Find the last two digits of $3^{375}$

Question

I tried writing out the first few powers of 3 to see if I could find any pattern in the last two digits:

$3^0=01$

$3^1=03$

$3^2=09$

$3^3=27$

$3^4=81$

$3^5=43$, etc.

However, I came up empty-handed, and I am unsure how to approach this.

Any ideas? Many thanks.

@DietrichBurde I am interested in many areas of maths, and do not feel I am at a level in calculus to start studying analysis at the moment... — Jamminermit, Feb 28 '20 at 19:10
Hint: expand $,3^{\large 375} = 3(-1+10)^{\large 187},$ via Binomial Theorem. Only first two terms survive $!\bmod 100\ \ \ $ — Bill Dubuque, Feb 28 '20 at 19:40

score 0 · Answer 1 · answered Feb 28 '20 at 19:18

0

Euler's Totient function $\varphi(x)$ counts the number of integers less than or equal to $x$ relatively prime to $x$.

One consequence (Euler's Theorem) is that $a^{\varphi(n)} \equiv 1 \mod{n}$ (more elaboration on Wikipedia).

You are looking for the remainder mod $100$ (last two digits), and $\varphi(100)=40$.

Can you proceed?

answered Feb 28 '20 at 19:18

Saketh Malyala

13,637

1

@DonThousand The multiplicative order of $3$ in the ring $\Bbb Z/100$ is $20$, since $3^5=243$, $43^2=1849$, $49^2=(50-1)^2=1$ modulo $100$. – dan_fulea Feb 28 '20 at 19:22
@dan_fulea Oops, you are right, I goofed. – Rushabh Mehta Feb 28 '20 at 19:23

score 0 · Accepted Answer · answered Feb 28 '20 at 19:22

0

Hint:

The Carmichael function is even better than Euler's totient function for this.

It says $3^{20}\equiv1\bmod100$, since $3$ and $100$ are relatively prime.

answered Feb 28 '20 at 19:22

J. W. Tanner

60,406

So just to make sure I get this right, the last two digits repeat in a cycle of 20, taking the remainder of 375/20 gives 15, so the last two digits are $3^{15}=07$? – Jamminermit Feb 28 '20 at 19:37
@Jamminermit: that is correct. In this case, Euler's totient function would give the same calculation, but for $3^{355}$, Carmichael would give $3^{15}$ whereas Euler would give $3^{35}$ – J. W. Tanner Feb 28 '20 at 19:49
I would check the math on that one. Euler's totient theorem would give $3^{375}\equiv3^{15}$ in this case. And although the Carmichael function can reduce the exponent further, it lacks a nice way to evaluate it like Euler's totient function. @Jamminermit – Simply Beautiful Art Mar 21 '20 at 21:19

score 0 · Answer 3 · answered Feb 28 '20 at 19:45

You are on the correct approach,but it will take you too long to know the cycle. Now you might be wondering why I said it's actually correct because it obvious you reach a cycle in atmost $\phi(100)$ tries as by Euler Totient function(this implies this process of continuation of multiplying 3 must terminate and you will reach a cycle). Consider equality $\mod(100)$ However since you have found that $3^{5}=43$ you can use The result $43^{5}=43$ the $3^{125}=43^{25}=43^{5}=43$ and then $3^{375}=43^{3}=7$. So the last digit is $07$

score 0 · Answer 4 · answered Feb 28 '20 at 19:46

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By Euler's theorem, it reduces to $3^{15}\pmod {100}$.

To finish it, $3^5\cong{43}\pmod{100}$. So it's $43^3\cong{49\cdot43}\cong2107\cong7\pmod{100}$.

answered Feb 28 '20 at 19:46

Find the last two digits of $3^{375}$

4 Answers4