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 re: Solutions by Joe)

Please ignore previous post, the whole last half was screwed up -

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. The proof is unfinished for one case.

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

Now it matters whether 0 is divisible by 6 or not. This determines whether 8005 is a valid solution or not.

> Suppose that 0 is not divisible by 6 (8005 is valid):

> Suppose there is a 0 in the number. Then there must be a 6, since the product will be 0, which we is not divisble by 6. With a 6 in the number, there can be no 7, because there is no 8, and 6 and 7 differ by 1. This leaves 0, 6, and 9. But any sum will be divisible by 3, and so not prime. So there can be no 0 in the number.

> This leaves 6, 7, and 9

> There can be no 6, since any number with a 6 will now be divisible by 6, so we have only 7's and 9's, but there can't be 7's since no such number is even, and there can't be just 9's because 9 + 9 is not equal to 9. So if 0 is not divisible by 6, we're out of digits, and no number with 10+ digits can satisfy the conditions.

Okay, now suppose that 0 is divisible by 6, and we're back to having 0, 6, 7, and 9.

There can't be both a 6 and a 7, because there is no 8, so the two remaining cases are:

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

Look at {0,6,9} first. There can't be a 0, because all sums will be divisible by 3, so there are only 6's and 9's, but then any digit product is divisible by 6, so no 6's, and then there's only 9's, which can't be the case because 9 + 9 is not equal to 9.

So now there is {0,7,9}

This is where I'm stumped. There can't be just 7's, because the number won't be even. There can't be just 9's because 9 + 9 is not equal to 9. There can't be just 0's because 0 is not considered prime. There can't be just 0's and 9's because the sum is divisible by three. There can't be just 7's and 9's because the number won't be even. There can't be just 0's and 7's because either the sum won't be prime, or, in the case of exactly one 7, 7 + 0 = 7 and there is no 9.

But if there is at least one of all three numbers 0, 7, and 9, I haven't been able to disprove it yet.

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

Sincerely,

Joe

Posted by Joe on 2004-09-17 17:17:21