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March 12

Project Euler 418

This is not a request for help in solving problem 418. It is a request for help in understanding what the problem means.

I quote

"Let n be a positive integer. An integer triple (a, b, c) is called a factorisation triple of n if:

1 <= a <= b <= c
a*b*c = n

Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n which minimises c / a. One can show that this triple is unique.

For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872. "

Now 165 factorises as 3 * 5 * 11 (we can include some extra "1 *" if we need them).

f(165) = 3+5+11 = 19. This is good.

But 100100 factorises as 2 * 2 * 5 * 5 * 11 * 91.

And the best I can manage for f(100100) is 25,44,91 which is 160. I can't get 142.

If I can't get 142 then I'm not understanding the problem...

Any ideas? -- SGBailey (talk) 11:38, 12 March 2013 (UTC)[reply]

91 is not prime - it is 7 x 13. So you have 100100 = 35 x 52 x 55 and 35 + 52 + 55 = 142. Gandalf61 (talk) 11:58, 12 March 2013 (UTC)[reply]

Doh! Thanks. -- SGBailey (talk) 12:43, 12 March 2013 (UTC)[reply]

Can you modify any of a normal number's first digits without affecting its normality?

Suppose pi is normal. If I remove the initial 3, is it still normal? What if I went to its trillionth digit and inserted a billion consecutive 8s? Basically, my intuition is that a normal number can be modified in any "finite" way (working within a finite number of digits from the beginning, removing a finite number of digits there, and inserting a finite number of digits) and it will remain normal, but I'm wondering if there's a mathematical proof of this. ± Lenoxus (" *** ") 13:09, 12 March 2013 (UTC)[reply]

Oops, I just took another look at the article and it appears my intuition was right. "The set of normal sequences is closed under finite variations: adding, removing, or changing a finite number of digits in any normal sequence leaves it normal." I presume this also applies to base-n normality; you can make any finite variations to a base-n normal number and it remains base-n normal. ± Lenoxus (" *** ") 13:15, 12 March 2013 (UTC)[reply]

Yes, the amount by which a finite number of occurrences (or non-occurrences) of a sequence affects its frequency tends to zero as the number of digits considered tends to infinity, and this is independent of the base. It's rather like the way a long run of heads would be "averaged out" in a much larger run of tosses of a fair coin. AndrewWTaylor (talk) 13:43, 12 March 2013 (UTC)[reply]

There is an additional argument necessary here to account for other bases. Adding a finite decimal number means adding a periodic expansion in any given base b. So the equidistribution property is preserved in that other base too. 96.46.206.197 (talk) 12:02, 13 March 2013 (UTC)[reply]

Nicely put! I hadn't noticed that. --Trovatore (talk) 03:39, 14 March 2013 (UTC)[reply]