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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If m is the square of integer n and m is divisible by 98, m must also
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10 Jun 2021, 22:07
10 Jun 2021, 22:07
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Question Stats:
61% (01:46) correct
38% (02:03) wrong based on 113 sessions
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If m is the square of integer n and m is divisible by 98, m must also be divisible by:
I. 28
II. 196
III. 343
(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III
I. 28
II. 196
III. 343
(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III
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Re: If m is the square of integer n and m is divisible by 98, m must also
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13 Jun 2021, 02:12
13 Jun 2021, 02:12
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m is the square of integer n and m is divisible by 98
Let's do prime factorization of 98 we get, 98 = 2*\(7^2\)
Since, m is a square of an integer and m is divisible by 98
That means that m must be a multiple of 98
=> m = 98*k (where k is an integer)
=> m = 2*\(7^2\) * k
Now, for m to be square of a number minimum value of k should be 2. (As this will make m a perfect square)
=> m = \(2^2\) *\(7^2\) = \(14^2\) = 196
So, m will be divisible by 28 and 196 and not by 343
So, Answer will be C
Hope it helps!
Let's do prime factorization of 98 we get, 98 = 2*\(7^2\)
Since, m is a square of an integer and m is divisible by 98
That means that m must be a multiple of 98
=> m = 98*k (where k is an integer)
=> m = 2*\(7^2\) * k
Now, for m to be square of a number minimum value of k should be 2. (As this will make m a perfect square)
=> m = \(2^2\) *\(7^2\) = \(14^2\) = 196
So, m will be divisible by 28 and 196 and not by 343
So, Answer will be C
Hope it helps!
General Discussion
Re: If m is the square of integer n and m is divisible by 98, m must also
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24 Aug 2021, 19:26
24 Aug 2021, 19:26
1 is an integer too, so if n=1 then m=1 so logically m should be divisible by all right?
