patelv 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 square?

kindly provide the answer option
and what is the source???

Options are :


a) 3
b) 4
c) 2
d) 1


In simple words we can say that "how many 4 digit no.s can be perfect squares with its firs two no. equal and last two no.are also equal.

means let the no. be xxyy and it should be perfect square.

Are you sure this is a GMAT question?
This is how I approach the problem

The given number is : aabb & the requirement is sqrt(aabb) = an integer.

aabb = aa00 + bb = aa*100 + bb = a*11*100 + b*11 = 11* (100a + b)

Because 11 is not a perfect square, then (100a + b) must have the factor 11 in them
11* (100a+b) = 11* 11 * ( 100a+b)/11

Now sqrt(aabb) will be equal to 11* sqrt((100a+b)/11)
In order for aabb to be a perfect square, (100a+b)/11 must also be a perfect square, and of course it must be an integer as well.
(100a + b)/11 = 9*a + (a+b)/11 is an interger , then a+b must be divisible to 11
a+ b =<18 ( as a&b =<9), so (a+ b)/11 = 1
Let's plug in all the possible a & b:
a=1 b =10, eliminate
a=2 b=9, then 9*a +(a+b)/11 = 2*9 + 1 = 19, not a perfect square, eliminate. Call 9*a + (a+b)/11 a name f(a,b) for short.
a=3 b=8, f(a,b) = 3*9 +1 = 28, eliminate for the same reason above.
a=4,b=7, f(a,b) = 37
a=5,b=6, f(a,b) = 46
a=6, f(a,b) = 55
a=7, f(a,b) = 64, that's it, 64 = 8^2, a perfect square.
a=8, f = 73
a=9, f= 82
In conclusion, there's only one 4-digit no. that matches the requirement.
It's when a =7, b = 4, the no. is 7744, and sqrt(7744) = 88
D is my answer.