1

I know how to get $99^{2} \mod 32$ from the $(99\mod32)(99\mod32)\mod32$ and get the answer. But the solution guide I have, has converted $$99^{2}\mod32$$ to $$3^{2}\mod32$$. Any idea how?

Amir Qasemi
  • 13

3 Answers3

3

We know that if $a \equiv b \mod n$, then $a^2 \equiv b^2 \mod n$. If you haven't learned this yet, maybe attempt a proof.

Since $99 \equiv 3 \mod 32$ , $99^2 \equiv 3^2 = 9 \mod 32$.

Sean Haight
  • 1,368
  • 1
    If you want I can add a short proof of the fact stated above. – Sean Haight Aug 27 '16 at 00:09
  • thanks. I just didn't know the proposition. The proof seems straightforward. – Amir Qasemi Aug 27 '16 at 00:21
  • 4
    More generally, if $a\equiv a'$ and $b\equiv b'$ modulo $n$, then $a+b\equiv a'+b'$ and $ab\equiv a'b'$, both modulo $n$. I frequently found, when teaching Algebra to college juniors, that it took a good while for the power of these facts to impress itself on them. – Lubin Aug 27 '16 at 01:31
0

Just as a more concrete representation, note that

\begin{align} 99^2 & = (3 \times 32 + 3)^2 \\ & = (3 \times 32)^2 + 2 (3)(3 \times 32) + 3^2 \\ & = (3 \times 3 \times 32 \times 32) + (2 \times 3 \times 3 \times 32) + 3^2 \\ & = (288 \times 32) + (18 \times 32) + 3^2 \\ & = (306 \times 32) + 3^2 \end{align}

and hence $99^2 \bmod 32 = 3^2 \bmod 32 = 9 \bmod 32 = 9$.

Brian Tung
  • 34,160
0

$$99^2 \equiv (33)^23^2\equiv 1^23^2 \equiv 3^2 \mod 32$$

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149