple of 77, then a+b+ca + b + c cannot be a multiple of 77, since 1b61 \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)(a, c), which equals 4×8=324 \times 8 = 32, minus the number of cases for which a+ca + c is a multiple of 77. The latter possibility occurs in 4 ways, namely, (a,c)=(1,6),(2,5),(3,4),(4,3)(a, c) = (1, 6), (2, 5), (3, 4), (4, 3). Therefore, the answer we seek is 324=2832 - 4 = 28 ways.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.