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Q.)The remainder, when $({15}^{23}+{23}^{23})$ is divided by 19, is

a)4

b)15

c)0

d)18

[CAT 2004 Question, Remainder Question, Number System, CAT Question]

Solution :

We know that $({ a }^{ n }+{ b }^{ n })=(a+b)({ a }^{ n-1 }+{ a }^{ n-2 }{ b }^{ 1 }+{ a }^{ n-3 }{ b }^{ 2 }+………….+{ b }^{ n-1 })$

So, We can conclude from the above formulae that $({ a }^{ n }+{ b }^{ n })$ is always divisible by (a+b).

Hence , $({15}^{23}+{23}^{23})$ is always divisible by (15+19) = 38

Now, if any number is divisible by 38 , then that number will also be divisible by 19 as 38 is a multiple of 19.

Hence, the remainder obtained , when $({15}^{23}+{23}^{23})$ is divided by 19 = 0.

Hence, OPTION(c)

This can also be solved in another way :

${ 15 }^{ 23 }+{ 23 }^{ 23 }={ (19-4) }^{ 23 }+{ (19+4) }^{ 23 }=19(X+Y)+{ 4 }^{ 23 }+{ (-4) }^{ 23 }=19(X+Y)$

This way also, we see that , it is divisible by 19. Hence, the remainder obtained is 0.