JEE Mains Maths · Binomial Theorem

Remainders and Divisibility

Using the binomial theorem to find the remainder of a large power: write the base as a multiple of the divisor plus or minus a small number, and keep only the terms the divisor does not swallow.

All Binomial Theorem notes JEE Mains Maths strategy

Why this matters

Twenty-four PYQs, twelve of them numerical answer, and 2023 alone has ten. Twenty-one ask for a remainder or a count of exponents with a given remainder; three ask what an expression is divisible by. Two ideas cover the page.

Concept 1 of 2: Remainders of large powers

Find a power of the base that is close to a multiple of the divisor

m

a^k=mq\pm1

mq\pm c

. Then

a^{kn}=(mq\pm1)^n

, and every term of the expansion except the last contains

m

. So the remainder comes from

(\pm1)^n

, times whatever power of

a

is left over. Remainders repeat in a cycle, and a combined divisor like 35 splits into 5 and 7.

Definition

$(mq+c)^n\equiv c^n\pmod m$ .
Look for $a^k\equiv\pm1$ : e.g. $2^3=8\equiv1\pmod7$ , $3^2=9\equiv-1\pmod{10}$ .
A negative remainder $-r$ means $m-r$ .
$(mq+c)^n\equiv c^n+n\,c^{n-1}mq\pmod{m^2}$ : two terms survive modulo $m^2$ .

Base near a multiple

(mq\pm1)^n=m(\dots)+(\pm1)^n

Worked example

Find the remainder when

2^{100}

is divided by 7.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q62Moderate

Example 1 · Binomial Theorem · Remainders and Divisibility

The remainder when

3^{2022}

is divided by 5 is

Leftover factors

If the exponent is not a multiple of the cycle, a factor remains outside the bracket:

2^{100}=2\cdot8^{33}

. Forgetting that factor gives remainder 1 instead of 2.

Drill 20 more on remainders of large powers

Concept 2 of 2: What an expression is divisible by

a^n-b^n

is always divisible by

a-b

, and by

a+b

when

n

is even. Pair the terms of a four-term expression two ways to find factors. For expressions like

9^n-8n-1

, expand

(1+8)^n

: the first two terms cancel the

8n+1

, leaving a multiple of

64

Definition

$(a-b)\mid(a^n-b^n)$ ; $(a+b)\mid(a^n-b^n)$ for even $n$ .
$(1+m)^n-mn-1=\binom n2m^2+\binom n3m^3+\dots$ , divisible by $m^2$ .
To show 'not divisible by $d$ ', compute the remainder modulo $d$ .

Removing the first two terms

(1+m)^n-mn-1=m^2\left[\binom n2+\binom n3m+\dots\right]

Worked example

Show that

4^n-3n-1

is divisible by 9.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q163Moderate

Example 2 · Binomial Theorem · Remainders and Divisibility

25^{190}-19^{190}-8^{190}+2^{190}

is divisible by

Pair the terms to match the signs

a^n-b^n-c^n+d^n

can be grouped as

(a^n-c^n)-(b^n-d^n)

(a^n-b^n)-(c^n-d^n)

. Each grouping gives a different common factor; try both.

Drill 2 more on what an expression is divisible by

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Remainders of large powers
Base near a multiple
$(mq\pm1)^n=m(\dots)+(\pm1)^n$
What an expression is divisible by
Removing the first two terms
$(1+m)^n-mn-1=m^2\left[\binom n2+\binom n3m+\dots\right]$

Watch out for (2)

Leftover factors
→ Remainders of large powers
Pair the terms to match the signs
→ What an expression is divisible by

Test yourself on Binomial Theorem

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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