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$\dfrac{33^{100}}{50}$

I have done

$\dfrac{2×{33^{100}}}{100}$

But it is still complex. How can we calculate the remainder of a number divided by 100?

user596843
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    See also : https://math.stackexchange.com/questions/742341/find-the-last-two-digits-of-the-number-999 , https://math.stackexchange.com/questions/390685/the-last-2-digits-of-7777/390689 , https://math.stackexchange.com/questions/1337375/what-are-the-last-two-digits-of-7717 https://math.stackexchange.com/questions/390685/the-last-2-digits-of-7777 https://math.stackexchange.com/questions/1844558/how-to-find-last-two-digits-of-22016 https://math.stackexchange.com/questions/1385396/last-two-digits-of-145532 – lab bhattacharjee Sep 26 '18 at 05:27
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    https://math.stackexchange.com/questions/1724246/find-last-two-digits-of-33100 https://math.stackexchange.com/questions/2493141/solving-for-the-last-two-digits-of-a-large-number-31000/2497007 – lab bhattacharjee Sep 26 '18 at 05:29

3 Answers3

2

The fundamental idea you're overlooking is:

Let $m,n,d$ be positive integers. You have an equation

$$ m \equiv n \bmod d $$

if and only if the following two quantities are the same:

  • The remainder when dividing by $m$ by $d$
  • The remainder when dividing by $n$ by $d$

So, you can turn your "find a remainder" problem into a "do modular arithmetic" problem which leads to an easier "find a remainder" problem.

(incidentally, the theorem above works for all integers when stated appropriately — but it can be tricky to find the right statement because people don't all agree on what "remainder" means when negative numbers are involved)

1

$$33^2=(30+3)(30+3)\equiv9+2\cdot30\cdot3\equiv190-1$$

$$33^{4m}=(33^2)^{2m}\equiv(190-1)^{2m}\equiv(1-190)^{2m}\equiv1-\binom{2m}1190\pmod{100}$$

Here $4m=100$

lab bhattacharjee
  • 274,582
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Use Carmichael Function,,

as $(33,50)=1,$ and $\lambda(50)=20$

$$33^{20}\equiv1\pmod{50}$$

$$33^{20m}=(33^{20})^m\equiv1^m\pmod{50}$$

Here $20m=100\iff m=?$

lab bhattacharjee
  • 274,582