Today I ran a prize puzzle contest in combinatorics (I’ll explain that some other time). One of the puzzles, old but good, was:

Is there a positive integer, $n$, such that $n^3$ and $n^4$ taken together contain each of the digits 0, 1, 2,… , 9 exactly once?

Clear? To me, sure. To you? If English is your native language, yup. But one of my students, out of ESL but not a native speaker, took the phrase “taken together” to mean “added together.” Not the intent. But I will argue that there was at least some (haha) ambiguity. My problem asked for two powers of ten, with 10 digits altogether. The student’s problem was to

find a number of the form $a = n^3 + n^4$ such that $a$ contains all ten digits exactly once.

1. Can you solve the problem as originally intended?
2. Can you solve the student’s (arguably more interesting) problem?
September 12, 2007 pm30 5:03 pm 5:03 pm

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’.

2. September 12, 2007 pm30 8:57 pm 8:57 pm

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.

3. September 12, 2007 pm30 11:40 pm 11:40 pm

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

4. September 13, 2007 am30 3:03 am 3:03 am

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.

5. September 22, 2007 pm30 2:24 pm 2:24 pm

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

6. September 23, 2007 pm30 3:05 pm 3:05 pm

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.)

7. September 23, 2007 pm30 4:59 pm 4:59 pm

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