$2. \text{Prove that no number in this sequence } 11,111,1111,11111,........... \\ \text{ is a perfect square}$

Give different methods of solving this. This was a problem to the OPC 2. The OPC 2 has already ended.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestDon't get me wrong, but I feel like most of the problems are from "an excursion in mathematics"

Log in to reply

Well I have heard a lot about that book. But these problems I have thought up and one is a famous one from IMO.

Log in to reply

OK Allright :)

Log in to reply

write

$\underbrace {1 \cdots 1} _ {n \textrm{ times}}= a^2$

Of course $a$ must be odd (and greater than $1$ since our sequence begins with $11$, in other words $n \geq 2$ ). Consider the even $b = a -1$ . We have

$\underbrace {1 \cdots 1} _ {n \textrm{ times}}= b^2 -2b +1$

$\underbrace {1 \cdots 1} _ {n-1 \textrm{ times}}\times 10= b^2 - 2b$

$b$ is even so $4$ divides the LHS, but clearly not th RHS. Contradiction.

Log in to reply

The numbers in the sequence can be written as $111....108+3$ now since the last two digits, i.e. $08$, is divisible by $4$, the numbers are of the form of $4k+3$. Now note that a perfect square, say $a^2$ when written in modulo $4$$\implies a^2=0,1,2\pmod4$ but not $=3\pmod4$ which contradicts to the form of the numbers in the sequence.

Log in to reply

any no will have last two digits 11 and therefore cannot be a perfect square .i hope u get it .if u dont just ask

Log in to reply

You mean any number with last 2 digits 11, 31, 51, 71, 91 cannot be a perfect square?

Log in to reply

Yup, It can't be.

Log in to reply

absolutely.

Log in to reply