OPC 2 Problem 2

2.Prove that no number in this sequence 11,111,1111,11111,........... is a perfect square 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.

Note by Rajdeep Dhingra
4 years, 2 months ago

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1 vote

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

Vaibhav Prasad - 4 years, 2 months ago

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

Rajdeep Dhingra - 4 years, 2 months ago

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OK Allright :)

Vaibhav Prasad - 4 years, 2 months ago

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write

11n times=a2 \underbrace {1 \cdots 1} _ {n \textrm{ times}}= a^2

Of course aa must be odd (and greater than 11 since our sequence begins with 1111, in other words n2n \geq 2 ). Consider the even b=a1b = a -1 . We have

11n times=b22b+1 \underbrace {1 \cdots 1} _ {n \textrm{ times}}= b^2 -2b +1

11n1 times×10=b22b \underbrace {1 \cdots 1} _ {n-1 \textrm{ times}}\times 10= b^2 - 2b

bb is even so 44 divides the LHS, but clearly not th RHS. Contradiction.

Andrea Palma - 4 years, 2 months ago

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The numbers in the sequence can be written as 111....108+3111....108+3 now since the last two digits, i.e. 0808, is divisible by 44, the numbers are of the form of 4k+34k+3. Now note that a perfect square, say a2a^2 when written in modulo 44    a2=0,1,2(mod4)\implies a^2=0,1,2\pmod4 but not =3(mod4)=3\pmod4 which contradicts to the form of the numbers in the sequence.

Marc Vince Casimiro - 4 years, 2 months ago

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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

Kaustubh Miglani - 3 years, 5 months ago

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

Andrea Palma - 3 years, 5 months ago

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Yup, It can't be.

Rajdeep Dhingra - 3 years, 5 months ago

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absolutely.

Kaustubh Miglani - 3 years, 5 months ago

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