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Exercise 223Number theory › CongruencesHigh schoolAdded 3 Nov 2026

Arithmetic modulo $n$

#modular arithmetic#operations#compatibility

Statement

Compute the following in Z/7Z\mathbb{Z}/7\mathbb{Z}:
(a) 3+53+5; (b) 4⋅64\cdot6; (c) 252^5; (d) 3−63-6.

Hint

Reduce modulo 77 at each step.

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Solution

(a) 3+5=8≡1(mod7)3+5=8\equiv1\pmod7.

(b) 4⋅6=24≡3(mod7)4\cdot6=24\equiv3\pmod7.

(c) 25=32≡4(mod7)2^5=32\equiv4\pmod7 (since 28=4⋅728=4\cdot7).

(d) 3−6=−3≡4(mod7)3-6=-3\equiv4\pmod7.

Conclusion. The results in Z/7Z\mathbb{Z}/7\mathbb{Z} are 1ˉ,3ˉ,4ˉ,4ˉ\bar1,\bar3,\bar4,\bar4.