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Worked Examples · Example 2

Q.If X=7X = 7 and Y=3Y = 3, then verify that X mod Y=(X+kY) mod YX \bmod Y = (X + kY) \bmod Y for k=6k = 6.

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✓ Free question

Both sides reduce to remainder 11, verifying X mod Y=(X+kY) mod YX\bmod Y=(X+kY)\bmod Y for X=7, Y=3, k=6X=7,\,Y=3,\,k=6.

a mod m=r where a=qm+r,  0≤r<m;(a+km) mod m=a mod ma\bmod m = r \text{ where } a=qm+r,\; 0\le r<m; \qquad (a+km)\bmod m = a\bmod m

Adding any whole multiple kYkY of the modulus leaves the remainder unchanged.

  • X=7X=7 (dividend), Y=3Y=3 (modulus), k=6k=6.
  1. Left side — X mod YX\bmod Y: 7=2×3+17 = 2\times 3 + 1, so 7 mod 3=17\bmod 3 = 1.
  2. Compute X+kYX+kY: X+kY=7+6×3=7+18=25X+kY = 7 + 6\times 3 = 7 + 18 = 25.
  3. Right side — (X+kY) mod Y(X+kY)\bmod Y: 25=8×3+125 = 8\times 3 + 1, so 25 mod 3=125\bmod 3 = 1.
  4. Compare: left side =1=1 and right side =1=1; since 1=11=1, the identity holds.
✓Final answer

X mod Y=7 mod 3=1=25 mod 3=(X+kY) mod YX\bmod Y = 7\bmod 3 = 1 = 25\bmod 3 = (X+kY)\bmod Y. Verified — both sides equal 1\mathbf{1}.

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