Comments on: Puzzler puzzled

By: Guess who’s a winner? « JD2718

Guess who’s a winner? « JD2718 — Sun, 01 Feb 2009 02:25:55 +0000

[...] prize: some sort of gift certificate, I will make it a prize for a puzzle solution in my combinatorics class. I might even use #25… Think a high school kid could do [...]

By: Dave Churchill

Dave Churchill — Thu, 11 Oct 2007 16:25:36 +0000

Is there a positive integer, n, such that the digits from both the numbers n^3 and n^4 contain (0…9) exactly once?

or:

Is there a positive integer, n, such that the concatenation of the string representation of n^3 and n^4 contains (0…9) exactly once?

There is a unique answer for both of these questions… although I cheated and brute forced.

By: jd2718

jd2718 — Sun, 23 Sep 2007 12:59:23 +0000

It may not be well-worded. Rule here: make a suggestion. I think all of us are always looking to improve, alter, modify, etc.

By: Maria Miller

Maria Miller — Sun, 23 Sep 2007 11:05:40 +0000

I was born in Finland and I could not understand the original question. I didn’t think it necessarily meant adding the powers, but the wording just left me confused. I guess it is a fairly rare phrasing? (And I’m a math teacher; I’ve spent plenty of my life reading and talking in English.)

By: randomwalker

randomwalker — Sat, 22 Sep 2007 10:24:17 +0000

for the first question i had exactly jd2718’s solution above :)

for the second i gave up and used a script

[x for x in xrange(350) if len(set(str(x*x*x*(1+x)))) == 10]

//also surprised there’s a unique answer

By: jd2718

jd2718 — Wed, 12 Sep 2007 23:03:41 +0000

Analytic for my puzzle? I am not certain. But here’s my thinking:

n^3 must have 3 or 4 digits, n^4 must have 6 or 7. That places us between 5 and 21, then between 18 and 56, so our candidates are between 18 and 21, inclusive. Ending in 0 or 1 is a no-no (both powers will end in 0 or 1, giving a repeated digit), leaving only 2 candidates to check.

Analytic for the girl’s puzzle. Hmm. After Clueless I will narrow things down. Ignore the 3rd power. We are between 178 and 316, inclusive. But I don’t want to trial and error 139 candidates.

Can’t end in 0. (both would end in 000, sum would have repeats.)

Can’t be one more than a multiple of 3 (sum would not be divisible by 3, but all permutations of 0 – 9 are divisible by 3)

Narrowing along the same lines, since the sum of the digits is 45, the number is divisible by 9. numbers congruent to 2 or 5 mod 9 do not produce 3rd and 4th powers whose sums are multiples of 9.

Ugly, but xa5, where a is odd, gives 3rd and 4th powers whose sum ends in two zeros.

I think I got from 139 candidates down to 40 or so.

Better ideas? brute force is allowed, but I prefer it as a later resort.

By: Mike

Mike — Wed, 12 Sep 2007 19:40:15 +0000

As the others have said, the solution is easy to find, but I wonder how to do it analytically?

By: vlorbik

vlorbik — Wed, 12 Sep 2007 16:57:35 +0000

1. easily (on a TI-83).
2. in principle, yes (there’s a 10-digit display).
but i’ll just take clueless’s word for it since
it could involve scrolling through over 100 #’s.

By: Clueless

Clueless — Wed, 12 Sep 2007 13:03:50 +0000

Maybe ‘concatenated’ is a better word than ‘taken’ in the statement of the first problem. Will middle schoolers be familiar with it, though?

It is interesting that the first problem has a unique solution. It is even more interesting that a solution to the second problem exists.

I just did a search, would be interested to know if there are more analytical methods possible. I guess it is possible to eliminate many choices after narrowing down the range of possible ‘n’.