1
$\begingroup$

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

$\endgroup$
2
  • 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
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks @Jyrki Lahtonen for your answer and valuable comments :) $\endgroup$ – Anushi Maheshwari Nov 10 '19 at 12:18
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.