JEE Maths›Binomial Theorem›Visual Solution

Visual SolutionPYQ 2024 · Jan 29 Shift 2Easy

Question

Remainder when

64^{32^{32}}

is divided by 9 is equal to _____.

Solution Path

64 = 8^2

→

8 = 9-1

→ binomial expansion → only

(-1)^{2t} = 1

survives → remainder

= 1

01Read the Problem

1/4

Find the remainder when

64^{32^{32}}

is divided by 9. A numeric answer question.

64^{32^{32}} \pmod{9} = \;?

02Rewrite the Base

2/4

Let

t = 32^{32}

. Then

64^t = (8^2)^t = 8^{2t}

. The key observation:

8 = 9 - 1

8^{2t} = (9 - 1)^{2t}

03Binomial TheoremKEY INSIGHT

3/4

Expand

(9-1)^{2t}

using the binomial theorem. Every term with

k \geq 1

has a factor of 9. Only the

k=0

term survives mod 9.

(-1)^{2t} = 1

since

2t

is even

04Final Answer

4/4

64^{32^{32}} \equiv (-1)^{2t} \equiv 1 \pmod{9}

\text{Remainder} = \boxed{1}

Concepts from this question3 concepts unlocked

★

Remainder via Binomial Expansion

EASY

Write the base as (divisor +/- small number), expand, and all terms except the last vanish mod divisor

(m \pm 1)^n \equiv (\pm 1)^n \pmod{m}

Solves 90% of JEE remainder problems in under 60 seconds. The entire expansion collapses to a single term.

Remainder problemsDivisibility proofsLast digit problems

Practice (5 Qs) →

★

Base Rewriting Strategy

EASY

Express the base in terms of the divisor: find the nearest multiple and write base = multiple +/- remainder

8 = 9 - 1,\quad 64 = 65 - 1,\quad 10 = 11 - 1

The first step in every binomial remainder problem. Wrong rewriting = wrong answer. Always check: is base close to a multiple of the divisor?

Remainder problemsModular arithmeticPower simplification

Practice (5 Qs) →

Parity of Exponent

EASY

(-1)^n = 1 if n is even, -1 if n is odd. Determines the sign of the surviving term.

(-1)^{2k} = 1, \quad (-1)^{2k+1} = -1

The final step in remainder problems. Students often forget to check if the exponent is even or odd, leading to sign errors.

Remainder problemsAlternating seriesSign determination

Practice (2 Qs) →

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