Hi,

We have two responses to your question

Let the two digit number "x" (base 10) have the tens digit "a" and the ones digit "b". Thus we have $x=10a+b.$ We also have $x=a+b^2$ from the information in the question. In addition we have the restrictions that "1<=a<=9" and "0<=b<=9". Such a number x exists only when $10a+b=a+b^2,$
i.e. $9a=b \times (b-1).$ We can test the digits 0 through 9 for b.

if b=0 then we have $9a=0 \times -1$, i.e. $a=0$ which does not result in a two digit number for x.
if b=1 then we have $ 9a=1 \times 0$, i.e. $a=0$ which does not result in a two digit number for x.
if b=2 then we have $ 9a=2 \times 1$ which has no integer solution a.
if b=3 then we have $ 9a=3 \times 2$ which has no integer solution a.
if b=4 then we have $ 9a=4 \times 3$ which has no integer solution a.
if b=5 then we have $ 9a=5 \times 4$ which has no integer solution a.
if b=6 then we have $ 9a=6 \times 5$ which has no integer solution a.
if b=7 then we have $ 9a=7 \times 6$ which has no integer solution a.
if b=8 then we have $ 9a=8 \times 7$ which has no integer solution a.
if b=9 then we have $ 9a=9 \times 8$ which has one integer solution a=8.

thus the answer is to have $a=8$ and $b=9$ giving the two digit number 89. As a check, $8 + 9^2=8+81=89$ also.

Cheers,
Dr. L. Dame

Another approach

Hi,

I just want to point out something else you could do with this problem. When you follow Dr. Dame's solution and get to

\[9a = b \times (b - 1)\]

you can use a divisibility argument rather than the case by case argument. From

\[9a = b \times (b - 1)\]

I can see that the left side is divisible by 3 and hence the right side must be divisible by 3. One way this can happen is that $b$ is divisible by $3 \times 3$ which gives Dr. Dame's solution. Also $b - 1$ could be divisible by 9 but then $b$ is 10 or larger which is impossible since $b$ is a digit. What if $b$ is divisible by 3 but not by 9 then $b - 1$ is also divisible by 3. But this is impossible, you can't have two consecutive integers divisible by 3. Likewise id $b - 1$ is divisible by 3 but not by 9. Hence Dr. Dame's answer is the only possible correct answer.

Harley