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Problem

Finding the last two digits of a base with a large exponent: $237^{222222212202}$.

Method

  1. If the number ends on 1:
    • the the second last digit is the second last base digit times the last exponent digit
    • the last digit is 1
  2. If the number ends on odd:
    • rewrite it until it is a number that ends on one and the first rule can be used $((n^4)^{e/4})$
  3. If the number ends on an even number:
    • $..76^{n} = ..76$
    • $..24^{n} = $ ..76 for even $n$ and 24 for uneven $n$
    • $..2^{10} = ..24$

Solution

$$(7^4)^{55555553050}\cdot 7^2 = (2401)^{..0} \cdot 7^2 = ..01 \cdot 7^2 = 49$$

Question

  • Is this method correct / could it be improved?
  • Is there an easier method?
  • Can someone please take my example and demonstrate it with Euler and explain it?

Thank you so much!

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. I got $69$, using $37$, not $7$ $\endgroup$
    – J. W. Tanner
    Apr 2, 2020 at 21:38
  • $\begingroup$ are you familiar with congruence? $\endgroup$
    – Arnaldo
    Apr 2, 2020 at 21:39
  • $\begingroup$ @Arnaldo yeah, that is modulo calculation, right? $\endgroup$
    – Heinrich Jensen
    Apr 2, 2020 at 21:40
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    $\begingroup$ Right, so you're essentially asking for $237^{222222212202}\mod 100$ $\endgroup$
    – J. W. Tanner
    Apr 2, 2020 at 21:42
  • $\begingroup$ Are you familiar with Eulers formula? That $\phi(100) = 40$ and so $237^{40}\equiv 1\pmod {100}$? The last two digits will cycle in a pattern the pattern will repeat every $40$ times. But to know that, you have to know Euler's formula. If you don't just try powering $237$ until you get the pattern. $\endgroup$
    – fleablood
    Apr 2, 2020 at 22:05

2 Answers 2

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Hints to compute $237^{222222212202}\mod 100$:

If $a\equiv b \mod 100$ then $a^n\equiv b^n\bmod 100$.

$a^{20}\equiv1\mod 100$ if $a$ is relatively prime to $100$.

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  • $\begingroup$ Shouldn't that be $a^{40}$, since $\varphi(100)=40$? $\endgroup$
    – Luke Collins
    Apr 2, 2020 at 21:46
  • $\begingroup$ @LukeCollins: The Carmichael function of $100$ is $20$ $\endgroup$
    – J. W. Tanner
    Apr 2, 2020 at 21:47
  • $\begingroup$ @JW Ah, I see, thanks. So $\lambda(100) \neq \varphi(100)$ since $(\mathbb Z/100\mathbb Z)^\times$ is not cyclic. $\endgroup$
    – Luke Collins
    Apr 2, 2020 at 21:49
  • $\begingroup$ The only real question is what level do we assume the OP is at? $\endgroup$
    – fleablood
    Apr 2, 2020 at 22:06
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There is a very simple way to go about this, you basically want to evaluate $237^{222222212202}\bmod 100$. Now in any group $G$ with $n$ elements, we have $g^n = 1$ for all $g$. Thus if we look at your number as a member of the group $(\mathbb Z/100\mathbb Z)^\times$, we get that $$237^{222222212202}=(237^{100})^{2222222122}\cdot 237^2 = 1^{2222222122}\cdot 237^2 = 237^2,$$ thus you just need to look at the last two digits of $237^2$, which are $69$.

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