Number Theory #2

We consider a few cases.

Case 1 : odd numbers

We let $a = 1$ , $b = 2$ and $c = N - 3$ . Since N is odd, $b | c$

Also, since $c > b$ , $N \ge 7$

Case 2a: even powers of 2

We let $a = 1$ , $b = 3$ and $c = N - 4$ . Since N is an even power of 2, $N \equiv 1 (mod\quad 3)$ thus $b | c$

However, since $c > b$ , $N \ge 16$ as N is a even power

Case 2b: odd powers of 2

We just take the values in Case 2a and multiply all of them by 2 to obtain the values in Case 2b. Thus, we have $N \ge 32$

$a = 2$ , $b = 6$ and $c = N - 8$

Case 2c: all the other even numbers

We consider the largest odd factor of N. If it is at least 7, we are done as we just multiply the values in Case 1 by a suitable power of 2.

Now, we are only left with numbers of the form $3 \times { 2 }^{ k }$ and $5 \times { 2 }^{ k }$ , as well as the smaller powers of 2.

For $k \ge 4$ , we can use Case 2a or 2b and multiply by 3 or 5 to obtain the desired result.

Finding the answer

We now only need to manually verify if the remaining numbers (1, 2, 3, 4, 5, 6, 8, 10, 12, 20 and 24) are possible values.

Clearly, anything less than 7 is impossible as 7 is the minimum possible value of N.

$10 = 1 + 3 + 6$

$20 = 2 + 6 + 12$

This gives us gives us $1 + 2 + 3 + 4 + 5 + 6 + 8 + 12 + 24 = \boxed { 65 }$ as our answer.