Strange numbers

Number Theory Level 3

Some, but not all, of the first $11$ positive integers is called strange. Here are some details:

If a number $n$ is strange ( $1\leq n\leq 11$ ), then $12-n$ is also strange.
If a number is strange, then the positive divisors of the numbers are strange numbers.
At least one of the strange numbers is even.
If there are $k$ number of strange numbers, then $11-k$ is not a strange number.
Not all integers between $1$ and $11$ are strange.

What is the sum of the strange numbers?

The answer is 48.

1 solution

Zee Ell
Aug 30, 2017

Since at least one of the strange numbers is even, that number has factors 1 and 2. Therefore, 1 and 2 are strange numbers.

This means, that 12 - 1 = 11 and 12 - 2 = 10 have to be strange numbers as well. Since 10 has the factor 5, it has to be strange, along with 12 - 5 = 7.

Now, we have at least 6 (k ≥ 6) strange numbers (1, 2, 5, 7, 10, 11). Since 11 - 6 = 5 (which is a strange number), therefore k ≠ 6, meaning that there are additional strange numbers.

At this point, we check all those integers between 1 and 11, which were not considered as strange yet.

3 implies 9 (and vice versa), 8 strange numbers, 11 - 8 = 3 , which had to be both strange and not strange at the same time, not a solution.

6 implies 3 and 9; 9 strange numbers, 11 - 9 = 2 , which had to be both strange and not strange at the same time, not a solution.

4 implies 8 (and vice versa), 8 strange numbers, 11 - 8 = 3, which is not strange number (yet), it is a solution.

If we take, the two pairs (3 and 9, 4 and 8), it would give us 10 strange numbers, 11 - 10 = 1, strange, hence it is not a solution L

If we presume, that not all numbers are strange, then we are only left with one solution:

$1+ 2 + 4 + 5 + 7 + 8 + 10 + 11 = \boxed {48}$

Strange numbers

The answer is 48.

1 solution

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