Problem of the Day

Solution

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If 3 distinct integers in the set are divisible by 3, this means the set contains 3 multiples of 3. Every third positive integer is a multiple of 3. This is possible, but it will depend on what the first integer in the set is.  Let’s explore this further by looking at some examples (multiples of 3 marked in red):

1, 2, 3, 4, 5, 6, 7 contains exactly 2 multiples of 3.

2, 3, 4, 5, 6, 7, 8 contains exactly 2 multiples of 3.

3, 4, 5, 6, 7, 8, 9 contains exactly 3 multiples of 3.

4, 5, 6, 7, 8, 9, 10 contains exactly 2 multiples of 3.

5, 6, 7, 8, 9, 10, 11 contains exactly 2 multiples of 3.

6, 7, 8, 9, 10, 11, 12 contains exactly 3 multiples of 3.

As we keep increasing the starting value by 1, the number of multiples of 3 will continue to repeat in this pattern: 2, 2, 3, 2, 2, 3, 2, 2, 3….  Write some more out if you don’t see this clearly. You may also notice that all of the sets of numbers that have exactly 3 multiples of 3 start with a multiple of 3.  This should make sense.  If there are only 7 numbers in the sequence, and multiples of 3 occur every third number, we will run out of numbers before we make it to the third multiple of 3 unless we start with one.

Statement (1): INSUFFICIENT. This statement is necessary but not sufficient.  If we have 3 multiples of 3 in this set, then each multiple of 3 will contain at least one factor of 3. The product of the 7 integers would then have at least 3 factors of 3, and this would make it divisible by 27. However, we don’t know for sure that the factors of 3 are distributed among 3 different numbers. They could all be concentrated in 2 numbers. For example, if the set is 25, 26, 27, 28, 29, 30, 31, the product of these numbers is certainly divisible by 27, but the list only contains 2 multiples of 3.

Statement (2): SUFFICIENT. From the analysis above we know that we cannot have 3 distinct multiples of 3 in the set unless the first number is a multiple of 3.  This allows us to say the answer is definitely no.