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

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Manager
Joined: 21 Jul 2009
Posts: 208
Location: New York, NY
If z is a multiple of 24, what is the remainder when z^2 is  [#permalink]

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Updated on: 17 Sep 2009, 08:07
2
00:00

Difficulty:

15% (low)

Question Stats:

79% (01:10) correct 21% (01:17) wrong based on 215 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

Originally posted by mendelay on 04 Sep 2009, 07:24.
Last edited by mendelay on 17 Sep 2009, 08:07, edited 2 times in total.
Director
Joined: 04 Jan 2008
Posts: 775

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04 Sep 2009, 07:44
is it
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
Joined: 27 Aug 2009
Posts: 26

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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?
Manager
Joined: 25 Aug 2009
Posts: 162

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04 Sep 2009, 13:44
1
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)
Senior Manager
Joined: 22 Dec 2009
Posts: 309

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31 Jan 2010, 06:21
1
1
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 = 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
Joined: 17 Nov 2009
Posts: 33
Schools: University of Toronto, Mcgill, Queens

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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

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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Posts: 746
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GMAT 2: 680 Q50 V32
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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!
Senior Manager
Joined: 10 Jul 2013
Posts: 314

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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

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

So remainder is 0 (A)
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Intern
Joined: 31 Aug 2013
Posts: 14
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.

SVP
Joined: 06 Sep 2013
Posts: 1746
Concentration: Finance
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
Math Expert
Joined: 02 Sep 2009
Posts: 50580
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
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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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 &nbs [#permalink] 05 Dec 2017, 06:29
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