In short

The binomial theorem is not just an expansion formula — it is a tool for computing remainders (write the base as a convenient sum, expand, and most terms vanish mod the divisor), approximating powers (keep the first two or three terms when x is small), proving that expressions are divisible by a given number, and finding the last one or two digits of large powers.

Your friend bets you cannot find the remainder when 7^{100} is divided by 48. Computing 7^{100} directly is out of the question — it has 85 digits. But rewrite 7 as 1 + 6, expand (1 + 6)^{100} using the binomial theorem, and suddenly every term except the first two contains 6^2 = 36 as a factor. Since 48 = 36 + 12 and 36 divides cleanly into 48... you are very close to the answer.

That is the core trick behind every application in this article: rewrite the base as a sum where the binomial expansion isolates what you need.

Finding remainders

To find the remainder when N^k is divided by d, write N = q + r where q is chosen so that a power of q is divisible by d. Then expand (q + r)^k. Most terms will be divisible by d, and the remainder comes from the few surviving terms.

Strategy for finding remainders using binomial expansionA diagram showing the steps: rewrite N as q + r, expand (q+r) to the k, identify terms divisible by d (all terms with q squared or higher), and collect the remaining terms to find the remainder. Remainder strategy Write N = q + r → Expand (q + r)ᵏ → Drop multiples of d Choose q so that q² (or q itself) is divisible by d Then only the first 1 or 2 terms of the expansion survive mod d Remainder = the surviving terms (reduced mod d)
The three-step strategy for remainder problems. The key is choosing the split $N = q + r$ so that higher powers of $q$ are absorbed by the divisor $d$.

Example. Find the remainder when 7^{100} is divided by 48.

Write 7 = 1 + 6. Then:

(1 + 6)^{100} = \binom{100}{0} + \binom{100}{1} \cdot 6 + \binom{100}{2} \cdot 6^2 + \binom{100}{3} \cdot 6^3 + \cdots + 6^{100}

Every term from r = 2 onward contains 6^2 = 36 as a factor. And 36 divides 48? Not exactly — 48 = 36 + 12. So try a different approach: write 7 = -1 + 8. Then:

(8 - 1)^{100} = \sum_{r=0}^{100} \binom{100}{r}\, 8^{100-r}\, (-1)^r

Every term with 100 - r \geq 2 (that is, r \leq 98) contains 8^2 = 64, which is divisible by 48? Again not exactly. The clean choice here is: 48 = 48, and 7^2 = 49 = 48 + 1, so 7^2 \equiv 1 \pmod{48}.

Therefore 7^{100} = (7^2)^{50} \equiv 1^{50} = 1 \pmod{48}.

The remainder is \mathbf{1}.

This example shows that the binomial approach works best when you can find a split where a small power is congruent to something simple modulo d. Sometimes the direct binomial expansion is the cleanest route; sometimes a preliminary observation (like 7^2 \equiv 1 \pmod{48}) makes it even shorter.

Approximations using the binomial theorem

When |x| is much smaller than 1, the expansion of (1 + x)^n converges rapidly — the later terms become tiny. Keeping just the first few terms gives a useful approximation:

(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots
Levels of binomial approximation for (1+x) to the nThree nested boxes showing increasing accuracy. Level 1: 1 + nx (linear). Level 2: 1 + nx + C(n,2)x squared (quadratic). Level 3: includes the cubic term. Each level is more accurate for small x. (1 + x)ⁿ — levels of approximation Level 3: + C(n,3)x³ + ... Level 2: + C(n,2)x² Level 1: (1 + x)ⁿ ≈ 1 + nx accurate when |x| ≪ 1 Each level adds one more term and improves accuracy
The simplest approximation $1 + nx$ is often good enough for back-of-the-envelope calculations. Adding $\binom{n}{2}x^2$ makes it much more precise. Each additional term costs one more computation but buys another order of accuracy.

Example. Approximate (1.02)^{10} to four decimal places.

