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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
(A) 22
(B) 28
(C) 45
(D) 49
(E) 89
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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
Sent from my A0001 using GMAT Club Forum mobile app
Sum of distinct primes= 2+3+5+7+11 = 28
Option B
Sent from my A0001 using GMAT Club Forum mobile app
05 Nov 2017, 11:10
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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
Positive multiples of 11 less than 100:(A) 22
(B) 28
(C) 45
(D) 49
(E) 89
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
