( 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
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