Unmarked numbers!

1, 16 , 31, .... , 991 = 67 marked
991 + 15 = 1006 - 1000 = 6
6 , 21, 36, ,,, , 996 = 67 marked
996 + 15 = 1011 -1000 = 11
11 , 26, 41 , ... 986 = 66 marked
986 + 15 = 1001 - 1000 = 1
total marked = 67 + 67 + 66 = 200
unmarked = 1000 - 200 = 800

Let me separate the marked numbers into three phases.

Phase $1$ : $1, 16, 31, ..., 991$

After the number $991$ , the next number should be $1006$ . But there is no number $1006$ , so it is $6$ .

Phase $2$ : $6, 21, 36, ..., 996$

Phase $3$ : $11, 26, 41, ..., 986$

Why is it $986$ ? Because if I wrote it $1001$ , it the same as $1$ (the first number).

Ok, what do we have here? $67$ numbers on Phase $1$ , $67$ numbers on Phase $2$ , and $66$ numbers on Phase $3$ . In total, there are $\boxed{800}$ unmarked numbers. (Rarely see such nice number.)

After considering the loop around the circle, we count integers congruent to 1,6, or 11 modulo 15. This is because $1000 \equiv 10 \mod{15}$ , and so adding multiples of 15 increases the residue of the next "circuit" by 5. Note that after 11 would follow the residue class $k \equiv 16 \mod{15} \equiv 1 \mod{15}$ and so these are the only three cases to check.

There are 67 integers congruent to 1 or 6 modulo 15 and 66 congruent to 11 modulo 15 which are less than 1000. Thus, there are $67+67+66=200$ integers marked, and so there are $1000-200 = \boxed{800}$ unmarked integers.

We have that after adding 15 n times, we would have gone around the circle i times. Thus we get that: $15n=1000i \Rightarrow 3n=200i$ Thus the smallest quantity for n will be: $n = \frac{LCM(3, 200)}{3} = \frac{600}{3} = 200$ Thus the total amount of unnumbered slots will be $1000-200=\boxed{800}$

If 1000 is a cycle 3 whole cycles is divisible by 15. In 3 cycles: there are 3000 numbers, out of which 200 will be marked, and 800 not.

Not trying to be rude, but how does this question have such a high rating ?

One can observe that we can convert the circle effect by expanding the 1000 slots to the least common multiple of 1000 and 15, which is 3000. The number of distinct slots that are marked is simply 3000 / 15 = 200.

This leaves 1000 - 200 slots = 800 unmarked.

The generalization of the problem is: given N slots arranged in a circle, we'll mark every K -th slot until there are no additional slots marked. The number of remaining unmarked slots = N - (LCM(N, K) / K)

Note that $1 + 15 \cdot 67 = 1006$ $1 + 15 \cdot 67 \cdot 2 = 2011$ $1 + 15 \cdot 200 = 3001$ Therefore only numbers congruent to one of $1,6,11 \pmod{15}$ get marked. Rest are unmarked, that leaves $12$ residues for each multiple of $15$ , note that $990 = 15 \cdot 66$ , hence there are $12 \cdot 66 + 8 = \boxed{800}$ unmarked numbers.

Sol1 (Think without much calculating): When we reached number 1000 (actually 991), just extend it to 2000 and 3000. It will start all over again when it reached 3000. ( $3001 \equiv 1 ( mod 1000)$ ). So there will be a loop that other numbers are not marked. The marked numbers will be 1, 6, 11, 16, ...., 991 -> 200 numbers.

Therefore: Numbers not marked $= 1000 - 200 = \boxed{800}$ .

Sol2 : Numbers from 1 to 1000 that are in the form of $15n + 1$ for all $n\in I - I^{-} = \lceil\frac{1000}{15}\rceil\ = 67.$

Find the other 2 cases that are in the form of $15n+6$ and $15n+11$ we get $67$ and $66$ .

Therefore: Numbers not marked $=1000 - (67+67+66) = \boxed{800}$ .

Note: We use ceiling because we also include $n = 0$ . I'm too lazy to say floor + 1. This is the formula. $\lfloor x \rfloor + 1 = \lceil x \rceil$ for all $x$ in real numbers.

Sol3 : Use the same technique as solution 1, but just solving equation.

We extend it to 3000 and we know that there'll be a loop.

The numbers that are marked is 1,16,31,46,...,2971,2986, 1,16,....

And we know that...

$15n+1 \leq 2986$ (for n = 0,1,...)

Solve for $n$ we get $n = 199.$ And we know that there'll be 200 numbers are marked (including n = 0).

Therefore: Numbers not marked $= 1000 - 200 = \boxed{800}$ .