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Show that no set of nine consecutive integers can be partitioned into two sets with the product of the elements of the first set equal to the product of the elements of the second set.

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platinum5 ·

Assume that such numbers do exist, and let us look at their prime factorizations.

For primes p greater than 7, at most one of the numbers can be divisible by p, and the partition cannot exist.

Thus the prime factors of the given numbers can be only 2, 3, 5, and 7.

repeated prime factors : Because the difference between two numbers divisible by 4 is at least 4, at most three of the nine numbers are divisible by 4.

Also, at
most one is divisible by 9,

at most one by 25,

and at most one by 49.

Eliminating these
at most 3+1+1+1 = 6 numbers, we are left with at least three numbers among the nine
that do not contain repeated prime factors.

They are among the divisors of 2 · 3 · 5 · 7 ,and so among the numbers 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210.

Because the difference between the largest and the smallest of these three numbers is at most 9, none of them can be greater than 21.

We have to look at the sequence
1, 2, 3, . . . , 29.

Any subsequence of consecutive integers of length 9 that has a term greater than 10 contains a prime number greater than or equal to 11, which is impossible.

And from 1, 2, . . . , 10 we cannot select nine consecutive numbers with the required
property. This contradicts our assumption.

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