# Two math professors problem

My friend asks me a question from internet. The question is as follows

Two math professors, professor Uno and professor Dos, play chess at the park while reminiscing about their past.

Prof. Uno says, "It just crossed to mind that when we first met, the square of your age contains the same three digits as the square of my age but in a different order."

Prof. Dos replies, "If you take the square of the sum of our ages when we first met and split it into two 2-digit numbers, you will have my age then and your age now."

If the sum of their current ages can be expressed as a 3-digit number $\overline{abc}$, what are the values of $a$, $b$, and $c$? Using the given information above, is it possible to determine their ages?

Using spreadsheet by trial and error, I got the current Prof. Uno's age is $72$ years old and Prof. Dos is $81$ years old. Because $72+81=153$, then $a=1$, $b=5$, and $c=3$.

My questions are

1. How do we solve this question using a formal mathematical way?
2. Is there an intuitive approach to solve this question without using brute force method?
3. Is it true that $72$ and $81$ is the only solution? How do we know that?

## 2 Answers

First clue: "the square of your age contains the same three digits as the square of my age but in a different order", thus their age is between 10 and 31 (this is the range of numbers for which the square contains three digits).

Looking of the squares of numbers in range $[10,31]$ we have the set $$[100,121,144,169,196,225,256,289,324,361, \\400,441,484,529,576,625,676,729,784,841,900,961]$$

Now, by observation we see that the pairs which have the same digits are $$(169,196),(169,961),(196,961),(144,441),(256,625)$$ Thus we have five optins.

Second clue: ""If you take the square of the sum of our ages when we first met and split it into two 2-digit numbers, you will have my age then and your age now."

The square of the sum of ages yield 4-digit number, thus it has to be bigger than 31. Our five options for ages are $(13,14),(13,31),(14,31),(12,21),(16,25)$, so the option $(13,14)$ is irrelavant.

Now we have $4$ options to check and find that the only option that fits is $(16,25)$, as $(16+25)^2=1681$.

If so, we found that the age of Prof. Dos age back then was $16$ and Prof. Uno was $25$.

Today, Prof. Uno is $81$, thus $66$ years have passed and Prof. Dos is $72$.

As you said, there is no option of finding the solution with no trial and error and it seems like another clue is needed.

Here's how I solved it:

First, take a look at what professor Uno said. He said that both his age and Dos's age when they first met yielded a 3-digit number when they first met. This means that their ages when they first med were somewhere between $10$ and $31$, since $32^2>1000$.

The second piece of information I took was the fact that the square of the sum of their ages when they first met was a $4$ digit number as it can be split into two $2$-digit ones. The first two digits of this $4$ digit number will be Dos's age then, so they will be at most $31$, meaning that the sum of their ages then will be at most $56$, as $57^2>3200$.

This limits the question down a bit, but still leaves quite a lot of candidates. Checking then using a program solves the problem, and I don't really think there exists a more elegant solution to the problem...

• Are you saying that in the end we must still use trial & error to solve this problem? – L Lawliet Oct 23 '14 at 10:30
• @LLawliet Unfortunatelly, that is exactly what I think, yes. I see no way in which a nicer solution can be obtained. I may be wrong, of course. – 5xum Oct 23 '14 at 12:59