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mendelay
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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z = 24K = 3*2^3K where K is a integer
than
z^2 = 24^2*k^2/9 = 3^2 * (2^6 *k^2) / 3^2 --> the 3^2 in the numerator and denominator simplifies
z^2 = 2^6*K^2
given that k is a integer --> the remainder must be zero
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z=18+6 whole square=18^2+6^2+2*18*6 ..leaves no remainder when divided by 9.
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