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If m is the square of integer n and m is divisible by 98, m must also
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11 Oct 2012, 16:51
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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 would like to know why the answer (see spoiler) and not B here we go: 98 has factors : 2 7 and 7 so : \(2^1\) and \(7^2\) But at the same time \(n\) \(=\) \(m^2\) so m is a perfect square so we need \(2^2\) and \(7^2\) at least28 = \(2^2\) and \(7\) 196 =\(2^2\) and \(7^2\) 343 = \(7^3\) Please explain i'm struggling a lot the last days with this concept
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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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11 Oct 2012, 17:03
carcass wrote: 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 would like to know why the answer (see spoiler) and not B here we go: 98 has factors : 2 7 and 7 so : \(2^1\) and \(7^2\) But at the same time \(n\) \(=\) \(m^2\) so m is a perfect square so we need \(2^2\) and \(7^2\) at least28 = \(2^2\) and \(7\) 196 =\(2^2\) and \(7^2\) 343 = \(7^3\) Please explain i'm struggling a lot the last days with this concept Given: \(m=n^2\) (not \(n=m^2\) as you've written) and \(m\) is a multiple of \(98=2*7^2\). Now, since \(m\) is a perfect square than it must have even powers of its primes, thus \(m\) must be divisible by \(2^2*7^2=196\) and naturally by all the factors of 196, which are: 1, 2, 4, 7, 14, 28, 49, 98 and 196. Answer: C. Hope it's clear.
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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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11 Oct 2012, 17:54
Bunuel wrote: carcass wrote: 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 would like to know why the answer (see spoiler) and not B here we go: 98 has factors : 2 7 and 7 so : \(2^1\) and \(7^2\) But at the same time \(n\) \(=\) \(m^2\) so m is a perfect square so we need \(2^2\) and \(7^2\) at least28 = \(2^2\) and \(7\) 196 =\(2^2\) and \(7^2\) 343 = \(7^3\) Please explain i'm struggling a lot the last days with this concept Given: \(m=n^2\) (not \(n=m^2\) as you've written) and \(m\) is a multiple of \(98=2*7^2\). Now, since \(m\) is a perfect square than it must have even powers of its primes, thus \(m\) must be divisible by \(2^2*7^2=196\) and naturally by all the factors of 196, which are: 1, 2, 4, 7, 14, 28, 49, 98 and 196. Answer: C. Hope it's clear. Sorry for typo Of course, factor foundation rule
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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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30 Nov 2016, 08:55
prime factorization of 98: (2)(7^2) > to make this into a perfect square we need to multiply by another 2, which will give us the overall number of 196.
m= 196 n=14 (i.e. 14^2 = 196)
so, m must be divisible by 196 and 28 because it has prime factors that can be combined to create those values.
Thus, C is the answer.



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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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30 Nov 2016, 10:11
carcass wrote: 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 \(m = n^2\) \(\frac{m}{98} = \frac{n^2}{98}\) Remainder is 0 Least possible values of \(n^2 = 196\) , so \(m = + 7\) Further \(196 = 7^2*2^2\) We find 196 is divisible by both 28 & 196 , but not divisible by 343 ( Coz 343 = 7^3 ) Hence our answer will be (C) I & II only
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If m is the square of integer n and m is divisible by 98, m must also
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06 Oct 2018, 06:42
carcass wrote: 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
if m is dvisibe by 98, it means (in my oinion ) that m must be at least 98*2 = 196 and square root of which, is 14 196 is divisible by 28 and 196 i wonder if my reasoning is correct




If m is the square of integer n and m is divisible by 98, m must also
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06 Oct 2018, 06:42






