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# If z is a multiple of 24, what is the remainder when z^2 is

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Manager
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If z is a multiple of 24, what is the remainder when z^2 is [#permalink]

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04 Sep 2009, 07:24
1
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00:00

Difficulty:

15% (low)

Question Stats:

76% (00:51) correct 24% (00:48) wrong based on 198 sessions

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If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6
[Reveal] Spoiler: OA

Last edited by mendelay on 17 Sep 2009, 08:07, edited 2 times in total.

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Director
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04 Sep 2009, 07:44
is it
[Reveal] Spoiler:
A--0?

z=24(K)=2^3(3)(k)

z^2=2^6(9)(k^2)

now z^2/9= 2^6(k^2)==multiple of 2^6

mendelay wrote:
If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

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Intern
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04 Sep 2009, 10:52
24= (8*3)n
the resultant number will be multiplied by 3^2 for sure... hence z^2/9 will have its remiander as 0...

wats the OA?

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Manager
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04 Sep 2009, 13:44
1
KUDOS
this question will hardly take more than 30 seconds..

this is not data sufficiency problem, in which you have to test the answer by plugging different values..

this is problem solving question asking for a definite answer..So, the answer would be same whether z is 24 or any other multiple of 24.

Let z = 24 ; =>$$z^2 = 24*24$$

=>$$z^2 = 9*64$$ ( divisible by 9)

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Senior Manager
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31 Jan 2010, 06:21
2
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1
This post was
BOOKMARKED
mendelay wrote:
If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

[Reveal] Spoiler:
A

z = 24 * K = 8 * 3 * k
z^2 = 8^2 * 9 * k^2

Therefore z^2 Mod 9 = 0 ( as 9 is a factor in the same)

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Intern
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12 Feb 2010, 08:27
jeeteshsingh wrote:
mendelay wrote:
If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

[Reveal] Spoiler:
A

z = 24 * K = 8 * 3 * k
z^2 = 8^2 * 9 * k^2

Therefore z^2 Mod 9 = 0 ( as 9 is a factor in the same)

nice explanation but also you can do the other way by calculating square of 24 and then divide by 9.
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Kudos [?]: 118 [0], given: 9

Director
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09 Aug 2013, 05:51
Easy one:

Z=24A

Z^2 = 24*24* A^2

REM(24/9) = 6

Rem (Z^2/9) = REM (6*6* A^2)/9
= REM(36*A^2)/9
= REM(0*A^2)/9

=0

Rgds,
TGC!
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Senior Manager
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09 Aug 2013, 10:12
mendelay wrote:
If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

[Reveal] Spoiler:
A

........
z=24k
or, z = 8×3×k
or, z^2 = 64 × 9 × k^2

So remainder is 0 (A)
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Re: If z is a multiple of 24, what is the remainder when z^2 is [#permalink]

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17 Nov 2013, 12:15
If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

z =24k

z^2= 24*24
ie 24*24/9 = 64

So,the remainder is 0.

Kudos [?]: 2 [0], given: 1

Current Student
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Re: If z is a multiple of 24, what is the remainder when z^2 is [#permalink]

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27 Jan 2014, 04:31
For this problem I just did the following

Took z = 24

Then 24/9 remainder is 6
Do this twice and you will get 6*6 / 9

Remainder of 36/9 is zero

Cheers!
J

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Re: If z is a multiple of 24, what is the remainder when z^2 is [#permalink]

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27 Jan 2014, 05:11
1
KUDOS
Expert's post
mendelay wrote:
If z is a multiple of 24, what is the remainder when z^2 is divided by 9?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

z is a multiple of 24, means that z is a multiple of 3, therefore z^2 must be a multiple of 9.

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Current Student
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Re: If z is a multiple of 24, what is the remainder when z^2 is [#permalink]

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27 Jan 2014, 20:22
24 = 2^3 * 3. Any multiple of this could be 2^4 * 3, 2^3 *3^2, etc.

Let's assume z = 24 for now. So z = 2^3 * 3. Then z^2 = 2^6 * 3^2. If we divide this by 9 or 3^2 we will end up with no denominator, meaning there is no remainder.

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Re: If z is a multiple of 24, what is the remainder when z^2 is [#permalink]

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05 Dec 2017, 06:29
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Re: If z is a multiple of 24, what is the remainder when z^2 is   [#permalink] 05 Dec 2017, 06:29
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