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This question can easily be solved by using binomial theorem.
22^56 can be written as (21+1)^56. Now applying binomial theorem
56C0 21^56+56C1 21^55+56C2 21^54+...........+56C56 21^0
=21(56C0 21^55+56C1 21^54+56C2 21^53+.............+56C56 21^-1)
let 56C0 21^55+56C1 21^54+56C2 21^53+..............+56C56 21^-1)=a
=21a which is divisible by 7 so the remainder is zero
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