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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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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.

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