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If z is a multiple of 24, what is the remainder when z^2 is
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Updated on: 17 Sep 2009, 08:07
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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
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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.



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Re: Another remainder question....
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04 Sep 2009, 07:44
is it 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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Re: Another remainder question....
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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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Re: Another remainder question....
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04 Sep 2009, 13:44
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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Re: Another remainder question....
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31 Jan 2010, 06:21
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) Answer A
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Re: Another remainder question....
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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) Answer A nice explanation but also you can do the other way by calculating square of 24 and then divide by 9.
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Re: Another remainder question....
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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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Re: Another remainder question....
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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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Re: If z is a multiple of 24, what is the remainder when z^2 is
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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.
Answer is (a).



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Re: If z is a multiple of 24, what is the remainder when z^2 is
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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 That's your answer Cheers! J



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Re: If z is a multiple of 24, what is the remainder when z^2 is
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27 Jan 2014, 05:11



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