Multiple of 10

All numbers end with $7$ and have power of multiple of $4$ , so they will have last digits as $1$ . On subtracting them the last digits of all of them will be $1-1=0$ , so all are multiples of $10$ .

We know that if the power of any number that ends in 7 is divisible by $4$ , it ends in a $1$ . We can use this to our advantage.

• Let us look at C first. $4 \equiv 0 \; \; \pmod{4}$ , so the number ends in a $1$ . since we subtract $1$ from it, it must end in a $0$ and is therefore divisible by $10$ !
• Next, onto D. $500 \equiv 0 \; \; \pmod{4}$ and the same principle can be applied here from C, so it is divisible by $10$ !

Now that we have cleared up C and D, we can move onto A and B.

• For B, we know that $37^{500}$ and $37^4$ both end in a $1$ . Since the last digits will subtract to make $0$ , it is divisible by $10$ !
• Finally, for A, we must see if these both end in $1$ so that we can draw the same conclusion as B. $100 \equiv 0 \; \; \pmod{4}$ and $20 \equiv 0 \; \; \pmod{4}$ so they both end in a $1$ and by using the same reasoning as B, it is divisible by $10$ !

So overall, $\green{\boxed{\text{A, B, C and D}}}$ are all multiples of $10$ !

Use Euler's Theorem and you will find all the values are equivalent to 0 (mod 10).