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If P is the product of all of the positive multiples of 11 less than 1

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If P is the product of all of the positive multiples of 11 less than 1 [#permalink]

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New post 05 Nov 2017, 02:36
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If P is the product of all of 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
[Reveal] Spoiler: OA

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Re: If P is the product of all of the positive multiples of 11 less than 1 [#permalink]

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New post 05 Nov 2017, 03:03
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P= 11^9 * (1*2*3*4*5*6*7*8*9)

Sum of distinct primes= 2+3+5+7+11 = 28

Option B

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If P is the product of all of the positive multiples of 11 less than 1 [#permalink]

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New post 05 Nov 2017, 11:10
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Bunuel wrote:
If P is the product of all of 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

Positive multiples of 11 less than 100:
11, 22, 33, 44, 55, 66, 77, 88, 99

P = the product of all these numbers. So each factor in each multiple will be a factor of P. No need to worry about the actual product; factors are the key.

Multiples of 11 whose other factor is a prime number:

22 = 2 * 11
33 = 3 * 11
55 = 5 * 11
77 = 7 * 11

Other multiples of 11? Except for 11 (which = 11 * 1, where 1 is not prime), their other factors are already-used single-digit primes:

11 = 11 * 1, 1 is not prime
44 = 11 * 2\(^2\)
66 = 11 * 2 * 3
88 = 11 * 2\(^3\)
99 = 11 * 3\(^2\)
Prime factors 2 and 3 have already been "used" in 22 and 33. Not distinct.

Sum of P's distinct prime factors?
11 + 2 + 3 + 5 + 7 = 28

Answer B

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Re: If P is the product of all of the positive multiples of 11 less than 1 [#permalink]

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New post 05 Nov 2017, 12:05
Bunuel wrote:
If P is the product of all of 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


Answer is B as follows
Positive multiple of 11 less than 100 are 11,22,33,44,55,66,77,88,99
These numbers can be written as
11
22=2*11
33=3*11
44=4*11 = \(2^2\)*11
55=5*11
66=6*11 = 2*3*11
77=7*11
88=8*11 = \(2^3\)*11
99=9*11 = \(3^2\)*11

P = 11*22*33*44*55*66*77*88*99 = 11*(2*11)*(3*11)*(\(2^2\)*11)*(5*11)*(2*3*11)*(7*11)*(\(2^3\)*11)*(\(3^2\)*11) = \(2^7*3^4*5*7*11^9\)
Prime factors are 2,3,5,7,11
Sum of these prime factors are 2+3+5+7+11 = 28

B is the answer.
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Re: If P is the product of all of the positive multiples of 11 less than 1 [#permalink]

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New post 08 Nov 2017, 17:36
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Expert's post
Bunuel wrote:
If P is the product of all of 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


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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Kudos [?]: 978 [1], given: 5

Re: If P is the product of all of the positive multiples of 11 less than 1   [#permalink] 08 Nov 2017, 17:36
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