A number theory problem by A Former Brilliant Member

Number Theory Level 3

Find the sum of all positive integers whose squares are of the form $\overline {ABABA}$ , a five digit number with $A$ and $B$ as digits.

The answer is 264.

3 solutions

Chris Lewis
May 21, 2019

There are only $90$ numbers of the form $n=\overline{ABABA}$ , so it's easy to check all of these with a few lines of code; however, we can narrow down the possibilities a bit, and actually solve this by hand.

The last digit of a square must be one of $0,1,4,5,6,9$ , so $A$ is one of these. We can eliminate $A=0$ as this would not give a five-digit number.

Now note that $n=10101A+1010B$ , and that $3$ , $7$ and $13$ are all factors of $10101$ .

$n \equiv 2B \pmod3$ . All squares are congruent to either $0$ or $1 \pmod3$ , so this means $B \in \left\{ 0,2,3,5,6,8,9 \right\}$ .

Likewise, $n \equiv 2B \pmod7$ . All squares are congruent to either $0$ , $1$ , $2$ or $4 \pmod7$ , so this means $B \in \left\{ 0,1,2,4,7,8,9 \right\}$ .

By looking at quadratic residues $\pmod{13}$ we find that $B \in \left\{ 0,1,3,4,9 \right\}$ .

Checking which numbers are in all of these sets, we find $B \in \left\{ 0,9 \right\}$ .

If the last digit of a square is $5$ , its penultimate digit must be $2$ ; but this isn't an option for $B$ , so we have $A \in \left\{ 1,4,6,9 \right\}$ .

This only leaves eight possibilities; we can reduce these even further by looking at each of them $\pmod8$ (the quadratic residues are $0,1,4$ ). This only leaves two viable options, $40404$ and $69696$ . But $40404=4\cdot10101$ and $10101 \equiv 5 \pmod8$ , so $40404$ is not a square.

Finally there is only one possibility, $69696$ ; checking this we find it is the square of $\boxed{264}$ .

Nice solution.

A Former Brilliant Member - 2 years ago

Kyle T
May 21, 2019

<?php
$t = 0;
for($a=1;$a<=1000;$a++){
    $a2 = pow($a,2);
    $arr = str_split($a2);
    if(strlen($a2)==5 && $arr[0]==$arr[2] && $arr[0]==$arr[4] && $arr[1]==$arr[3]){
        echo $a.' '.$a2.'<br>';
        $t+=$a;
    }
}
echo 'answer: '.$t;exit;
?>
/* 
prints: 
264 69696
answer: 264
*/

A Former Brilliant Member
May 21, 2019

It turns out that A can be 0 without that affecting the result as the only square with A being 0 is 0.

0, 1010, 2020, 3030, 4040, 5050, 6060, 7070, 8080, 9090, 10101, 11111, 12121, 13131, 14141, 15151, 16161, 17171, 18181, 19191, 20202, 21212, 22222, 23232, 24242, 25252, 26262, 27272, 28282, 29292, 30303, 31313, 32323, 33333, 34343, 35353, 36363, 37373, 38383, 39393, 40404, 41414, 42424, 43434, 44444, 45454, 46464, 47474, 48484, 49494, 50505, 51515, 52525, 53535, 54545, 55555, 56565, 57575, 58585, 59595, 60606, 61616, 62626, 63636, 64646, 65656, 66666, 67676, 68686, 69696, 70707, 71717, 72727, 73737, 74747, 75757, 76767, 77777, 78787, 79797, 80808, 81818, 82828, 83838, 84848, 85858, 86868, 87878, 88888, 89898, 90909, 91919, 92929, 93939, 94949, 95959, 96969, 97979, 98989 and 99999 are the five digit numbers in the prescribed format with A being 0. The only squares in that list are 0 and 69696.

Moderator: The question posed by Alak B does not fit the answer.The answer is confusing and unhelpful.

E Koh - 2 years ago

I beg to differ. The only unclarity is whether A could be 0. The language is such that A probably may not be allowed to 0. It turns out that even if A were permitted to be 0, the same answer results. The problem clearly requests the sum of positive integers whose squares are of the required $\overline{ABABA}$ form.

Your complaint is more properly done using a report.

A Former Brilliant Member - 2 years ago

Randolph Herber: Are you a moderator?

E Koh - 2 years ago

No, I am not a moderator.

To submit a report, in the lower right corner of the problem is "... More". Click that. A pop-up that says "View reports" should come up. Click that. At that point, you can enter a report.

A Former Brilliant Member - 2 years ago

A number theory problem by A Former Brilliant Member

The answer is 264.

3 solutions

0 pending reports