Open Contests

Let $n$ be the number of digits of $b$ and let $b=ka$. Then by the conditions of the problem, $10^n\cdot a+ka=(a+ka{)}^2$, or $$a=\frac{10^n+k}{(k+1{)}^2}(1)$$ If $k$ were odd, then the numerator on the r.h.s. of (1) would be odd and the denominator even, so $a$ could not be an integer. Hence $k$ is even.

If $k=2$ then the cross-sum of $10^n+2$ is 3, which is not divisible by $(2+1{)}^2=9$. The case $k=4$ also leads to a contradiction, since $10^n+4$ ends with 4, hence cannot be divisible by $(4+1{)}^2=25$. Thus $k\ge 6$. In the following we show first that $k\le 8$ and finally that $k\ne 8$. The assumptions $ka=b\ge 10^{n-1}$ give $10ka\ge 10^n$. Equality (1) implies $$10^n=(k+1{)}^2\cdot a-k=k^{2}a+2ka+a-k=(k+2)\cdot ka+a-k.$$ Thus $10ka\ge (k+2)\cdot ka+a-k$, whence $$(8-k)\cdot ka\ge a-k.$$ As $a$ is positive, $(8-k)\cdot ka>-k$. As both sides of this inequality are divisible by $k$, this implies $(8-k)\cdot ka\ge 0$. Consequently $8-k\ge 0$, i.e., $k\le 8$. If $k=8$, the inequality (2) implies $a\le 8$ whereas the equality (1) reduces to $a=\frac{10^n+8}{81}$. Hence $a$ ends with digit 8, leaving $a=8$ and $b=8\cdot 8=64$ as the only possibility. But $864\ne (8+64{)}^2$, contradicting the conditions of the problem.