## Wednesday, January 16, 2013

### Math Teachers at Play #58...

... is up at Let's Play Math!

Denise starts it off with a puzzle:
A Smith number is a [composite, that is, not prime] integer the sum of whose digits is equal to the sum of the digits in its prime factorization.
Got that? Well, 58 will help us to get a better grasp on that definition. Observe:
58 = 2 × 29
and
5 + 8 = 13
2 + 2 + 9 = 13
And that’s all there is to it! I suppose we might say that 58′s last name is Smith. [Nah! Better not.]
• What is the only Smith number that’s less than 10?
• There are four more two-digit Smith numbers. Can you find them?

Then she adds sweet little riddles and jokes. The posts she links to look great. I'm looking forward to spending more time with them.

1. Oh, it is bound to decimal system.
And a definition should say "composite integer".
If not so, there are 2, 3, 4, 5, and 7 satisfying "less than 10" challenge.
Three-digit: 666 (and 121).
Martin

2. Nice catch. [Prime factorization of 2 (or 3 or 5 or 7) is just 2 (or 3 or 5 or 7), so it has to be equal. Composite means there will be something to add.]

4 = 2 x 2
and 2 + 2 = 4 (which is the sum of the digits of 4).

666 = 6 x 111 = 2x3 x 3x37
6+6+6 = 18
2 + 3 + 3 + 3+7 = 18

121 = 11 x 11
1 + 2 + 1 = 4
1+1 + 1+1 = 4

Got it. Thanks.

3. Just to make a new problem...
If we change decimal notation into e.g., binary, it is obvious that Smith numbers do not hold their property any more. (I find "number relations" that are not bound to numbers but to "number representation" -- how we write them, somewhat inferior...)

Can we find some Smith numbers in binary/diadic system?
Three minutes scratching on paper shows
1111(2) = 11(2) × 101(2) (four ones on left and right side of 15 = 3 × 5)
110011(2) = 11(2) × 100001(2) (four ones on left and right side of 51 = 3 × 17)
Scrath, scratch... a pattern is evolving...
Why not
11011(2) = 11(2) × 10001(2)? (because 10001(2) is a composite number, i.e. 11(2) × 11(2) -- 9 = 3 × 3)
So, it is easy to spot (and not so easy to prove) that every number in binary system of form
3 × (2^n + 1) is a Smith number(2) whenever 2^n+1 is a prime. Fermat claims that 65537 is a prime, so
110 000 000 000 000 011(2) = 11(2) × 10 000 000 000 000 001(2)

4. It was not very polite of me to send an "anonymous anonymous" post. It was the same Martin again.
Reason to drop into your blog was searching for a drawing of egyptian knots -- I have found one in you former articles. I need it for a math textbook... (and I downloaded it without prior permission).
Regards
Martin

5. It seems like binary makes this relationship especially likely. Perhaps there are things to prove in binary that wouldn't work in other bases. I'd find that interesting, even though it's 'bound to the number representation', because of what it would say about why binary is special.

I wonder if 11 x 11 can be generalized: 11 in base 6 is the same as 7. So 11x11 = (49 base ten, which is...) 121 (base six).

Base two: 11x11 (means 3x3) = 1001. The digits of the factors add to (4, which is written as...) 100 and the digits of the product add to (2 which is written as...) 10. So binary works on this, but differently.

Base three: 11x11 (means 4x4) = 121. Yep. Now can it be proved to work in any base, so it's not 'bound to the number representation'? And are there any other Smithish numbers that are strong like this?

6. And I have sworn to finish Pythagora theorem section today...

"And are there any other Smithish numbers that are strong like this?" -- They are not so strong and they are not numbers; they are number representations or strings.

Let's try with
121(8) = 3(8)^4 | 81 = 9 × 9 (promising) = 3 × 3 × 3 × 3 (ahhh)
Left sum: 4; right sum: 12.
121(5) = 11(5)^2 = 2 × 2 × 3 ×3
Left sum: 4; right sum: 10.

Equality 121(n) = 11(n)^2, however, holds for any number system where n > 2.
It just says that n^2 + 2n + 1 = (n + 1)^2. And it is Smith number when n + 1 is a prime. 