Divisibility Rule for Algebraic Function

( aⁿ – bⁿ ) is always divisible by (a – b) whereas ( aⁿ + bⁿ ) is divisible by (a + b) only when n is odd. See the following image for various expansion of the terms.

Question: ( xⁿ – aⁿ ) is completely divisible by (x – a) when
(a) n is any natural number
(b) n is even natural number
(c) n is odd natural number
(d) n is prime

Ans: (a) n is any natural number

Explanation:

( xⁿ – aⁿ ) is always divisible by (x – a).

Question: ( 2⁶ⁿ – 4²ⁿ ) is always divisible by (n is a natural number)
(a) 15
(b) 18
(c) 36
(d) 48

Ans: (d) 48

Explanation:

2⁶ⁿ – 4²ⁿ = (2⁶)ⁿ – (4²)ⁿ is always divisible by
2⁶ – 4² = 64 – 16 = 48

Question: \displaystyle{\frac{2222^{9999}+5555^{9999}}{7}} , Remainder?
(a) 0
(b) 1
(c) 3
(d) 5

Ans: (a) 0

Explanation:

( aⁿ + bⁿ ) is divisible by (a + b) when n is odd

Hence, 2222⁹⁹⁹⁹ + 5555⁹⁹⁹⁹ is divisible by
2222 + 5555 = 7777, this is multiple of 7

Hence, remainder = 0

Question: \displaystyle{\frac{2^{70}+3^{70}}{13}} , Remainder?
(a) 0
(b) 1
(c) 3
(d) 5

Ans: (a) 0

Explanation:

2⁷⁰ + 3⁷⁰ = (2²)³⁵ + (3²)³⁵ = 4³⁵ + 9³⁵

We know, for n = odd
( aⁿ + bⁿ ) is always divisible by (a + b)

Now, 4 + 9 = 13

So, remainder = 0

Aliter:

Question: \displaystyle{\frac{16^3+17^3+18^3+19^3}{70}} , Remainder?
(a) 0
(b) 1
(c) 3
(d) 5

Ans: (a) 0

Explanation:

( aⁿ + bⁿ + cⁿ + dⁿ ) is divisible by (a + b + c + d) when n is odd

Here, n = 3 (odd)

16 + 17 + 18 + 19 = 70

Hence, remainder = 0