Consider four digit numbers for which the first two digits are equal and the last two digits are also equal

I accept 7744 is one perfect square satisfying the above condition, but what about the 4-digit numbers like 1600,2500,3600,4900,6400,8100
These are also perfect square numbers.
9 years agoHelpfull: Yes(5) No(1)
we conider the numbers of the series 0000,0011,0022,0033,0044,0055,...,0099 and 1100,1111,1122...,1199 and so on untill 9900,9911,...,9999
the only number which is perfect square is 7744(square of 88)
so answer is (4)1
13 years agoHelpfull: Yes(4) No(9)
38,68 and 88 last digits in their squares equal
9 years agoHelpfull: Yes(1) No(3)
any four digit number in which first two digit are equal and last two digits are equal will be of the form
11*(100a+b) i.e multiple of 11 like 1122 ,3366 etc
let required no. be aabb
since aabb is a perfect square only value is 7 and 4 therefore no. is 7744 hence option 4
9 years agoHelpfull: Yes(1) No(0)

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?
(1) 2 (2) 4 (3) 0 (4) 1 (5) 3