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perfect squares

by harsh.champ » Sun Feb 07, 2010 7:23 am
Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?


(A)3
(B)2
(C)4
(D)0
(E)1

The ans. [spoiler]is (E).[/spoiler]

Solving approach needed !!!
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by shashank.ism » Sun Feb 07, 2010 7:23 am
harsh.champ wrote:Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?


(A)3
(B)2
(C)4
(D)0
(E)1

The ans. [spoiler]is (E).[/spoiler]

Solving approach needed !!!
The ans. [spoiler]is (E).[/spoiler] The number is 7744

If we let the four-digit number be XXYY, then this number can be expressed as:
1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k^2 (since it's a perfect square)
In order for this to be true, 100X + Y must be the product of 11 and a perfect square, and looks like X0Y. So now our question is "which product of 11 and a perfect square looks like X0Y?" We can test them:
11 x 16 = 176; 11 x 25 = 275; 11 x 36 = 396; 11 x 49 = 593; 11 x 64 = 704; 11 x 81 = 891
The only one that fits the bill is 704. This means there is only one four-digit number that works, and it's 7744
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by ajith » Sun Feb 07, 2010 7:37 am
harsh.champ wrote:Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?


(A)3
(B)2
(C)4
(D)0
(E)1

The ans. [spoiler]is (E).[/spoiler]

Solving approach needed !!!
All the four digit perfect squares have a 2 digit square root.
now the number format is
c(1100) +11d = 11(100c+d)

if the number is divisible by 11, the square root should also be divisible by 11

now the possible numbers are = 33^2, 44^2,55^2,66^2, 77^2,88^2, 99^2

Now I do a trial and error and find that 88^2 = 7744 (nothing else works out too)
So E
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