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If "t" is not divisible by 5 then it can take one of the following forms:-
(5n+1) or (5n+2) or (5n+3) or (5n+4).

Now let's take first case
1. t = 5n+1

So for checking divisibility of (5n+1)^4 by 5, we need to expand this expression. All the terms of its expansion will have some power of 5 except one!
It is:- 4C0 x (5n)^0 x (1)^4 = 1.
So the remainder is "1".
2. For (5n+2)^4:-
That one exceptional term is:- 4C0 x (5n)^0 x (2)^4 = 16.
So here also remainder is "1".
3. For (5n+3)^4:-
That one exceptional term is:- 4C0 x (5n)^0 x (3)^4 = 81
So here also remainder is "1"
And 4. For (5n+4)^4:-
That one exceptional term is:- 4C0 x (5n)^0 x (4)^4 = 256
So here also remainder is "1".

So for all cases the remainder is "1", so there is only one possible remainder.

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If a positive integer t is not divisible by 5, how many possible different remainders can t^4 have when it is divided by 5?

Remainder of a number by 5 = Remainder of Units' digit of the number by 5

=> We need to find possible value of units' digit of t^4 where t can be any positive integer but not a multiple of 5.

Now, for t possible value of units's digit = 1, 2 , 3, 4, 6, 7, 8, 9 (it cannot be 0 and 5 as then it will become a multiple of 5)
=> Units' digit of t^4 when
  • Units' digit of t = 1 will be = Units' digit of 1^4 = 1
  • Units' digit of t = 2 will be = Units' digit of 2^4 = 6
  • Units' digit of t = 3 will be = Units' digit of 3^4 = 1
  • Units' digit of t = 4 will be = Units' digit of 4^4 = 6
  • Units' digit of t = 6 will be = Units' digit of 6^4 = 6
  • Units' digit of t = 7 will be = Units' digit of 7^4 = 1
  • Units' digit of t = 8 will be = Units' digit of 8^4 = 6
  • Units' digit of t = 9 will be = Units' digit of 9^4 = 1

=> Possbile units' digits are 1 and 6
And Remainders of both 1 and 6 by 5 is 1


So, Answer will be A
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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