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Worked Examples · Example 1

Q.Find

(i) 21 mod 821 \bmod 8
(ii) −31 mod 7-31 \bmod 7
(iii) 18 mod 1818 \bmod 18
(iv) 7 mod 127 \bmod 12
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✓ Free question

Using a mod m=ra\bmod m = r where a=qm+ra=qm+r and 0≤r<m0\le r<m: the four remainders are 5,  4,  0,  75,\;4,\;0,\;7.

a mod m=rwherea=qm+r,    0≤r<ma \bmod m = r \quad\text{where}\quad a = qm + r,\;\; 0 \le r < m

  • aa = dividend, mm = modulus (m>0m>0), qq = quotient, rr = remainder (always non-negative and less than mm).
  1. (i) 21 mod 821 \bmod 8: 21=2×8+521 = 2\times 8 + 5, so q=2q=2, r=5r=5. Hence 21 mod 8=521\bmod 8 = 5.
  2. (ii) −31 mod 7-31 \bmod 7: we need rr with 0≤r<70\le r<7. Take q=−5q=-5: −31=(−5)×7+4=−35+4-31 = (-5)\times 7 + 4 = -35+4, so r=4r=4. Hence −31 mod 7=4-31\bmod 7 = 4.
  3. (iii) 18 mod 1818 \bmod 18: 18=1×18+018 = 1\times 18 + 0, so r=0r=0. Hence 18 mod 18=018\bmod 18 = 0.
  4. (iv) 7 mod 127 \bmod 12: since 7<127<12, 7=0×12+77 = 0\times 12 + 7, so r=7r=7. Hence 7 mod 12=77\bmod 12 = 7.
✓Final answer

(i) 5\mathbf{5} (ii) 4\mathbf{4} (iii) 0\mathbf{0} (iv) 7\mathbf{7}.

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