If \(N\) is the product of all integers from \(1\) to \(100,\) both inclusive, then what is the remainder when \(N + 100\) is divided by \(7^{16}?\)
A. 0
B. 7
C. 49
D. 100
E. 1343
Answer: D
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If \(N\) is the product of all integers from \(1\) to \(100,\) both inclusive, then what is the remainder when
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Between 1 and 100, there are 14 multiples of 7 (since 7*14 = 98), so 100! is divisible by at least 7^14. But two of those multiples of 7, namely 49 and 98, are multiples of 7^2, so they give us one extra 7, and 100! is actually divisible by 7^16.
So 100! is a multiple of 7^16, and 100! + 100 is thus exactly 100 larger than a multiple of 7^16, which is another way of saying that the remainder is 100 when we divide 100! + 100 by 7^16 (here using the fact that 100 < 7^16, so 100 is a legitimate remainder when dividing by 7^16).
This is not a realistic GMAT question though.
So 100! is a multiple of 7^16, and 100! + 100 is thus exactly 100 larger than a multiple of 7^16, which is another way of saying that the remainder is 100 when we divide 100! + 100 by 7^16 (here using the fact that 100 < 7^16, so 100 is a legitimate remainder when dividing by 7^16).
This is not a realistic GMAT question though.
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