Visual SolutionPYQ 2024 · Apr 9 Shift 1Tricky

Question

The remainder when

428^{2024}

is divided by 21 is _____.

Solution Path

428 \equiv 8 \pmod{21}

8^2 = 64 \equiv 1

. So

428^{2024} = (8^2)^{1012} \equiv 1

01Question Setup

1/4

Find the remainder when

428^{2024}

is divided by

21

428^{2024} \pmod{21}

02Reduce mod 21

2/4

428 = 21 \times 20 + 8

, so

428 \equiv 8 \pmod{21}

. Then

8^2 = 64 = 63 + 1 \equiv 1

8^2 \equiv 1 \pmod{21}

03Final ComputationKEY INSIGHT

3/4

428^{2024} \equiv 8^{2024} = (8^2)^{1012} \equiv 1^{1012} = 1

(8^2)^{1012} \equiv 1

04Final Answer

4/4

The remainder is

1

\boxed{1}

Concepts from this question2 concepts unlocked

★

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

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

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