Problem 879

Every integer can be written in the form $10a + b$, where $a, b$ are integers and $0 \leq b \leq 9$. The twentieth power of this, according to the binomial theorem, is

$$(10a)^{20} + \binom{20}{1} \cdot (10a)^{19} \cdot b + \ldots + \binom{20}{19} (10a) \cdot b^{19} + b^{20}$$

Here, the first 19 terms are clearly divisible by 100, and even the twentieth term is: $\binom{20}{19} \cdot (10a) \cdot b^{19} = 200 \cdot a \cdot b^{19}$. Therefore, the last two digits of $(10a + b)^{20}$ are the same as the last two digits of $b^{20}$, and $0 \leq b \leq 9$. Let's calculate these!

Clearly, the last two digits of $0^{20}$ are 00, and $1^{20}$ ends in 01. Furthermore, $2^{20} = 1048576$, and its last two digits are 76. It is also evident that $5^{20}$ ends in 25, because $5^2 = 25$, and from this, by successively multiplying the last two digits by 5, we get the last two digits of higher powers of 5; but $5 \cdot 25 = 125$ also ends in 25, so $5^{20}$ does as well.

From the binomial theorem, $(10 - b)^{10} - b^{10} = 10^{10} - \binom{10}{1} \cdot 10^9 \cdot b + \binom{10}{2} \cdot 10^8 \cdot b^2 - \ldots - \binom{10}{9} \cdot 10 \cdot b^9$, and similarly to the above, this is divisible by 100. Therefore, $b^{10}$ ends in the same two digits as $(10 - b)^{10}$, and thus the last two digits of $b^{20}$ and $(10 - b)^{20}$ are the same. From this, it immediately follows that $9^{20}$ also ends in 01, and $8^{20}$ ends in 76. For $b = 3$, $3^{20} = 9^{10}$, and the last two digits of $9^{10}$ are the same as those of $1^{10}$, which means $3^{20}$ also ends in 01. Therefore, the same applies to $7^{20}$.

Let's consider the case $b = 4$. For this, $4^{20} = (2^{20})^2$, but the last two digits of $2^{20}$ are 76, so $4^{20}$ ends in the same two digits as $76^2$, and $76^2 = 5776$, so the last two digits of $4^{20}$ are 76, and the same applies to $6^{20}$.

In summary: the last two digits of the twentieth power of an integer can be $00, 01, 25, 76$.