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If \(P\) is the product of all the positive multiples of 11

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by Ian Stewart » Sun Aug 04, 2019 5:07 am
"the distinct primes of P" doesn't mean anything in math, if P is a number. I'm not sure if the question has been correctly transcribed, but they surely mean to say "the distinct prime divisors of P".

The product of the positive multiples of 11 less than 100 is:

(11)(22)(33) ... (99) = (1)(11)(2)(11)(3)(11) ... (9)(11) = 11^9 * 9!

and since 9! is divisible by the primes less than or equal to 9, so by 2, 3, 5, and 7, this product is divisible by the primes 2, 3, 5, 7 and 11, which sum to 28.
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by Scott@TargetTestPrep » Sun Aug 11, 2019 6:15 pm
swerve wrote:If \(P\) is the product of all the positive multiples of 11 less than 100, then what is the sum of the distinct primes of \(P\)?

A. 22
B. 28
C. 45
D. 49
E. 89

The OA is B

Source: Magoosh
We can create the following expression:

11 x 2(11) x 3(11) x 4(11) x 5(11) x 6(11) x 7(11) x 8(11) x 9(11)

We see that the prime factors of P are 2, 3, 5, 7, and 11. So, the sum of those distinct primes is 2 + 3 + 5 + 7 + 11 = 28.

Answer: B

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