Time Bomb

It turns out that the bomb is evenly distributed when n is a power of 2. So, $1+2+4+...+64=128-1=\boxed{127}$

Proving this is harder.

Part 1: Powers of two work

Let $n=2^k,k \ge 1$ . Suppose the bomb is held by the same person twice. Then there are two integers $0 \le a < b \le 2^k-1$ for which the bomb is on the same person at a seconds and b seconds. At s seconds, the bomb is $1+2+3+...+s = \dfrac {s(s+1)}{2}$ places counter-clockwise from the ending space.

Thus we have $\dfrac {a(a+1)}{2} \equiv \dfrac{b(b+1)}{2} \pmod{2^k}$

$2^{k+1} \mid (a(a+1)-b(b+1)$

$2^{k+1} \mid (a-b) (a+b+1)$

Note (a-b) and (a+b+1) have opposite parity.

If (a-b) is even, we must have $2^{k+1} \mid (a-b) \implies a=b$ due to the restricted range. Contradiction!

If (a+b+1) is even, we must have $2^{k+1} \mid (a+b+1)$ .

However $1 \le a+b+1 \le (2^k-1)+(2^k-1)+1 = 2^{k+1}-1$ so this is also impossible.

Therefore no such a and b exist by contradiction. $\square$

Part 2: Everything else fails

Let $n=2^km$ for some odd m greater than one and some non-negative integer k.

Consider $s \in S=\left\{0 , m-1, 2m-1, 3m-1, ... , 2^km-1\right\}$

We have $m \mid \dfrac{s(s+1)}{2}$ since m is odd.

Therefore $\dfrac{s(s+1)}{2}$ is congruent to one of $R=\left\{0,m,2m,3m,...,(2^k-1) m \right\}$ modulo $2^km$ .

As $|S| =2^k+1>2^k = |R|$ we will have some repeat by the pigeonhole principle.

Therefore for some two values in S, the bomb will be the same number of places counterclockwise from the ending space, that is, they will be in the same place. $\square$

Note that when m=1, $m-1=0$ , so $|S|=|R|$ and there is no problem.

Sorry for my poor English. If there are any mistakes, please let me know!

The bomb is moving counter-clockwise 1 step, 2 steps, 3 steps and so on. If there are people, for the bomb to return to the place where it was, the sum of some consecutive numbers that are smaller than should equal to or its multiple.

Therefore, any number that can be expressed by the sum of consecutive numbers do not satisfy the condition. At the same time, numbers that cannot be expressed in such a way will be the numbers that satisfy the condition.

If , where and are positive integers, .

Let be and be . Then, and

Since and are positive integers, and are both even numbers or both odd numbers.

If and are both even numbers, and ( and are positive integers)

Therefore, by adjusting the value of and , can be expressed into any number with the exception of numbers that do not have an odd number as a divisor, which are

If and are both odd numbers, and ( and are positive integers)

Like the case above, can be expressed into any number but .

Therefore, it is evident that numbers that are not are disqualified.

However, we have to see whether all of satisfy the condition.

If the multiple of these numbers can be expressed by the sum of consecutive numbers, those too are disqualified.

Let , where are consecutive numbers that are smaller than .

If is an odd number and is an odd number, .

Since is a positive integer, must be a positive integer, since both of them are odd numbers.

Since is always larger than , cannot be expressed by the sum of consecutive numbers that are smaller than .

If is an even number, , and since and are consecutive numbers, must be an odd number.

Therefore, .

Because there are positive integers that are smaller than , it cannot be expressed by consecutive positive integers.

If is an even number, , where is an odd number, and the same can be applied as above.

Therefore, it is true that satisfy the conditions.

Since ,