Here is a numbered list of statements, some true, some false, which refer to a specific number (unique positive integer, base 10).

It just so happens that if a statement is true then its index number appears among the number's digits, and if a statement is false then its index number does not appear among the number's digits.

0. The sum of the number's digits is a prime.
1. The product of the number's digits is odd.
2. Each of the number's digits is less than the next digit (if there is one).
3. No two of the number's digits are equal.
4. None of the number's digits is greater than 4.
5. The number has fewer than 6 digits.
6. The product of the number's digits is not divisible by 6.
7. The number is even.
8. No two of the number's digits differ by 1.
9. At least one of the number's digits is equal to the sum of two other digits. (Any of the digits may be equal, as long as all 3 digits are distinct... for example: {2, 2, 4} or {2, 3, 5} )

Find the number.

(In reply to Solutions by Rafal)

I don't think that 80555, 85055, or 85505 count, because 5 + 0 = 5, but there is no 9.

However, I also found 8005 and 97532. Originally I assumed the decreasing condition for 2 was in the reverse order, so I only found 8005. Here is a convoluted first attempt at proving that no number with more than 9 digits satisfies.

Suppose there is a number with 10 digits or more that satisfies all conditions.

There can be no 5, because the number has more than 5 digits.

This means that only 9 digits are left for 10 slots, so by the pigeonhole principle there can be no 3, since two digits must be equal.

By similar argument there can be no 2, since the sequence of digits must be strictly monotonic and at least 2 are equal, violating that condition.

So we are left with 0,1,4,6,7,8,and 9

Suppose there is a 4. Then there can only be 1's and 0's other than that, by the 4 condition. But there can then be no 1's, since no product of these would be odd if there is at least one 4, so we have only 4's and 0's. But there can be no 0's, since any sum will be even, which is not prime. And with only 4's, the number is even, but does not have a seven. So there can be no 4's in the number.

Now down to 0,1,6,7,8, and 9

Suppose there is an 8 in the number. Then there is no 9 or 7, by the condition for 8, so only 0,1,6,8's are in the number. No product of these digits will be odd, so there are no 1's, and without 1's all sums will be even, and non-prime, so there are no 0's. With only 6's and 8's we will have an even number with no 7, thus there can be no 8's in the number.

Left are 0,1,6,7,9

Suppose there is a 1 in the number. Then there is no 0 and no 6, since including these would make an even digit product. Without 0's and 6's, there can be no 7's, since the number can't be even, and we are left with 1's and 9's. But there can be no 9's, since there is no pair of digits that will sum to another digit in the number, so we have only 1's. But having only 1's, we would need a 4 as well, so there can be no 1's.

Now we have 0,6,7,9

ASIDE: Here's where the proof diverges depending on whether or not you assume that 0 is divisible by 6, and 8005 is a valid solution. If it is, then the end is simple, as there can be no zero's or sixes (could have done this at the beginning) because any such digit product will be either 0, or divisible by 6, so we have only 7's and 9's, and then we can't have any 7's because the number won't be even, and we can't have just 9's because 9 + 9 does not equal 9, so we have no digits, and no 10-digit or greater number exists that satisfies these conditions.

If, however, you say that 0 does not count as divisible by 6, and 8005 is not a valid solution, then it gets more complicated, as far as I can see.

You can't have 6's AND 7's, because that would require an 8, so we have two subgroups.

{0,6,9} and {0,7,9}

Tacking {0,6,9} first, there can't be any 0's in such a number because the sum will always be divisble by 3, and so not prime, and then with just 6's and 9's to choose from, there can't be any 6's because the digit product will be divisible by 6 and we have just 9's, but 9 + 9 is not equal to 9, so we've run out of digits on this path.

{0,7,9}

This is the one I haven't figured out. I can show that there must be all three numbers present, but I can't show that with all three an contradiction is reached.

Can't be 0, can't be all 7's (not even), can't be all 9's (9 + 9 not equal to 9)

Can't be 0's and 9's because sum of the number will be divisble by 3.

Can't be 7's and 9's (not even).

Can't be 7's and 0's because 7 + 0 = 7 and there is no 9.

So we have a 10+ digit number with at least one 0,7, and 9.

So far I haven't been able to disprove this one.

You can probably knock this down to 9+ by switching the order of things, and now that I look at it, if you assume that 0 is divisible by 6 (8005 is valid), you can probably get it down to 8+ or maybe even lower. I like it better when you assume that 0 is not divisible by 6, though, since otherwise there can't be a 6 ever, so why not just say no 6's?

Anyways, if anyone figures out that last step or a better way to prove uniqueness of solution, let me know.

Sincerely,

Joe

Posted by Joe on 2004-09-17 16:33:08