1980 AHSME Problems/Problem 30

Problem

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?

$\text{(A)} \ 0 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 8 \qquad \text{(E)} \ 9$

Solution

If $N$ is a squarish number, then $N = 10000a^2 + 100b^2+ c^2$ , where $a$ , $b$ , and $c$ are all integers between $4$ and $9$ inclusive. The only way that $N$ could be a perfect square would be if $N = (100a + c)^2$ . Expansion yields $N = 10000a^2 + 200ac + c^2$ , so $2ac = b^2$ , and therefore $b$ must be even.

If $b = 4$ , then $ac = 8$ , which is impossible.

If $b = 6$ , then $ac = 18$ , which is impossible.

If $b = 8$ , then $ac = 32$ , which is possible if $(a,c)$ is either $(8,4)$ or $(4,8)$ . This yields squarish numbers $641616 = 804^2$ and $161664 = 408^2$ .

Therefore, there are $2$ squarish numbers. $\fbox{(B)}$

Wwei.yu (talk)Wei

-Edited by j314andrews

1980 AHSME (Problems • Answer Key • Resources)
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1980 AHSME Problems/Problem 30

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