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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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carcass
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If m is the square of integer n and m is divisible by 98, m must also [#permalink] New post   11 Oct 2012, 17:51
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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

Show Spoiler
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 least

28 = \(2^2\) 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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C
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Bunuel
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Re: If m is the square of integer n and m is divisible by 98, m must also [#permalink] New post   11 Oct 2012, 18:03
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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 least

28 = \(2^2\) 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 [#permalink] New post   11 Oct 2012, 18:54
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 least

28 = \(2^2\) 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 :pray :roll:
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Re: If m is the square of integer n and m is divisible by 98, m must also [#permalink] New post   30 Nov 2016, 09:55
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 [#permalink] New post   30 Nov 2016, 11:11
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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 [#permalink] New post   06 Oct 2018, 07:42
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
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If m is the square of integer n and m is divisible by 98, m must also [#permalink] New post   13 Jun 2021, 03:13
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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!

To learn more about Divisibility Rules, watch the following video

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Re: If m is the square of integer n and m is divisible by 98, m must also [#permalink] New post   15 Sep 2022, 05:18
98 = 2*7*7

Now, since m is a square of a number and is divisible by 98, m will be in the form (2^2n)*(7^2m), where m and n are natural numbers.

Thus, m will always be divisible by 2, 4, 7, 14, 28, 49, 98, and 196.

Thus, the correct option is C.
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Re: If m is the square of integer n and m is divisible by 98, m must also [#permalink]
15 Sep 2022, 05:18
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