, 2022

We start by looking for patterns in the rightmost two digits of
powers of 4, powers of 5 and powers of 7.

The first few powers of 5 are $$5^1 = 5 \qquad 5^2 = \mathbf{25} \qquad 5^3 = 1\mathbf{25} \qquad 5^4 = 6\mathbf{25} \qquad 5^5 = 31\mathbf{25}$$ It appears that,
starting with $5^2$, the rightmost
two digits of powers of 5 are always 25.

To see this, we want to understand why if the rightmost two digits of a
power of 5 are 25, then the rightmost two digits of the next power of 5
are also 25.

The rightmost two digits of a power of 5 are completely determined by
the rightmost two digits of the previous power, since in the process of
multiplication, any digits before the rightmost two digits do not affect
the rightmost two digits of the product.

This means that the rightmost two digits of every power of 5 starting
with $5^2$ are 25, which means that
the rightmost two digits of $5^{129}$
are 25.

The first few powers of 4 are $$4^1 = 4 \qquad 4^2 = \mathbf{16} \qquad 4^3 = \mathbf{64} \qquad 4^4 = 2\mathbf{56} \qquad 4^5 = 10\mathbf{24} \qquad 4^6 = 40\mathbf{96} \qquad 4^7 = 16\,3\mathbf{84}$$ $$4^8 = 65\,5\mathbf{36} \qquad 4^9 = 262\,1\mathbf{44} \qquad 4^{10} = 1\,048\,5\mathbf{76} \qquad 4^{11} = 4\,194\,3\mathbf{04} \qquad 4^{12} = 16\,777\,2\mathbf{16}$$ We note that the rightmost two digits
repeat after 10 powers of 4. This means that the rightmost two digits of
powers of 4 repeat in a cycle of length 10.

Since 120 is a multiple of 10 and 127 is 7 more than a multiple of 10,
the rightmost two digits of $4^{127}$
are the same as the rightmost two digits of $4^7$, which are 84.

The first few powers of 7 are $$7^1 = 7 \qquad 7^2 = \mathbf{49} \qquad 7^3 = 3\mathbf{43} \qquad 7^4 = 24\mathbf{01} \qquad 7^5 = 16\,8\mathbf{07} \qquad 7^6 = 117\,6\mathbf{49}$$ We note that the rightmost two digits repeat
after 4 powers of 7. This means that the rightmost two digits of powers
of 7 repeat in a cycle of length 4.

Since 128 is a multiple of 4 and 131 is 3 more than a multiple of 4, the
rightmost two digits of $7^{131}$ are
the same as the rightmost two digits of $7^3$, which are 43.

Therefore, the rightmost two digits of $4^{127} + 5^{129} + 7^{131}$ are the
rightmost two digits of the sum $84 + 25 + 43
= 152$, or 52. (This is because when we add integers with more
than two digits, any digits to the left of the rightmost two digits do
not affect the rightmost two digits of the sum.)