



 
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 $10a+b=a+b^2,$
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, 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  


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