Using the chinese remainder theorem to find the last two digits of $49^{19}$

Question

I am trying to find the last 2 digits of $\ 49^{19}$ but I am having some trouble. So far I know that

x = $\ 49^{19}$ mod 100
x = $\ 49^{19}$ mod 25
x = $\ 49^{19}$ mod 4

I can then apply the CRT and set u = 4 $\ *$ ( $\ 4^{-1}$) mod 25 and v = 25 $\ *$ $\ 25^{-1}$ mod 4. However I am having trouble finding x.

Duplicate of Specific steps in applying the Chinese Remainder Theorem to solve modular problem splitting modulus — Bill Dubuque, Aug 20 '19 at 21:35
You are making it too hard. $49\equiv 1 \pmod 4$ and $49 \equiv -1\pmod {25}$ so $x=49^{19}\equiv 1^{19}\equiv 1\equiv 49\pmod 4$ and $x=49^{19}\equiv (-1)^{19}\equiv -1 \equiv 49\pmod 4$. so $49^{19}\equiv 49\pmod{100}$.... But you don't need CRT. $49^{19}=\sum {19\choose k}50^{k}(-1)^{19-k}$. As $50^k\equiv 0\pmod{100}; k\ge 2$ we have $(50-1)^{19}\equiv 1950-1\equiv 49\pmod{100}$ — fleablood, Aug 21 '19 at 00:09

Mohammad Riazi-Kermani · Answer 1 · 2019-08-21T01:16:50.390

2

We really do not need Chinese Remainder theorem for this problem.

Note that $$49^2=2401 \equiv 1 \pmod{100}$$

$$ 49^{18} = (49^2) ^ {9} \equiv 1^9 =1 \pmod{100}$$

$$ 49^{19} \equiv 1\times 49 =49 \pmod{100}$$ However with Chinese Remainder Theorem notice that $$100=4\times 25$$ and we are looking for a solution to the system $$ x\equiv -1 \mod 25\\x\equiv 1 \mod 4$$

Since $$1(25)-6(4)=1$$ the solution to the system is $x=1(25)-(-6)(4)=49$

edited Aug 21 '19 at 01:16

answered Aug 20 '19 at 20:54

Mohammad Riazi-Kermani

69,891

1

You didn't use CRT (see title) and this is exactly the same answer you posted 2 days ago. – Bill Dubuque Aug 20 '19 at 21:29
@BillDubuque I have edited my answer thanks to your comment. – Mohammad Riazi-Kermani Aug 21 '19 at 01:21

Using the chinese remainder theorem to find the last two digits of $49^{19}$

1 Answers1