Here x = 0.02 and n = 10:

\begin{aligned} (1.02)^{10} &= 1 + 10(0.02) + \binom{10}{2}(0.02)^2 + \binom{10}{3}(0.02)^3 + \cdots \\ &= 1 + 0.2 + 45 \times 0.0004 + 120 \times 0.000008 + \cdots \\ &= 1 + 0.2 + 0.018 + 0.00096 + \cdots \\ &\approx 1.2190 \end{aligned}

The exact value (to six places) is 1.218994. The three-term approximation gives 1.2190, accurate to 0.01\%.

Proving divisibility

The binomial theorem is a clean way to prove that a given expression is always divisible by some integer. The technique: expand using the binomial theorem, and show that every term contains the required factor.

Using binomial expansion to prove divisibilityA diagram showing the expansion of (1+a) to the n minus 1 minus na. The first two terms of the expansion are 1 and na. Subtracting 1 and na leaves only terms with a squared or higher, each divisible by a squared. Proving (1+a)ⁿ − 1 − na is divisible by a² (1+a)ⁿ = 1 + na + C(n,2)a² + C(n,3)a³ + ... + aⁿ ↑ subtract 1 ↑ subtract na ↑ every term has a² as factor (1+a)ⁿ − 1 − na = a²[C(n,2) + C(n,3)a + ... + aⁿ⁻²] The bracketed expression is an integer when a is an integer → divisible by a²
After subtracting $1$ and $na$ from $(1+a)^n$, every surviving term has $a^2$ or higher as a factor. So $(1+a)^n - 1 - na$ is always divisible by $a^2$ for any positive integer $a$ and $n \geq 2$.

Claim. For any positive integer n \geq 2 and any integer a, the expression (1 + a)^n - 1 - na is divisible by a^2.

Proof. Expand (1 + a)^n:

(1 + a)^n = 1 + na + \binom{n}{2}a^2 + \binom{n}{3}a^3 + \cdots + a^n

Subtract 1 + na:

(1 + a)^n - 1 - na = \binom{n}{2}a^2 + \binom{n}{3}a^3 + \cdots + a^n = a^2\left[\binom{n}{2} + \binom{n}{3}a + \cdots + a^{n-2}\right]

The expression in brackets is an integer (each \binom{n}{r} is an integer, and a is an integer). So the whole expression is divisible by a^2.

This technique extends: to show 6^n - 5n - 1 is divisible by 25, write 6 = 1 + 5 and expand (1 + 5)^n. The first two terms are 1 + 5n. Subtracting gives terms each containing 5^2 = 25. Done.

Finding last digits

The last digit of N^k is the remainder when N^k is divided by 10. The last two digits give the remainder modulo 100. The binomial theorem handles both.

Last digit equals remainder modulo 10A number N to the k is shown. Its last digit is circled and equals N to the k mod 10. Its last two digits equal N to the k mod 100. Last digits and remainders N^k = ...d₃d₂d₁ last digit = d₁ = Nᵏ mod 10 last two digits = d₂d₁ = Nᵏ mod 100 Write N = (multiple of 10) ± small number, then expand
The last digit of a number is its remainder mod $10$. The last two digits are its remainder mod $100$. The binomial expansion isolates these remainders efficiently.

Example. Find the last two digits of 11^{25}.

Write 11 = 10 + 1:

(10 + 1)^{25} = \sum_{r=0}^{25} \binom{25}{r} \cdot 10^{25-r} \cdot 1^r

For the last two digits, you need this sum modulo 100. Every term with 10^{25-r} where 25 - r \geq 2 (i.e., r \leq 23) is divisible by 100. The surviving terms are:

Total: 50 + 1 = 51 \pmod{100}.

The last two digits of 11^{25} are \mathbf{51}.

Combining the techniques

Many problems use more than one of these ideas at once. Proving that 3^{2n+2} - 8n - 9 is divisible by 64 for all n \geq 1 requires both a rewrite and a divisibility argument:

Write 3^{2n+2} = 9^{n+1} = (1 + 8)^{n+1}. Expand:

(1 + 8)^{n+1} = 1 + (n+1) \cdot 8 + \binom{n+1}{2} \cdot 64 + \cdots

So 3^{2n+2} - 8n - 9 = (1 + 8n + 8) - 8n - 9 + [\text{terms with } 64] = 0 + 64[\ldots].

