2021 AAMC 10 #19

Problem:

How many sequences of five nonzero digits are such that the product of any two consecutive digits is a perfect square?

$\mathbf{\text{(A) }}279\mathbf{\text{(B) }}280\mathbf{\text{(C) }}281\mathbf{\text{(D) }}282\mathbf{\text{(E) }}283$

Solution:

We perform casework on the first digit of the sequence.

If $a$ is a perfect square, meaning that it is one out of $1,4,9$, then the other $4$ elements in the sequence must also be perfect squares. So, there are $3^4$ possibilities for this case, but there are $3$ such cases, so there are $3\cdot 3^4=3^5$ possibilities. 

If $a=2$, then the other elements each must be one out of $2,8$, so there are $2^4$ possibilities. 

If $a$ is one out of $3,5,6,7$, the other elements must be $a$ itself, so there is only $1$ possibility for each of these $4$ cases, giving us $4$ possibilities. 

If $a=8$, then the other elements must be one out of $2,8$, so there are $2^4$ possibilities. 

Lastly, summing up all of these, we get that the total number of possibilities is $$3^5+2^4+4+2^4=\boxed{{\displaystyle \mathbf{\text{(A) }}279}}$$