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The textbook gives the answer as:

By Fermat’s little theorem, we know that $7^{10} ≡ 1 \pmod{11}$, and so $(7^{10})^k ≡ 1 \pmod{11}$, for every positive integer $k$.

Therefore, $7^{222} = 7^{22\cdot10 + 2} = (7^{10})^{22}7^2 ≡ (1)^{22}\cdot 49 ≡ 5 \pmod{11}$.

Hence, $7^{222}\mod{11} = 5$.

The part that I do not understand is $(7^{10})^{22}7^2 ≡ (1)^{22}\cdot 49 ≡ 5 \pmod {11}$.

Can someone please explain to me why can we say: $(7^{10})^{22}7^2 ≡ (1)^{22}\cdot49$

Bernard
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Julia
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    I'm sure you have seen the result: if $a\equiv b$ and $c\equiv d$, then $ac\equiv bd$ (the same modulus throughout). A consequence of that is $a\equiv b\implies a^n\equiv b^n$ (proven by induction on $n$), Here $7^{10}\equiv1$ so $(7^{10})^{22}\equiv 1^{22}$. – Jyrki Lahtonen Jul 09 '15 at 20:42
  • @JyrkiLahtonen.Thank you, I think that answers my question. Rookie question, but is it not necessary to write the (mod 11) every time? – Julia Jul 09 '15 at 20:46
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    The context is clear enough: there is only one modulus. – Bernard Jul 09 '15 at 20:55
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    If you make an umbrella statement that all the congruences are modulo $11$, then you IMO don't need to repeat it. I won't allow my students to drop it until I'm sure they can handle it :-). Here space is also a concern. As Bernard says, context usually makes it clear. – Jyrki Lahtonen Jul 09 '15 at 20:55
  • See this answer for proofs of the Congruence Sum, Product, Power and Polynomial Rules. – Bill Dubuque Jul 09 '15 at 22:12

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If we have $$a \equiv b \pmod{c}$$ then we can also state that $$a^n \equiv b^n \pmod{c}$$ in your case, take $a= 7^{10}$, $b=1$ and $c=11$. Noting that you can write $$7^{222} = (7^{10})^{22 + 2}$$ the result follows. Since $$(7^{10})^{22} = 1^{22} \pmod{11}$$

The reason that we can say $a \equiv b \pmod{c} \implies a^n \equiv b^n \pmod{c}$ is from using the well known result that if $a \equiv b \pmod{c} \implies ad \equiv be \pmod{c}$ then taking $d = a$ and $e=b$ yields $a^2 \equiv b^2\pmod{c}$ and we continue by induction on $n$ to get the result.

Zain Patel
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  • Your "well-known result" is completely wrong. $7 \equiv 3 \pmod {2}$ but $7\cdot 3 \equiv 3 \cdot 10 \pmod{2}$ is certainly not true. The correct proof is as follows. If $a \equiv b \pmod{c}$, then $\frac{b-a}{c}$ is an integer. It follows that $\frac{(b-a)(b^{n-1} + b^{n-2}a + \cdots + a^{n-1})}{c}$ is an integer as well, so $a^n \equiv b^n \pmod{c}$. – user217285 Jul 15 '15 at 22:27
  • @Nitin, I have to head out. You can either edit my answer (I'll be grateful) to include your helpful information or I can do it later, when I get back! :-) – Zain Patel Jul 15 '15 at 22:32