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Worked Examples · Example 5

Q.Find the remainder when (127×137×23×50×235×15)(127 \times 137 \times 23 \times 50 \times 235 \times 15) is divided by 7.

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Replace each factor by its remainder mod 77, multiply the remainders, and reduce once more — the final remainder is 44.

(∏ai) mod m=(∏(ai mod m)) mod m\Big(\textstyle\prod a_i\Big)\bmod m = \Big(\textstyle\prod (a_i\bmod m)\Big)\bmod m

where each aia_i is a factor of the product and m=7m=7 is the divisor.

  1. Reduce each factor modulo 77:

127 mod 7=1 (127=18×7+1)127\bmod 7 = 1\ (127=18\times7+1)

137 mod 7=4 (137=19×7+4)137\bmod 7 = 4\ (137=19\times7+4)

23 mod 7=2 (23=3×7+2)23\bmod 7 = 2\ (23=3\times7+2)

50 mod 7=1 (50=7×7+1)50\bmod 7 = 1\ (50=7\times7+1)

235 mod 7=4 (235=33×7+4)235\bmod 7 = 4\ (235=33\times7+4)

15 mod 7=1 (15=2×7+1)15\bmod 7 = 1\ (15=2\times7+1) …

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