That is: 3^{2n+2} - 8n - 9 = 64\left[\binom{n+1}{2} + \binom{n+1}{3} \cdot 8 + \cdots\right], which is divisible by 64 since the bracketed expression is a positive integer for n \geq 1.

Proving 3 to the 2n+2 minus 8n minus 9 is divisible by 64A diagram showing the key rewrite: 3 to the 2n+2 equals (1+8) to the n+1. The expansion gives 1 + 8(n+1) + terms with 64. Subtracting 8n+9 leaves exactly the terms with 64. 3^(2n+2) − 8n − 9 is divisible by 64 3^(2n+2) = 9^(n+1) = (1 + 8)^(n+1) = 1 + 8(n+1) + C(n+1,2)·64 + C(n+1,3)·8³ + ... Subtract (8n + 9): 1 + 8n + 8 − 8n − 9 = 0 Remaining: 64·[C(n+1,2) + C(n+1,3)·8 + ...] Every surviving term is a multiple of 64 → result divisible by 64
The rewrite $3^{2n+2} = (1 + 8)^{n+1}$ makes the first two terms cancel with $8n + 9$, leaving only terms divisible by $8^2 = 64$.

Two worked examples

Example 1: Find the remainder when $3^{256}$ is divided by $13$

Step 1. Look for a power of 3 close to a multiple of 13. Notice that 3^3 = 27 = 26 + 1 = 2 \times 13 + 1, so 3^3 \equiv 1 \pmod{13}.

Why: if 3^3 \equiv 1 \pmod{13}, then 3^{3k} \equiv 1^k = 1 for any k, and the problem reduces to finding 256 \mod 3.

Step 2. Divide the exponent by 3: 256 = 3 \times 85 + 1.

So 3^{256} = 3^{3 \times 85 + 1} = (3^3)^{85} \cdot 3 \equiv 1^{85} \cdot 3 = 3 \pmod{13}.

Why: split the exponent into a multiple of 3 (which gives 1 \pmod{13}) and a leftover.

Step 3. Alternatively, use the binomial theorem directly. Write 3^3 = 27 = 26 + 1:

3^{256} = 3 \cdot (3^3)^{85} = 3 \cdot (26 + 1)^{85}

(26 + 1)^{85} = 1 + 85 \cdot 26 + \binom{85}{2} \cdot 26^2 + \cdots

Every term except the 1 contains 26 = 2 \times 13, hence is divisible by 13.

So (26 + 1)^{85} \equiv 1 \pmod{13}, and 3^{256} \equiv 3 \cdot 1 = 3 \pmod{13}.

Why: the binomial expansion makes the divisibility by 13 visible term by term. Only the constant term 1 survives.

Step 4. Verify with a small case. 3^3 = 27, and 27 \div 13 = 2 remainder 1. Then 3^4 = 81, and 81 \div 13 = 6 remainder 3. Since 256 \equiv 1 \pmod{3}, the answer 3 is consistent.

Result. The remainder when 3^{256} is divided by 13 is \mathbf{3}.

Finding 3 to the 256 mod 13 using binomial expansionA diagram showing the chain of reasoning: 3 cubed equals 27 equals 26+1, 26 is 2 times 13, so (26+1) to the 85 expands to 1 plus multiples of 13. Multiply by 3 to get the remainder 3. 3²⁵⁶ mod 13 — step by step 3²⁵⁶ = 3¹ · (3³)⁸⁵ = 3 · 27⁸⁵ 27 = 26 + 1 = 2·13 + 1 (26+1)⁸⁵ = 1 + 85·26 + C(85,2)·26² + ... ≡ 1 (mod 13) 3²⁵⁶ ≡ 3 × 1 = 3 (mod 13) Check: 3⁴ = 81 = 6×13 + 3, so 3⁴ ≡ 3 (mod 13) ✓
The key insight: $3^3 = 27 = 26 + 1 \equiv 1 \pmod{13}$. Since $256 = 3 \times 85 + 1$, the remainder is $3^1 = 3$. The binomial expansion of $(26 + 1)^{85}$ confirms that all terms except $1$ are multiples of $13$.

