Comments on: Zimaths

By: Barry Cipra

Barry Cipra — Fri, 19 Dec 2008 17:13:19 +0000

Zac’s generating function is a *great* way to organize the calculation, much better than the ad hoc approach I originally took.

A little finagling with the various terms, using the identity x^10 = 1, shows that the final polynomial is
x^3 (1+x^2)^6 (1+x^4)^4
From Pascal’s triangle one gets
(1+x^2)^6 = 7 + 7x^2 + 15x^4 + 20x^6 + 15x^8
and
(1+x^4)^4 = 1 + 4x^2 + 4x^4 + x^6 + 6x^8
Everything else can now be done with fairly simple arithmetic. The coefficient of x^3 in the final result, for example, is
7 + 42 + 15 + 80 + 60 = 204.
However, this seems to only make it more mysterious as to why the final numbers come out so nearly equal!

By: Zac

Zac — Fri, 19 Dec 2008 08:56:54 +0000

The “easier way” that I would use is a generating series for each term’s “bag”.

1: x^1
2: (x^2 + x^4 + x^6 + x^8)
3: …

Multiply them all together, and set x^10 = 1 since you only care about the last digit. The coeffiecient of each number is the weighting for that value. Of course, this basically does the same as your “[Tally [ Flatten [ Outer [ PlusMod …”, but you can do it by hand, if you’re comfortable multiplying polynomials with 5 terms :)

204x + 204x^3 + 205x^5 + 206x^7 + 205x^9

By: Barry Cipra

Barry Cipra — Wed, 17 Dec 2008 20:37:00 +0000

If you’re interested, here’s a variation on the original problem that’s a little trickier:

Let a, b, c, etc. be randomly (and independently) chosen from the set {1,2,3,…,100}. What is the probability that
1^a + 2^b + 3^c + 4^d + 5^e + 6^f + 7^g + 8^h + 9^i
has units digit equal to 5? Is there some units digit that’s more likely than any other (and if so, which one)?

Note that the powers here are drawn between 1 and 100, not between 0 and 99. Your Mathematica simulation should have no problem estimating the probability, but it may have a hard time answering the follow-up question.

By: Barry Cipra

Barry Cipra — Wed, 17 Dec 2008 13:55:45 +0000

I see I should have read the other comments more carefully before making my own. I basically restated what svat had to say.

By: Barry Cipra

Barry Cipra — Wed, 17 Dec 2008 13:46:45 +0000

You could have realized that your answer 1/5 has to be wrong without ever computing what the correct answer is. Here’s how.

As soon as you determine that the units digit in the powers of 3 and 7 repeats every four steps, then, since 4 divides 100, it’s no longer necessary to choose at random from 0 to 99, it suffices to choose from 0 to 3 (or 1 to 4). This reduces the number of combinations from 10,000 to 16, all equally likely, some number of which — lets call it k — have the desired property. It doesn’t really matter what this desired property is, the probability of its occurrence is k/16. And here’s the rub: There’s no way a fraction with denominator 16 can reduce to one with denominator 5. (The fancy name for this is the Fundamental Theorem of Arithmetic.)

Note, this doesn’t tell you *where* you went wrong in your solution, it merely tells you that you *did* go wrong.

By: svat

svat — Wed, 17 Dec 2008 03:00:07 +0000

How do you think that error could have been avoided?

(Perhaps by being more systematic about making that list, or by writing that list with an “open mind” instead of using it to verify a conjecture, so that the fact that your answer was 2/10 instead of something/16 would have tipped you off?)

By: Nemo

Nemo — Wed, 17 Dec 2008 00:38:00 +0000

7 = -3 (mod 10)

By: Rob Blake

Rob Blake — Tue, 16 Dec 2008 23:48:43 +0000

There’s a mistake in your thinking regarding the list 1 digit sums mod 10. When you perform the final ones addition, you’re twice as likely to add a 1 and a 3 than you are to add a 1 and a 1. If you write out the full table of all combinations, you’ll see that the result 0 has 4 possible combinations and the other even numbers have 3 possible combinations, leading to a total of 16 combinations. Therefore the probability distribution should be:

0 = 4/16
2,4,6,8 = 3/16

Your graph confirms this.