Q:

1980 AHSME Problems/Problem 30

A six digit number (base 10) is squarish if it satisfies the following conditions:
(i) none of its digits are zero;
(ii) it is a perfect square; and
(iii) the first of two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two-digit numbers.
How many squarish numbers are there?
$\text{(A)} \ 0 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 8 \qquad \text{(E)} \ 9$

I cannot access soution for this problem. Can someone please explain how it can be done.
Link - https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_30

A:

Assume that a squarish number is $(100a+10b+c)^2$ with $a,b,c\in\{0,1,\ldots,9\}$.
Because $$10^4a^2\le (100a+10b+c)^2=10^4 a^2+2000ab+100b^2+200ac+20bc+c^2<10^4(a+1)^2$$ we see that:

For the top two digits to form a square, they must be $a^2$ specifically. Because zero is not allowed, we can deduce that $a\in[4,9]$.
Carry from $2000ab$ into the top two digits will ruin this whenever $ab\ge5$. Given that $a\ge4$ this forces $b=0$. Initially we still have $a=4, b=1$ as a possibility. As $\sqrt{17}=4.123\ldots$ we are to treat the cases $c\in\{1,2\}$.
But then $20bc=20c$ will change the two lowest digits ruining this possibility. 
Thus $b=0$ and the squarish number is
$$10^4a^2+100\cdot(2ac)+c^2.$$ Furthermore, also $c\in[4,9]$ for otherwise the next to last digit would be a zero.

We need $2ac$ to be a two digit square. It is relatively easy to see that the only possibilities are $\{a,c\}=\{4,8\}$ one way or the other. I did it by looking at the prime factors of $a$ and $c$ respectively, keeping in mind that one of them must be even. The pair $8,9$ would also make $2ac$ a square, but then $2ac>100$). The pairs $\{3,6\}$,$\{2,4\}$,$\{1,2\}$ and $\{1,8\}$ would also make $2ac$ a square, but earlier we saw that both $a$ and $c$ must be at least $4$, so these, too, must be excluded. 
This leaves
$$408^2=166464\quad\text{and}\quad804^2=646416$$
as the only squarish numbers.