The strategy is always the same: find a power of the base that is 1 more (or 1 less) than a multiple of the divisor. The binomial theorem then does the heavy lifting.

Example 2: Find the last two digits of $7^{82}$

Step 1. The last two digits of N are N \bmod 100. Write 7^2 = 49 = 50 - 1.

Why: 50 - 1 is a useful split because 50^2 = 2500 is divisible by 100. After the binomial expansion, most terms will vanish modulo 100.

Step 2. Express 7^{82} = (7^2)^{41} = (50 - 1)^{41}.

Expand using the binomial theorem:

(50 - 1)^{41} = \displaystyle\sum_{r=0}^{41} \binom{41}{r}\, 50^{41-r}\, (-1)^r

Why: here a = 50 and b = -1 in the binomial expansion.

Step 3. Identify terms that survive modulo 100. Any term with 50^{41-r} where 41 - r \geq 2 (i.e., r \leq 39) has 50^2 = 2500 as a factor, which is divisible by 100. So only r = 40 and r = 41 survive:

r = 40: \binom{41}{40} \cdot 50^1 \cdot (-1)^{40} = 41 \times 50 \times 1 = 2050

r = 41: \binom{41}{41} \cdot 50^0 \cdot (-1)^{41} = 1 \times 1 \times (-1) = -1

Why: for r = 40, (-1)^{40} = 1 (even power). For r = 41, (-1)^{41} = -1 (odd power).

Step 4. Combine: 2050 + (-1) = 2049. Take this modulo 100: 2049 \bmod 100 = 49.

Result. The last two digits of 7^{82} are \mathbf{49}.

Finding the last two digits of 7 to the 82A diagram showing the expansion of (50-1) to the 41. Terms with 50 squared or higher vanish mod 100. Only the r=40 term (2050) and r=41 term (negative 1) survive. Their sum mod 100 is 49. 7⁸² mod 100 — last two digits 7⁸² = (7²)⁴¹ = (50 − 1)⁴¹ Terms with 50² or higher → divisible by 100 → vanish r = 40: C(41,40)·50·1 = 41 × 50 = 2050 r = 41: 1·1·(−1) = −1 2050 + (−1) = 2049 2049 mod 100 = 49 → last two digits are 49
The split $7^2 = 50 - 1$ leaves only two terms alive modulo $100$. The last two digits of $7^{82}$ are $49$. Interestingly, $7^{82}$ ends in the same digits as $7^2 = 49$ — this is no coincidence, and [modular arithmetic](/wiki/modular-arithmetic) explains why.

The split 7^2 = 50 - 1 is the key insight. If you had written 7 = 10 - 3, the expansion (10 - 3)^{82} would require tracking terms with 10^1 (which does not vanish mod 100) all the way back to r = 81, making the calculation messy. Choosing a split where the large part squares to a multiple of 100 keeps the computation to just two terms.

Common confusions

Going deeper

If you can split a base as a convenient sum, expand using the binomial theorem, and identify which terms survive modulo a divisor, you have the toolkit for all standard application problems. The material below connects these ideas to more general settings.

The Frobenius endomorphism and prime moduli

When the divisor is a prime p, the binomial theorem yields a particularly clean result: (a + b)^p \equiv a^p + b^p \pmod{p}. This is because every interior binomial coefficient \binom{p}{r} for 1 \leq r \leq p - 1 is divisible by p (the prime p appears in the numerator p! but not in r!(p-r)!). This identity is the starting point of Fermat's Little Theorem and has deep consequences in algebra.

Error bounds for approximations

When you truncate the binomial expansion of (1 + x)^n after k terms, the error is bounded by the (k+1)-th term (when |x| < 1, since the terms decrease in absolute value from some point onward). For the linear approximation, the error is at most \binom{n}{2}x^2. This gives a rigorous bound: (1.01)^{10} = 1.1 \pm 0.0045, accurate to about 0.4\%.

Last digits and cyclicity

The last digit of N^k depends only on the last digit of N and follows a cycle of length at most 4: the last digits of 7^1, 7^2, 7^3, 7^4 are 7, 9, 3, 1, and then they repeat. The binomial theorem provides one proof; the theory of orders in modular arithmetic provides the general framework.

Where this leads next