I am trying to find the last two digits of 3745
I know I need to solve 3745 = x(mod 100), but I'm not sure how to go about doing that. I know it has something to do with Euler's
Any help is greatly appreciated
Asked
Active
Viewed 219 times
0
-
What is $\phi(100)$? – Henno Brandsma Apr 22 '20 at 18:10
-
1Do you know the Euler's theorem? Do you see $a=37,n=100$? $$ a^{\varphi(n)} \equiv 1 \quad(\bmod n) $$ Do you know how to find $\varphi(100)=40$ ? Do you know that: $$ a^{x+ y}=a^{x}\cdot a^{y} $$ This is all you need.. $$ 37^{45}=37^{40}37^5\equiv1\cdot37^5\pmod{100} $$ Where $37^5=69343957$ so the last two digits are $57$. – Vepir Apr 22 '20 at 18:27
-
3Does this answer your question? Last Two Digits Problem – Sander Christopoulos Apr 22 '20 at 18:50
-
@Vepir This makes the most sense to me. Thank you. – jmsCoder Apr 22 '20 at 19:01
2 Answers
0
$$100=4\cdot 25$$
$$37^{45}\equiv 1 \pmod 4$$
$$37^{45}\equiv 12^{45} \pmod {25}$$
$$12^{45}\equiv (6)^{22} \cdot 12 \equiv 11^{11}\cdot 12 \equiv (-4)^5\cdot 11\cdot 12 \equiv -1024\cdot 11\cdot 12 \pmod {25}$$
Finally we get
$$37^{45}\equiv 7 \pmod {25}$$
Now we can quickly apply the chinese remainder theorem to solve the linear congruences
$$x\equiv 1\pmod 4$$ $$x\equiv 7\pmod {25}$$
And finally,
$$37^{45}\equiv 57\pmod {100}$$
h-squared
- 1,333
- 6
- 8
-
-
I first used negative numbers to evaluate $11^{11} \pmod {25}$. It is easy to see that $11^2=121\equiv 21 \pmod {25}. Now since $21$ is large, it is convineient to use $-4$ as a negative remainder means $4$ less than a multilple of $5$. you could have used $21$ or you could have breaken up $11^{11}$ in a different way, maybe using cubes. – h-squared Apr 23 '20 at 02:46
0
Essentially we need $37^{45}\pmod{100}$
$$(40-3)^2\equiv3^2-2(3)40\equiv70-1\pmod{100}$$
$$37^{2n+1}=37(37^2)^n$$
$$\equiv37(70-1)^n$$
$$\equiv37(-1)^n(1-\binom n170^1)\pmod{100}$$
Here $n=22$
So, $$\binom n170^1=22\cdot70\equiv40\pmod{100}$$
lab bhattacharjee
- 274,582
-
I think you need to check you first line, where you deduced that $37^2 \equiv 9 \pmod {100}$ – h-squared Apr 22 '20 at 18:27
-