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Worked Examples · Example 10

Q.Verify a≡b(modm)a \equiv b \pmod m, if a=41a = 41, b=21b = 21 and m=5m = 5.

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✓ Free question

Since a−b=20a-b = 20 is a multiple of m=5m=5, the congruence 41≡21(mod5)41\equiv 21\pmod 5 holds.

a≡b(modm)  ⟺  m∣(a−b)a\equiv b\pmod m \iff m\mid (a-b)

i.e. aa and bb are congruent modulo mm exactly when mm divides their difference. Here a=41a=41, b=21b=21, m=5m=5.

  1. Compute the difference:

a−b=41−21=20a - b = 41 - 21 = 20

  1. Check divisibility by m=5m=5:

20÷5=4 (exact, remainder 0)20 \div 5 = 4\ \text{(exact, remainder }0)

so 5∣205\mid 20.

  1. Cross-check by remainders:

41 mod 5=1,21 mod 5=141\bmod 5 = 1,\qquad 21\bmod 5 = 1

equal remainders confirm the congruence.

✓Final answer

41−21=2041-21 = 20 is divisible by 55, hence 41≡21(mod5)41\equiv 21\pmod 5 is verified.

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