, 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 $\text{2 ), (0 ), (6 ), (2 ), then the number of integers (j}$ with $\$1\leq j\leq$ 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.