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Find the remainder when 7^103 is divided by 24 Solution) we know by the concept of mod that..... 49 is congruent to 1 mod 24(means if 1 is subtracted fom 49 u get 48 which is divisible by 24) so 7^2congruent to 1 mod 24 so 7^102congruent to 1 mod 24(raising to the power 51) hence the remainder when 7^102 is divided by 24 is 1 i.e. 7^102=24m+1 so multipling by 7 both sides we have....7^103=24.7m+ 7hence 7^103=24n + 7so remainder is 7 (ans)
log6 X + log6 (x-5) = 1
Need help, please anybody solve this: Consider the universal set T and its subsets A, B and C underneath as: T = {a, b, c, d e, f} A = {a, d} B = {b, c, f} C = {a, c
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