The binomial theorem elegantly solves remainder and divisibility problems by expressing the base as a sum involving the divisor.

Remainder of $a^n$ mod m: Write a = mq + r where r is small. Then $a^n$ = (mq+r)^n = sum C(n,k)(mq)^{n-k}$r^k$. All terms with k < n contain m as a factor. The remainder comes from the last few terms.

Example: 3^{256} mod 5. Write 3^4 = 81 = 80+1 = 16*5+1. So 3^{256} = (3^4)^{64} = (80+1)^{64}. All terms except C(64,0)*1 are divisible by 5 (they contain factor 80). Remainder = 1.

Divisibility proofs: To show p | f(n), expand f(n) using binomial theorem and verify each term has factor p. Classic example: 6^n - 5n - 1 is divisible by 25, since (1+5)^n - 5n - 1 = C(n,2)*25 + higher terms, all divisible by 25.

Last k digits: To find last two digits of $a^n$, compute $a^n$ mod 100. Express the base near a multiple of 100 (or 10) and expand. Example: 7^{20} mod 100. 7^2=49=50-1, so 7^{20}=(50-1)^{10}. Only the last two terms of the expansion matter mod 100: C(10,9)50(-1)^9 + (-1)^{10} = -500+1 = -499 = 01 mod 100.

These problems appear in JEE as both MCQ (find the remainder) and numerical answer type $\frac{find the last digit}{two digits}$.