# A neat coincidence with place value problems

Over the last few days I’ve been working through some pretty difficult place value problems with my younger son.  These problems come from the “Algebra with Integers” section of Art of Problem Solving’s Introduction to Number Theory book.  As I wrote yesterday part of the difficulty is that we haven’t really covered any algebra yet.

Despite the lack of familiarity with algebra, my son seems to really like these problems.  He is absolutely fascinated when we arrive at the answers and is amazed that some of the problems have more than one answer.  The problems are all like the one in the video below:

Tonight I tried to throw in a little curve ball – how many two digit integers are 11 times the sum of their digits?  I wasn’t sure if he would see that this situation was impossible right away.  He didn’t, which was lucky because I really wanted to see how he would react to the algebra telling him there were no solutions.  Initially that math gave him a little trouble, but I loved seeing him realize that something was wrong without being quite sure what it was:

Definitely a fun couple of days.  I like these problems both for the easy introduction to algebra and for the place value review.   They also serve as a surprise example of a “low entry / high ceiling” problem.   I hadn’t really thought about this feature of these problems until I saw this challenge problem involving digits and decimal representation from Christopher Long on Twitter yesterday:

as well as a few fun follow ups like this one (and check Chris’s twitter feed for plenty more):

Pretty sure that the first time I saw a problem of this type was in the book Lure of the Integers by Joe Roberts which shows this interesting result: The problems on Twitter are obviously a little harder than the ones I’m covering with my son, but I think older kids would probably like them a lot.   Even just showing some of the solutions might catch a kid’s interest since It is pretty surprising to see how large the solutions are.   I love simple to state problems that turn out to be a little more difficult than you might initially think!

Always happy to see the math I’m covering with the boys overlap in some small way with the math people are sharing on Twitter!

# 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?”

“510”

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

” . . . . fifty ten . . . ?”

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

“60.”

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

“15”

“What is 4 times 15?”

“60 . . . hey, it works!”