Problem 21 · 2018 Math Kangaroo
Stretch
Number Theory
factorizationcasework
Some whole numbers are written on a board, among them the number 2018. The sum of all of them is 2018, and the product of all of them is also 2018. Which of the following could be how many numbers are on the board?
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Answer: B — 2017
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Hint 1 of 2
The product is already 2018 from that one number, so every other number must be \(\pm 1\).
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Hint 2 of 2
The sum is already 2018, so the extra \(+1\)'s and \(-1\)'s must cancel; that forces an exact count.
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Approach: every other number is +1 or -1, then match the sum and product constraints
- The list already contains 2018, whose value equals both the required sum and the required product, so every other number must be a \(+1\) or a \(-1\) (these don't change a product and are the only integers that can).
- For the product to stay \(+2018\), the number of \(-1\)'s must be even; for the sum to stay 2018, the \(+1\)'s and \(-1\)'s must cancel, so there are equally many of each, say \(k\) of each.
- Then the board holds \(1 + k + k = 2k+1\) numbers, which is always odd, so among the options only 2017 is possible (take \(k = 1008\)).
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