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Exercise 1 · Q10

Q.Find 3128 mod 73^{128} \bmod 7.

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
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Since 36≡1(mod7)3^6\equiv 1\pmod 7, reduce the exponent: 128≡2(mod6)128\equiv 2\pmod 6, giving 3128≡32=2(mod7)3^{128}\equiv 3^2 = 2\pmod 7.

Find the cyclic period d with 3d≡1(mod7), then 3n≡3 n mod d(mod7)\text{Find the cyclic period } d \text{ with } 3^{d}\equiv 1\pmod 7,\ \text{then } 3^{n}\equiv 3^{\,n\bmod d}\pmod 7

Here n=128, m=7n = 128,\ m = 7.

  1. Compute the cycle of 3k mod 73^k \bmod 7:
kk3k3^k3k mod 73^k \bmod 7
133
292
3276
4814
52435
67291

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