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

  • 3
    $\begingroup$ What have you tried? $\endgroup$ – Don Thousand Nov 10 '19 at 5:03
  • 1
    $\begingroup$ I think there are too many permutation, so not getting a efficent way to solve this. Let's say no. is abcdef then ab,cd,ef = {x^2 where x:4 to 9} $\endgroup$ – Anushi Maheshwari Nov 10 '19 at 5:07

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.

  • $\begingroup$ Thanks @Jyrki Lahtonen for your answer and valuable comments :) $\endgroup$ – Anushi Maheshwari Nov 10 '19 at 12:18

The $2$ digit square numbers are $16,25,36,49,64,81$. Note that square numbers can't be congruent to $2\mod3$. Note that $16,25,49,64\equiv1\mod3$, while the rest are divisible by $3$.

Since a $6$ digit square must also be congruent to $0,1\mod 3$, and the sum of the digits of a number is congruent to the number modulo $9$, we have two possibilities for our $6$ digit prime.

  1. It is composed of numbers from $\{16,25,49,64\}$

  2. It is composed of numbers from $\{36,81\}$.

Next, we note that squares are congruent to $0,1\mod8$. Hence, so must the last three digits of the number.

Let's analyze case 2 first. Note that if the square ended in $81$, the middle square must be $36$, since $8\not\mid 180$. If the square ended in $36$, the middle square must be $81$ for similar reasons.

So for case $2$, we have $4$ remaining possibilities. Any square in front, followed by either $3681$ or $8136$.

Now we divide each of these possibilities by $9$ to get $90409,40409,40904,90904$. We can remove two of these cases as they are congruent to $2\mod3$. We are left with $90409$ and $90904$. We can remove $90904$ since it is not divisible by $8$. $90409$ is close to $90000$ which is $300^2$. $301^2=90000+600+1>90409$, showing that it too, isn't a square.

So, there were no successful results from case $2$. So, we solely need to consider squares in $16, 25, 49, 64$. Note that the middle square must have an even ones digit to ensure the six digit square is congruent to $0,1\mod8$. Since $10101$ is clearly not a square (too close to $10000$), there are no six-digit squares with any $2$ digit square repeated thrice.

You can finish the casework from here.

  • $\begingroup$ Why are you excluding for example the case of a single $64$ accompanied by two of $36,81$? That leaves a number congruent to $1$ modulo $9$, and hence a possible square. $\endgroup$ – Jyrki Lahtonen Nov 10 '19 at 5:41
  • $\begingroup$ What I'm saying that for example 811681 passes all your tests. It is $\equiv1\pmod3$, and $681\equiv1\pmod8$. $\endgroup$ – Jyrki Lahtonen Nov 10 '19 at 5:54
  • $\begingroup$ @JyrkiLahtonen I don't know how I missed that. I clearly need some coffee. Will fix in a bit, $\endgroup$ – Don Thousand Nov 10 '19 at 6:01
  • $\begingroup$ Thanks, @DonThousand for your answer :) $\endgroup$ – Anushi Maheshwari Nov 10 '19 at 12:17

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