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Exercise 1 · Q1

Q.Find the sum of 132 and 121 mod 23.

Andaman Nicobar CbseNCERTSubjective· 2mImportance★★★★★est
35% · 37/106 Questions
✓ Free question

132+121=253132+121 = 253, and 253253 is exactly 11×2311\times 23, so the sum modulo 2323 is 00.

(a+b) mod m=((a mod m)+(b mod m)) mod m(a + b) \bmod m = \big((a\bmod m) + (b\bmod m)\big)\bmod m

Here a=132, b=121, m=23a=132,\ b=121,\ m=23.

  1. Add the numbers. 132+121=253132 + 121 = 253.
  2. Divide by the modulus 2323. 23×11=25323 \times 11 = 253, so 253=23×11+0253 = 23\times 11 + 0.
  3. Remainder. 253 mod 23=0253 \bmod 23 = 0.
  4. Check via components: 132 mod 23=132−5×23=132−115=17132\bmod 23 = 132 - 5\times23 = 132-115 = 17; 121 mod 23=121−5×23=121−115=6121\bmod 23 = 121 - 5\times23 = 121-115 = 6; (17+6) mod 23=23 mod 23=0(17+6)\bmod 23 = 23\bmod 23 = 0. ✓
✓Final answer

(132+121)≡0(mod23)(132 + 121) \equiv 0 \pmod{23}.

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