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

Question

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 :)

Answer 1

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.

Answer 2

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)

Answer 3

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 $\, s = \color{#c00}{\pm 1}$

See also this tens digit logarithm law.