, 2026

Since we are given that the cycle length for the units digits of
powers of $3$ is equal to $4$, and $43=4\times10+3$, the units digit of $3^{43}$ is equal to the units digit of $3^3$, which is $7$. Each integer multiple of $10$ has units digit $0$, and so we begin by determining the repeating sequence of units digits for powers of $4$ and $8$. $4^1=4, 4^2=16, 4^3=64,
$\dots$$$8^1=8, 8^2=64, 8^3=512, 8^4=4096,
8^5=32\,768, $\dots$$For powers of$4, the consecutive units digits that repeat are $4$, $6$, with cycle length

2$. For powers of$8, the consecutive units digits that repeat are $8$, $4$, $2$, $6$, with cycle length

4. We can determine the units digit of

4^j+8^j by adding the corresponding units digits of the individual powers of

4$and$8, and then taking the units digit of that sum. Thus, the units digits of

4^j+8^j are the units digits of

4+8=12$,$6+4=10$,$4+2=6$,$6+6=12, which are $2$, $0$, $6$, $2$, and this sequence continues to repeat with cycle length

4$. So,$4^j+8^j$is a multiple of$10$(has units digit$0$) once every$4$consecutive values of$j$. Since$$2026=4\times$ 506 +2$, and$0 is the second digit in the repeating sequence

2\), $0$, $6$, $2$, then the number of integers $\text{j with 1 j 2026 for which 4 j+8 j is a multiple of 10 is 506+1=507 . For 2 k , the consecutive units digits that repeat are (2}$, $4$, $8$, $6$, with cycle length $4$. For $3^k$, the consecutive units digits that repeat are $\text{3 ), (9 ), (7 ), (1 ), with cycle length (4}$. Similar to how we worked with $4^j+8^j$ in part (b), the consecutive units digits of $2^k+3^k$ that repeat are $\text{5 ), (3 ), (5 ), (7 ), with cycle length (4}$. For $8^k$, the consecutive units digits that repeat are $\text{8 ), (4 ), (2 ), (6 ), with cycle length (4}$. When $k$ is even, that is when $k=2m$ for positive integers $m$, $2026k=2026\times2m=4052m$. Since $4052=4\times1013$, then $4052m$ is a multiple of $4$, and so $2026k$ is a multiple of $4$ for all even integers $k$. Thus for all even integers $k$, the units digit of $8^{2026k}$ is $6$ (the fourth units digit in the repeating sequence $8$, $4$, $2$, $6$). When $k$ is odd, that is, when $k=2m+1$ for positive integers $m$, we get the following equivalent equations $$\begin{align*} 2026k&=2026\times(2m+1)\\ &=4052m+2026\\ &=4052m+2024+2\\ &=4\times1013m+4\times506+2\\ &=4(1013m+506)+2\end{align*}$$ So, $2026k$ is $2$ more than a multiple of $4$ for all odd integers $k$. Thus for all odd integers $k$, the units digit of $8^{2026k}$ is $4$ (the second units digit in the repeating sequence $8$, $4$, $2$, $6$). For $9^k$, the consecutive units digits that repeat are $9$, $1$ with cycle length $2$. Thus when $k$ is odd, the units digit of $9^k$ is $9$, and when $k$ is even, the units digit is $1$. For all integers $k$, $2026k$ is an even integer, and so the units digit of $9^{2026k}$ is $1$ for all integers $k$. Summarizing, we determined that the units digit of $8^{2026k}$ is $4$ when $k$ is odd, and is $6$ when $k$ is even. Also, the units digit of $9^{2026k}$ is $1$ for all integers $k$. Therefore, $8^{2026k}+9^{2026k}$ has consecutive units digits $4+1=5$ and $6+1=7$ that repeat with cycle length $2$. Recall that $2^k+3^k$ has consecutive units digits $5$, $3$, $5$, $7$ that repeat with cycle length $4$. Thus, $2^k+3^k$ and $8^{2026k}+9^{2026k}$ have the same units digit, $5$, for all odd values of $k$. Also, $2^k+3^k$ and $8^{2026k}+9^{2026k}$ have the same units digit, $7$, for all values of $k$ equal to a multiple of $4$. For $1\leq k \leq 50$, there are $25$ odd values of $k$ and $12$ values of $k$ equal to a multiple of $4$ (since $50=4\times12+2$), and so there are $25+12=37$ integers $k$ for which $2^k+3^k$ and $8^{2026k}+9^{2026k}$ have the same units
digit.