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 Question from a teacher: Find a “2-digit number” where the sum of the 1st digit (on the Left) and the square of the 2nd digit equal the same number.

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

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