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pmal04
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(k+1)^3=k^3+3k^2+3k+1. So the remainder is always 1 ;)
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rashminet84
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There is a remainder theorem that states

If x/y gives remainder z

then ax/y will give remainder az, as long as az is < y

in addition

if p/y gives remainder q

then p*x/y will give remainder z*q........ as long as zq is < y


N = (k+1)^3

K+1/k will always give remainder of 1

so when (k+1)(k+1)(K+1) is divided by K, i will give remainder of 1*1*1.....ie 1

But we have to be carefull while using this theorem, its safe as long as remainder is 1, otherwise care has to be taken that the remainder doesn't exceed the divisor, because in that case, the rule cannot be applied.
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Quote:
What is the remainder when the positive integer n is divided by the positive integer k,
where k > 1?
(1) n = (k+1)^3
(2) k = 5

Answer is A
Stat 1

plug in numbers

n = (k+1)^3

Let K be 2
n=(2+1)^3
n=3^3 =9
n/k=9/2 ---the remainder is 1

n = (k+1)^3

Let K be 3
n=(3+1)^3
n=4^3 =64
n/k=64/3 ---the remainder is 1

n = (k+1)^3

Let K be 4
n=(4+1)^3
n=5^3 =125
n/k=125/4 ---the remainder is 1

A is suff

stat 2.
k = 5
we don't know about n..hence not suff.
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Quote:
(k+1)^3=k^3+3k^2+3k+1. So the remainder is always 1


I liked the tip.
Thanks!



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