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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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If m is the square of integer n and m is divisible by 98, m must also  [#permalink]

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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

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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Re: If m is the square of integer n and m is divisible by 98, m must also  [#permalink]

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11 Oct 2012, 18:03
1
1
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.

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]

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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.

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  [#permalink]

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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.

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Re: If m is the square of integer n and m is divisible by 98, m must also  [#permalink]

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30 Nov 2016, 11:11
1
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]

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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
If m is the square of integer n and m is divisible by 98, m must also   [#permalink] 06 Oct 2018, 07:42
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