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Find the last two digits of $2019^{2019}$

I know that you can typically find the last two digits of a number to any power by reducing the number to end with a one and so on (I will show an example of what I am talking about below).

However, $2019$ cannot be reduced such that I will get an even exponent required in this strategy of solving.

So, how do I figure out the last two digits of this equation?

* Example of the method I referred to *

Find the last two digits of $41^{2789}$

  • Multiply the tens digit of the number (4 here) with the last digit of the exponent (9 here) to get the tens digit. The unit digit will always equal to one.

  • $61 (4 × 9 = 36)$. Therefore, 6 will be the tens digit and one will be the unit digit

Keep in mind, I am an algebra 2 student, but my teacher, also a calc teacher, thought I might be able to figure this one out :)

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    $\begingroup$ Hint: What are the last two digits of 19$^{10}$ ? $\endgroup$ – J. W. Tanner Jan 23 at 0:37
  • $\begingroup$ Did you see congruences? $\endgroup$ – Bernard Jan 23 at 0:40
  • $\begingroup$ @Bernard I haven't learned about them, and from what I have been looking into, it seems like I should teach myself them in order to figure this out $\endgroup$ – Levi Kline Jan 23 at 0:41
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    $\begingroup$ Would you agree that the last two digits of $2019^n$ are the same as those of $19^n$? $\endgroup$ – J. W. Tanner Jan 23 at 0:44
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    $\begingroup$ Possible duplicate of Finding the last two digits of a number by binomial theorem $\endgroup$ – Bill Dubuque Jan 23 at 0:55
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Calculation shows that the last two digits of $19^5$ are $99$.

Therefore, the last two digits of $19^{10}$ are $01$.

Therefore, the last two digits of $19^{10n}$ are $01$ for $n \in \mathbb N$.

Therefore, the last two digits of $19^{10n+9}$ are same as those of $19^9$, which calculation shows to be 79 (same as those of $19^4 19^5$ or $19^4 99$).

Therefore, the last two digits of $2019 ^ {2019}$ are the same as the answer submitted earlier.

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Hint:

$$2019^{2019}=-(1-2020)^{2019}$$

$$=-(1-\binom{2019}12020+\text{terms divisible by}10^2)$$

$=2019\cdot2020-1+100u$(say)

$$=(2000+19)(2000+20)-1+100u$$

$=19\cdot20-1+100v$(say)

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Both rules work by applying the Binomial Theorem to $\,(\color{#c00}{\pm1} + 10a)^n$. The rule you mention is the first case below $(\color{#c00}{+1}).\,$ Your example needs the $(\color{#c00}{-1})$ case derived below it - with sign tweaks.

$ \begin{align}(\color{#c00}{+1}+10t+100j)^{\Large u+10j} &= 1 + (u+10j)(10t+100j) + 100(\cdots)\\ &= 1 + 10tu + 100(\cdots) \end{align}$

$\begin{align}(\color{#c00}{-1}+10t+100j)^{\Large u+10j} &= s - s(u+10j)(10t+100j) + 100(\cdots),\ \ \ s = (\color{#c00}{-1})^{\large u}\\ &= s - s 10tu + 100(\cdots) \end{align}$

$\,(-1+20\color{#0a0}20)^{\large 201\color{#a0f}9}$ has $t=\color{#0a0}2, u = \color{#a0f}9\,$ so $\,s = (-1)^u = -1\,$ therefore

$$ s - s10\color{#0a0}t\color{#a0f}u\, =\, -1 + 10(\color{#0a0}2)\color{#a0f}9\, =\, 179 $$

Therefore the last two digits are $79$.

Remark $ $ Unifying both cases yields $\ \bbox[5px,border:1px solid red]{s^{\large u} (1 + 10 stu)}\ $ where $\, s = \color{#c00}{\pm 1}$

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  • $\begingroup$ I really like this solution. I don't quite understand it right now but I will definitely review the theorem you referred to and will try working it out a bit more. Thanks for the help! $\endgroup$ – Levi Kline Jan 23 at 4:47

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