# Binomial Theorem: Divisibility Applications

> Binomial divisibility applications in JEE Mathematics: prove divisibility of 9^n − 8n − 1 type expressions, find remainders and expansion shortcuts.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/binomial-applications-divisibility
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Binomial Theorem: Divisibility Applications", PrepElephant, https://prepelephant.com/topics/jee/mathematics/binomial-applications-divisibility

## Direct answer

Split the base and most of the expansion dies: writing 9^n as (1 + 8)^n leaves 1 + 8n alive at the front and a multiple of 8² = 64 from the second term onwards, so 9^n − 8n − 1 is divisible by 64 for every positive integer n. That is the whole method — choose a and b so the divisor divides one addend, expand (a + b)^n, and observe that everything beyond the first two terms carries the divisor as a factor. Remainder questions run identically: expand the base as divisor ± small number and only the surviving constant or linear term matters.

## What you must remember

- **The workhorse line:** (1 + x)^n = 1 + nx + C(n,2)x² + C(n,3)x³ + ...; subtracting 1 + nx leaves only terms containing x² or higher powers — the basis of every "divisible by x²" argument.
- **Classic quotable results:** 9^n − 8n − 1 divisible by 64; 6^n − 5n − 1 divisible by 25; generally (1 + k)^n − nk − 1 is divisible by k² for any positive integer k.
- **Remainder by expansion:** 49^6 = (48 + 1)^6 leaves remainder 1 on division by 48, since every term beyond the first carries 48.
- **Last digits:** 11^n = (10 + 1)^n gives 11^n ≡ 10n + 1 (mod 100), so 11⁴ ends in 41 — last-two-digit questions are linear in n.
- **Bernoulli-type bound:** for x ≥ 0 and n ≥ 1, (1 + x)^n ≥ 1 + nx, with equality only when n = 1 or x = 0 — the two-term truncation is a lower bound.
- **Sign hygiene:** (1 − x)^n = 1 − nx + C(n,2)x² − ...; beyond the second term the signs wash out, so (1 − x)^n − 1 + nx is still divisible by x².
- **Choosing the split:** factor the divisor out of one addend — for divisibility by 25 write 6 = 5 + 1, not 6 = 4 + 2; the wrong split produces no clean factor.

## A divisibility proof done properly

Claim: 3^(2n) − 8n − 1 is divisible by 64 for all n ≥ 1. Rewrite 3^(2n) as 9^n = (1 + 8)^n and expand: 1 + 8n + C(n,2)·8² + C(n,3)·8³ + ... + 8^n. Subtract 8n + 1. Every surviving term contains at least 8²: the r = 2 term carries 8², and each later term carries a higher power. Factoring, the difference equals 64·[C(n,2) + 8·C(n,3) + ... + 8^(n−2)], an integer multiple of 64. The two moves that earn full credit are naming the rewrite 9^n = (1 + 8)^n and stating explicitly that all remaining terms share the factor 8² — the bracket being an integer closes the proof.

The remainder twin: the remainder of 49^6 on division by 48. Since 49 = 48 + 1, every term of (48 + 1)^6 beyond the first carries 48, leaving the constant term 1. Compare induction — base case, hypothesis, algebraic push for each statement; the binomial route is three lines and covers every n at once, which is why Main numerical questions (answers like 1, 4, 41) are built on it.

## Main versus Advanced

Main asks remainders of large powers modulo small numbers and straight divisibility of the 9^n − 8n − 1 type; both are numerical-answer friendly. Advanced wraps the same expansion inside series summation — sums like Σ C(n, r)·2^r evaluated as (1 + 2)^n — or asks which of several divisibility statements hold for all n. The recurring traps: forgetting the linear correction, i.e. claiming (1 + x)^n − 1 is divisible by x² when only (1 + x)^n − 1 − nx is; picking a split where the divisor fails to divide either addend cleanly; and losing a minus sign with bases like 49 = 50 − 1, where odd-r terms alternate.

## Frequently asked questions

### Why is 9^n − 8n − 1 always divisible by 64?

Because 9^n = (1 + 8)^n = 1 + 8n + (terms each containing 8²), so subtracting 8n + 1 leaves an integer multiple of 64.

### How do I find the remainder of a large power without a calculator?

Write the base as (divisor ± small), expand, and keep only the terms below the divisor's power — usually just the constant or linear term.

### What are the last two digits of 11^n?

(10 + 1)^n ≡ 10n + 1 (mod 100), so the last two digits are those of 10n + 1 — for n = 4, digits 41.

### Does the binomial argument replace induction in exams?

For fixed-base powers like 9^n it is shorter and fully accepted; induction remains the tool when the base itself changes with n.

### What exactly is divisible by x² in a (1 + x)^n expansion?

(1 + x)^n − 1 − nx is divisible by x²; dropping the − nx correction is the standard error.