How to find mod for $19^{33} \pmod{20413}$?

Question

Struggling to understand how to answer this question as the formula previously used required n to be a multiple of the previous number, which as 33 is involved this method is no longer effective. From the mark scheme the answer to the question should be 6338.

Appreciate the help

You should start with the question How do I compute $a^b \pmod c$ by hand? to see whether the answers there address this question. — Carl Mummert, May 01 '19 at 15:22
If you want to do stuff like this with paper and pencil, exponentiation by squaring is the way to go. If the exponent were bigger than the modulus then there would be well known short cuts. Here you can also use the factorization $20413=137\cdot 149$ and then use the Chinese remainder theorem. Finding that factorization OTOH... — Jyrki Lahtonen, May 01 '19 at 15:22
Another possibility is to note $20413 = 137 \times 149$ where both factors are prime. Now compute for both primes and use the Chinese Remainder Theorem? — gt6989b, May 01 '19 at 15:23
Your comments on your prior question reveal that you are using this in cryptography (RSA) so there is no special simplifying structure at hand, i.e. you need to resort to brute-force modular exponentiation by repeated squaring as described in the linked dupes. You might find helpful the little-known viewpoint of the method via a binary Horner polynomial, as I describe here. — Bill Dubuque, May 01 '19 at 17:32

score 2 · Answer 1 · edited May 01 '19 at 16:33

We factor $20413=137\cdot 149.$ Compute $19^{33}\bmod {137}$ and $19^{33}\bmod{149}$ then use Chinese Remainder Theorem to find $19^{33}\bmod{137\cdot 149}.$

We can find $19^{33}\bmod {137}$ by the method of repeated squaring.$$\begin{align}19^2&=361&\equiv -50\pmod{137}\\ 19^4&\equiv (-50)^2 = 2500 &\equiv 34\pmod{137}\\ 19^8&\equiv 34^2 = 1156 &\equiv 60\pmod{137}\\ 19^{16}&\equiv 60^2=3600 &\equiv 38\pmod{137}\\ 19^{32}&\equiv 38^2=1444&\equiv 74\pmod{137}\\ 19^{33}&=19\cdot 19^{32}\equiv 19\cdot 74=1406&\equiv 36\pmod{137} \end{align}$$

Similarly, you can compute $19^{33}\bmod 149=80.$ Then solve the equation:

$$\begin{align}x&\equiv 36\pmod{137}\\x&\equiv 80\pmod{149}\end{align}$$

using Chinese Remainder Theorem, and this will give you $x\equiv 6338\pmod{137\cdot 149}.$

Assuming you could not reasonably factor the modulus, $M=20413,$ you could have just done one case of repeated squaring, modulo $M.$ The squaring will be harder, but you'll only have to do it once, and you do not have to do Chinese Remainder Theorem, nor factor $M.$

score 1 · Answer 2 · edited May 01 '19 at 17:23

1

Square and multiply by $19$ the result given in your previous question on $\,19^{\large 16}\bmod 20413$.

edited May 01 '19 at 17:23

Bill Dubuque

272,048

answered May 01 '19 at 16:25

Not an answer, but a good remark: the author keeps asking almost the same question, without attempting to answer or even to understand the answers given. – Jean-Claude Arbaut May 01 '19 at 17:12
I think the hard part for most new to a topic, is relating it to a previous known topic. That and the multitude of methods used in modular arirhmetic get even me confused at times. – May 01 '19 at 17:39

score 0 · Answer 3 · answered May 01 '19 at 15:44

0

Using my 9-digit handheld calculator and following this answer, $$19^2=361$$ $$19^4=361^2=130321\equiv 7843 \pmod {20413}$$ $$19^8\equiv 7843^2 \equiv 8280 \pmod {20413}$$ $$19^{16}\equiv 8280^2\equiv 11546 \pmod {20413}$$ $$19^{32}\equiv11546^2\equiv13226\pmod{20413}$$ Therefore, $19^{33}\equiv 19\times13226 \equiv 251294\equiv6338\pmod {20413}. $

answered May 01 '19 at 15:44

J. W. Tanner

60,406

Thanks for the help once again. Now fully understand how to tackle this type of question. – Radar May 01 '19 at 15:51

How to find mod for $19^{33} \pmod{20413}$?

3 Answers3