if aabb is a 4 digit no. and also a perfect square then value of 2 + b is

Question

anonymous · Answer

sry

anonymous · Answer

there is a mistake 
it is a+b
ok

anonymous · Answer

any1

anonymous · Answer

a + b would have to be 11 i think >.>

anonymous · Answer

hou

anonymous · Answer

i'm not entirely sure, i'll post my pic and explain:

anonymous · Answer

ok

anonymous · Answer

i think your right

anonymous · Answer

basically i wrote this:
$$aabb = a(10^{3})+a(10^{2})+b(10)+b = a(10^{2})(11)+b(11) = 11(100a+b)$$
in order for this to  be a square, the term 110a+b must have 11 as a factor. so now i look at:
$$100a+b \equiv 0 \mod 11$$
but:
$$100 \equiv 1 \mod 11$$
so now i have $$a+b \equiv 0 \mod 11$$
and since we are dealing with a and b being digits (the numbers 0 - 9), that means a+ b can be bigger than 18, so the only multiple of 11 they could be is 11. Therefore a + b = 11

anonymous · Answer

sry for typo, it  should say, "...the term 100a+b must have..."

anonymous · Answer

er. another typo <.< curse these mornings >.>, " ...that means a + b cant* be bigger than 18..."

anonymous · Answer

One example of this is the number 7744, that is 88^2, and 7 + 4 = 11 Actually is the only example.

anonymous · Answer

OK