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Harvard-MIT Math Tournament

Number theory Difficulty 5.1 AIME, harder Prove it United States

Problem:

Matt has somewhere between 1000 and 2000 pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2,3,4,5,6,7$ , and $8$ piles but ends up with one sheet left over each time. How many piles does he need?

Solution

Solution:

The number of sheets will leave a remainder of $1$ when divided by the least common multiple of $2,3,4,5,6,7$ , and $8$ , which is $8 \cdot 3 \cdot 5 \cdot 7 = 840$ . Since the number of sheets is between $1000$ and $2000$ , the only possibility is $1681$ . The number of piles must be a divisor of $1681 = 41^{2}$ , hence it must be $41$ .

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