Waste less time on Facebook — follow Brilliant.
×

NT Problem

How many perfect squares can there be of the form N=\(\overline{abba}\) for some 4-digit positive integer N?

Note by Abhimanyu Swami
4 years ago

No vote yet
2 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Note that we can write \(N=\overline{abba}\) more mathematically rigorously as \(N=1000a+100b+10b+a=1001a+110b=11(91a+10b)\). Since \(N\) is a square, \(11|91a+10b\).

Therefore we want to find all \(a,b\) such that \(91a+10b\equiv 0\pmod{11}\implies 3a-b\equiv 0\pmod{11}\).

Now we can do casework on \(a\). Note that since the number of possible values of \(b\) is less than the number of possible numbers modulo \(11\), there must exist at most one unique solution to \(b\) for every \(a \), and there will not exist any solution for \(b\) when \(3a\equiv 10\pmod{11}\) which happens at \(a=7\).

We list out the solutions: \((a,b)=(1,3),(2,6),(3,9),(4,1),(5,4),(6,7),(8,2),(9,5)\). Plugging this back in gives \(1331,2662,3993,4114,5445,6776,8228,9559\). Factorizing gives that none of them are squares. Therefore there are \(\boxed{\textbf{no solutions}}\)

Daniel Liu - 4 years ago

Log in to reply

Can you please explain what is meant by your line '.................the number of possible values of ..............happens at 7) with explaining a bit about modular arithmetic because I know very less about the topic.

Krystal D'Souza - 4 years ago

Log in to reply

What I'm basically saying: we know that 3a-b has to be divisible by 11. Note that when we choose an "a", we might have some solutions for "b". However, notice that if we have a solution for "b", there can't be any more solutions because you can't have two single digit numbers both have the same remainder when divided by 11. Does that make sense?

Daniel Liu - 4 years ago

Log in to reply

@Daniel Liu Correct, but be careful: this doesn't occur ONLY because "the number of possible values of \(b\) is less than the number of possible numbers modulo \(11\)"; for example, if \(b\) could only be the numbers \(5, 1, 16,\) there are fewer possible values, but \(b\) can be \(5\) mod \(11\) in two ways. What you said occurs mostly because the possible digits, \[0, 1, 2, \ldots, 9,\] are consecutive, and there being only \(10\) of them also contributes to this fact.

Michael Tang - 4 years ago

Log in to reply

Checking my memory for a list of 4 digit perfect squares, there aren't any of that form... but let's just make sure :P

\(N = 1001a + 110b = 11(91a + 10b)\). So, \(11|91a + 10b \rightarrow 11|3a - b\), but we also have to consider the other factors of \(91a + 10b\), namely if they come in even pairs. I'll worry about later..

Since \(a, b < 10\) and \(a > 0\), the congruence \(3a - b \equiv 0 \pmod {11}\) has very countable solutions in \((1, 3), (2, 6), (3, 9), (4, 1), (5, 4), (6, 7), (8, 2), (9, 5)\). These can be checked to not be perfect squares.

Michael Tong - 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...