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Q.)If X = $({ 16 }^{ 3 }+{ 17 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 })$ , then X divided by 70 leaves a remainder of

(a)0

(b)1

(c)69

(d)35

[CAT 1995 Question , CAT Question, Number System , Remainder , Odd , Even , MBA Questions]

Solution :

X = $({ 16 }^{ 3 }+{ 17 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 })$

${ 16 }^{ 3 }$ = even

${ 17 }^{ 3 }$ = odd

${ 18 }^{ 3 }$ = even

${ 19 }^{ 3 }$ = odd

even + odd + even + odd = even + even + odd + odd = even

Hence, X is divisible by 2 as X is even.

Now , We know that $({ a }^{ 3 }+{ b }^{ 3 }) = (a+b)({a}^{2} – ab + {b}^{2})$

$({ 16 }^{ 3 }+{ 17 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 })$ can be regrouped as $({ 16 }^{ 3 }+{ 19 }^{ 3 }+{ 17 }^{ 3 }+{ 18 }^{ 3 })$

Now, $({ 16 }^{ 3 }+{ 19 }^{ 3 }) = (16+19)({16}^{2}-16*19+{19}^{2}) = 35({16}^{2}-16*19+{19}^{2})$

$({ 17 }^{ 3 }+{ 18 }^{ 3 }) = (17+18)({17}^{2}-17*18+{18}^{2}) = 35({17}^{2}-17*18+{18}^{2})$

So, X will also be divisible by 35.

Hence, X is divisible by 2 and 35 both. Hence, X is divisible by 70.

So, the remainder obtained when X is divided by 70 = 0

Hence, OPTION(I)