Question

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

Answer 1

sry

Answer 2

there is a mistake
it is a+b
ok

Answer 3

any1

Answer 4

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

Answer 5

hou

Answer 6

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

Answer 7

ok

Answer 8

i think your right

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

OK