Binomial Theorem: Divisibility Applications
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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.