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If the product of the integers from 1 to n is divisible by 490, what i
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02 Mar 2022, 04:00
02 Mar 2022, 04:00
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If the product of the integers from 1 to n is divisible by 490, what is the least possible value of n?
(A) 7
(B) 14
(C) 21
(D) 28
(E) 35
(A) 7
(B) 14
(C) 21
(D) 28
(E) 35
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
Re: If the product of the integers from 1 to n is divisible by 490, what i
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05 Mar 2022, 02:26
05 Mar 2022, 02:26
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Expert Reply
Breaking Down the Info:
We should break down 490 first to get \(49*10 = 7^2*10\). Then the key factor here is the 2 multiples of 7.
We will need two multiples of 7 so we need n to be at least 14.
Answer: B
We should break down 490 first to get \(49*10 = 7^2*10\). Then the key factor here is the 2 multiples of 7.
We will need two multiples of 7 so we need n to be at least 14.
Answer: B
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Re: If the product of the integers from 1 to n is divisible by 490, what i
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07 Mar 2022, 11:15
07 Mar 2022, 11:15
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GreenlightTestPrep wrote:
If the product of the integers from 1 to n is divisible by 490, what is the least possible value of n?
(A) 7
(B) 14
(C) 21
(D) 28
(E) 35
(A) 7
(B) 14
(C) 21
(D) 28
(E) 35
Take a moment to learn the following property; it will provide a useful way to think of divisibility questions.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-------------------------
Okay, now let’s use this to answer this question.
490 = (2)(5)(7)(7)
So, in order for n! to be divisible by 490, n! must have at least one 2, one 5, and two 7's in its prime factorization.
I'll just check the answer choices....
(A) 7
7! = (7)(6)(5)(4)(3)(2)(1)
Since we need two 7's, we can eliminate choice A
(B) 14
14! = (14)(13)(12)(11).....(7)(6)(5)(4)(3)(2)(1)
When we rewrite 14 as the product of 2 and 7, we get: 14! = (7)(2)(13)(12)(11).....(7)(6)(5)(4)(3)(2)(1)
Since 14! has at least one 2, one 5, and two 7's in its prime factorization, we know that 14! Is divisible by 490
Answer: B
