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If P is the product of all of the positive multiples of 11
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05 Nov 2017, 01:36
05 Nov 2017, 01:36
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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
Kudos for correct solution.
(A) 22
(B) 28
(C) 45
(D) 49
(E) 89
Kudos for correct solution.
Re: If P is the product of all of the positive multiples of 11
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05 Nov 2017, 06:28
05 Nov 2017, 06:28
The product of all the multiples of 11 below 100 would be 11*22*33*44*55*66*77*88*99 but it is quite long to compute the multiplication using the calculator and then looking for the prime factors.
Let's use a trick: since all the numbers are multiples of 11 we can factorize 11 from each term of the multiplication in order to get 11*(1*2*3*4*5*6*7*8*9).
Now, let's focus on those numbers of the multiplication which are prime numbers because they coincide with the prime factors of the number of interest. Thus, we have 2, 3, 5, 7 and 11.
Summing them we get 28 and answer B!
Let's use a trick: since all the numbers are multiples of 11 we can factorize 11 from each term of the multiplication in order to get 11*(1*2*3*4*5*6*7*8*9).
Now, let's focus on those numbers of the multiplication which are prime numbers because they coincide with the prime factors of the number of interest. Thus, we have 2, 3, 5, 7 and 11.
Summing them we get 28 and answer B!
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Re: If P is the product of all of the positive multiples of 11
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15 Mar 2021, 23:18
15 Mar 2021, 23:18
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
Kudos for correct solution.
(A) 22
(B) 28
(C) 45
(D) 49
(E) 89
Kudos for correct solution.
\(P = (11)(22)(33)(44)(55)(66)(77)(88)(99)\)
\(P = (11)(2)(11)(3)(11)(2^2)(11)(5)(11)(2)(3)(11)(7)(11)(2^3)(11)(3^2)(11)\)
\(P = (2^7)(3^4)(5)(7)(11^9)\)
Sum of Primes in \(P = 2 + 3 + 5 + 7 + 11 = 28\)
Hence, option B
