find a number $aabb$ such that is full square

Question

I know that I can write $aabb=1000a+100a+10b+b$, $a,b\in \{0,1,2,3,4,5,6,7,8,9\}, a\neq 0$, so $aabb=1100a+11b=m^2$, where $m\in \mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11\mid m^2$ and $11\mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?

score 1 · Answer 1 · edited Nov 15 '18 at 21:37

1

We have that

$$11\mid (100a+b) \implies (a+b)\equiv 0 \pmod{11},$$

and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.

edited Nov 15 '18 at 21:37

amWhy

209,954

answered Nov 15 '18 at 18:27

user

154,566

It's not immediately clear why you exclude cases of b = 2,3,7,8? – IanF1 Nov 15 '18 at 19:32
1

@IanF1 Because $n^2\mod 10$ can’t take those values. – user Nov 15 '18 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people. – IanF1 Nov 15 '18 at 19:35
"a divides b": a \mid b which renders $a\mid b$, when formatted in mathjax. Don't use a|b – amWhy Nov 15 '18 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye – user Nov 15 '18 at 21:49

find a number $aabb$ such that is full square

1 Answers1