A neat place value exercise: what about fifty ten?

My younger son and I started a new chapter in our Introduction to Number Theory book today – “Algebra with Integers.”  It is big step up in difficulty from the prior chapters, and even more of a step up because I have really not covered any algebra with him at all.  Even though the discussions are a little longer now, many of the problems and examples remain accessible to him.  Sadly, that’s not going to remain true for much longer, so our little tour through introductory number theory will come to an end soon.

One of the first problems that we looked at today focused on place value:

Find a two digit integer that is equal to three times the sum of its digits.

Even getting going with this problem is a challenge if you’ve not had algebra, but a few concrete examples like 34 and 87 helped get the ball rolling.  Eventually he was able to write down an equation that would help solve the problem.  If the 10’s digit of the number is A and the 1’s digit of the number is B, we know that:

(1) 10*A + B = 3*(A + B),

and so 7A must be equal to 2B.  It was really interesting for me to see the leap from “the number is AB” to “the number is equal to 10A + B” happen right in front of me.

Solving this equation is no small task  and really is only possible after you recognize that A and B have to be single digit positive integers.  Once you do recognize that important piece, though, it is not super hard to see that the solution to the original problem is A = 2 and B = 7.  Thus, the number we are looking for is 27.

With that problem as a warm up we were ready for another challenge: find all two digit positive integers that are equal to four times the sum of their digits.

If we proceed as above we find that the equation we need for this problem is:

(2) 10*A + B = 4*(A + B),

which simplifies (a little more than the first one) to 2A = B.  Instead of just one solution here, there are several:  12, 24, 36, and 48 come to mind quickly once you know that B = 2A.

At this point my son told me that he thought it was interesting that these numbers were all multiples of 12 but then dropped that thought to tell me that he thought that there were no more solutions.

“Why not?”

“Because the next one would start with 5, but double 5 is 10 and 5(10) isn’t a two digit number.”

“What would it be if it was a number?”


“But the 5 is the tens digit, not the hundreds digit.”

” . . . . fifty ten . . . ?”

“Interesting.  What do we usually call that number?”


“What is the sum of the digits of fifty ten?”


“What is 4 times 15?”

“60 . . . hey, it works!”

“What about sixty twelve.”

“. . . . yes, 18 X 4 is 40 + 32 = 72 = sixty twelve.”

I left it there, but thought it was a fun conversation because it got him thinking about place value.  Just like our conversation last week about the relationship between the rows of Pascal’s triangle and powers of 11:

A day in the life: Building and Extending Number Sense

I really like being able to sneak in some ways to think about number sense and place value while talking about problems that (to him) aren’t obviously related to those topics.

Made for a fun morning.

One thought on “A neat place value exercise: what about fifty ten?

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