Recently Dave Marain asked us to count certain palindromes.

And also recently, Sue Van Hattum posed a puzzle that involved certain properties (Fibonacci-ness, triangularity, semi-primality, don’t these words sound interesting?)

And somehow, thinking about one, then the other, I realized:

121 is a palindromic perfect square. And not only that, it is a palindromic perfect square in every base (except binary, too many symbols).

How do you pose the right question to get kids to explore this? If they already know algebra, we’re playing with $(x+1)^2$, but if they don’t know algebra, isn’t this cooler? $11^2 = 121$ regardless of the base, and we can play with different models of multiplication and regrouping?

And even before that, in base 3 121 represents 16 base 10, and in base 4 it represents 25 base 10, and in base 5 it represents 36 base 10, and isn’t that unexpected and very cool and very likely to get noticed by the kids themselves?

6 Comments leave one →
1. December 21, 2009 pm31 3:06 pm 3:06 pm

I think this has something to do with 11…. if you take 1331 it seems to have the same effect…. 1331 base four == 125 base ten which is 5^3…
1331 base five = 216 base five which is 6^3.. notice we keep getting the cube of the number one above… same as with the square…. (Ok, I only checked these two)… but I’m guessing 14641 base 7 will be 8^4…. and checking quickly I find… Eureka, that is true…. so 11^n expressed in any base that will accommodate seems to obey the property that 11^n base b = (b+1)^n in base ten… (did I say that right???) …. 11^2 base three = 4^2 in base ten … and 11^3 base four = 5^3 in base ten…

I wonder what happens with nines??? thanks for an interesting exploration..

• December 21, 2009 pm31 3:11 pm 3:11 pm

Instead of 9s, consider “10-1” – neat things happen.

2. December 22, 2009 am31 5:53 am 5:53 am

Ok, Let me say that more like algebra… in any base (1 n^2 + 2n + 1) = (n+1)^2… and the same idea for the cubes..
Ahh the difference a cup of coffee can make

3. December 29, 2009 pm31 11:27 pm 11:27 pm

I haven’t thought this through much yet, but I noticed the mention of 9, or rather base – 1. We get that add the digits thing for 3 because it’s a factor of 9. (I haven’t proved that, actually. It’s a hunch…)

If we had eight fingers and counted in base eight, we could use the sum of the digits to find multiples of 7, but we wouldn’t get a second freebie the way we do now. Ditto for base 12. Counting in base 5, we’d be able to add the digits to find multiples of 4 and also 2. (Good thing, since multiples of 2 would no longer always end in the same digits.) Counting in base 16 (scary!), we could find multiples of 3, 5 and 15 that way. (I think.)

[I want to take a number theory course! Or maybe it would be more fun to study this with friends online, if we just had one person to point us toward good problems that would push us toward the material that lets you go deeper.]

4. December 31, 2009 am31 2:36 am 2:36 am

So start with, when a number is divided by 10 – 1, the remainder is the same as when 10 times the number is divided by 10 – 1.

Notice, I did not specify the base.

5. January 23, 2015 pm31 12:43 pm 12:43 pm

Outstanding post however I was wondering if you could write a litte more on this subject?
I’d be very thankful if you could elaborate a
little bit more. Thanks!