Question

The number of perfect square in the list 11, 111, 1111, ………. are

Answer 1

Hint: We know that integers can be written in the form of 2n if integer is even and odd integers can be written in form of 2n + 1, Now, x be any integer which is even, then it can be written in form of 2n.
For example, 8 = 2$\cdot $(4)
Squaring x, we get
\[{{x}^{2}}=\text{ }{{\left( \text{ }2n\text{ } \right)}^{2}}=\text{ }4{{n}^{2}}\] which is of form 4m, where \[m\text{ }=\text{ }{{n}^{2}}\]……. ( i )
Let, x be any integer which is odd, then it can be written in form of 2n + 1,
For example: 31 = $2\cdot (15)+1$ .
Squaring x, we get,
\[{{x}^{2}}=\text{ }{{\left( \text{ }2n\text{ }+\text{ }1\text{ } \right)}^{2}}=\text{ }4{{n}^{2}}+\text{ }4n\text{ }+\text{ }1\text{ }=\text{ }4\text{ }\left( \text{ }{{n}^{2}}+\text{ }n\text{ } \right)\text{ }+\text{ }1\] , which is of form 4m + 1, where \[m\text{ }=\text{ }{{n}^{2}}+\text{ }n\] ….. ( ii )
So, from ( i ) and ( ii ) we can say that,
Every perfect square is of the form 4m or 4m + 1.
Now, 11 = 4$\cdot $2 + ( 4 - 1 ) = 4$\cdot $3 - 1
111 = 4$\cdot $27 + ( 4 - 1) = 4$\cdot $28 - 1
1111 = 4$\cdot $277 + ( 4 – 1 ) = 4$\cdot $278 – 1
And so on.
We see that every term of the list is of form 4m – 1.
But, we proved above that perfect square if of the form 4m or 4m + 1.
Thus, no term in the list 11, 111, 1111, ……. Is a perfect square.
Hence, zero numbers of perfect squares are in list 11, 111, 1111, ……

Note: You can write terms of given list in other general form also. Whenever you contradict any condition you just disprove the universal condition for that statement. For every perfect square there exist unique m such that the perfect square is equal to 4m + 1.