Show that 1110≡1(mod100)
here, I asked a question about this same problem and all the answers say that I prove this by binomial expansion.
But by doing binomial expansion to show this is the same as doing 1110 mod 100 in the calculator, no property is being showed to prove this.
But I think this proof might be right.
gcd
11^{φ(100)} \equiv 1 \pmod{100}
11^{40} \equiv 1 \pmod{100} \ ⇒
11^{10} *11^{10} *11^{10}* 11^{10}\equiv 1\pmod{100}
by Looking everybody mod \ 100, I can say that 11^{10} \equiv 1\pmod{100}
Answer
This doesn't work because the solutions to x^4 \equiv 1 \pmod{100} are x \equiv 1, 7, 43, 49, 51, 57, 93, 99 \pmod{100}. So from (11^{10})^4 \equiv 1 \pmod{100} the best you can say is that 11^{10} \equiv 1, 7, 43, 49, 51, 57, 93, \text{ or } 99 \pmod{100}.
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