71. What are the last two digits of the number

$$
n^{a}+(n+1)^{a}+(n+2)^{a}+\ldots+(n+99)^{a}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-018.jpg?height=46&width=972&top_left_y=892&top_left_x=89)
a) $a=4$
b) $a=8$ ?
71. It is clear that the last two digits of the numbers $n, n^{2}, n^{3}, \ldots$, where $n$ is a non-negative integer, depend only on the last two digits of the number $n$ - this follows from the rule of multiplying multi-digit numbers "in a column". On the other hand, the last digits of 100 consecutive non-negative integers necessarily run through the values $00,01,02, \ldots, 99 \cdot$ (although, generally speaking, not in this order); therefore, our problem reduces to finding the last two digits of the sum (for $a=4$ and $a=8$)

$$
N_{a}=0^{a}+1^{a}+2^{a}+\ldots+99^{a}
$$

a) If $n=10 x+y$ is a two-digit number, then

$$
n^{4}=(10 x+y)^{4}=10^{4} x^{4}+4 \cdot 10^{3} x^{3} y+6 \cdot 10^{2} x^{2} y^{2}+4 \cdot 10 x y^{3}+y^{4}
$$

where only the last two terms affect the last two digits of the number $n^{4}$, since all other terms end with two zeros. Therefore, we need to find only the last two digits of the sums

$$
\sum_{x} \sum_{y} 4 \cdot 10 x y^{3} \quad \text { and } \sum_{x} \sum_{y} x^{0} y^{4} \quad\left(=10 \sum_{y} y^{4}\right)
$$

where $x$ and $y$ independently take values from 0 to 9 and it is assumed that $0^{0}=1$ (so $\left.\sum_{x} x^{0}=0^{0}+1^{0}+\ldots+9^{0}=10\right)$.

Notice now that for a fixed $y$

$\sum_{x} 4 \cdot 10 x y^{3}=4 \cdot 10(0+1+2+3+4+5+6+7+8+9) \cdot y^{3}=1800 y^{3}$

ends with two zeros; therefore, $\sum_{x} \sum_{y} 4 \cdot 10 x y^{3}$ also ends with two zeros - and does not contribute to the last two digits of the number $N_{4}$. Thus, we are left to find only the last two digits of the number

$10 \sum_{y} y^{4}=10\left(0^{4}+1^{4}+2^{4}+3^{4}+4^{4}+5^{4}+6^{4}+7^{4}+8^{4}+9^{4}\right)=10 S_{4}$, or the last digit of the sum $S_{4}$.

But this task is easy to solve: in the following table, the last digits of the numbers $y, y^{2}$ and $y^{4}$ are sequentially listed:

$$
\begin{array}{c|cccccccccc}
y & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
y^{2} & 0 & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \\
\boldsymbol{v}^{4} & 0 & 1 & 6 & 1 & 6 & 5 & 6 & 1 & 6 & 1
\end{array}
$$

Therefore, the last digit of the sum $S_{4}$ coincides with the last digit of the sum

$$
0+1+6+1+\ldots+1=4(1+6)+5=33
$$

which means that the last two digits of the sum $N_{4}$ (and the number in the problem statement) are 30.

6) In exactly the same way as in part a), we find that for $a=8$ the last two digits of the sum $N_{8}$ (and thus the number of interest to us) coincide with the last two digits of the number $10 S_{8}$, where $S_{8}=0^{8}+1^{8}+\ldots+9^{8}$. But from the written table of the last digits of the numbers $y^{4}$, and the equality $y^{4 k}=\left(y^{4}\right)^{k}$, it follows that the number $y^{4 k}$ (where $y$ is a digit) ends with the same digit as $y^{4}$. Therefore, for $a=8$ the last two digits of the number in question will be the same as for $a=4$, i.e., they will be the digits 30.

Note. It is not difficult to see that we will get the same result for all multiples of 4 values of $a$, i.e., for $a=4,8,12,16, \ldots$