This is a problem in number theory and related to the Chinese remainder theorem.
I do not know at what level you're working on. However, this problem is probably made to be solved without the use of advanced theory. Here's an approach that you can apply to other problems of the same nature.
We have the following conditions:
1. if counted in pairs, one will remain
2. if counted in threes, two will remain
3. if counted in fours, three will remain
4. if counted in fives, four will remain
5. if counted in sixes, five will remain
6. if counted in sevens, nothing will remain
Conditions 1 to 5 "happen" to be one less than a multiple of the counting number. Therefore, we can find a solution to conditions 1 to 5 by finding the LCM of the corresponding numbers 2,3,4,5,6 and subtract 1.
Since LCM(2,3,4,5,6)=3*4*5=60, we see immediately that 59 satisfies all of conditions 1 to 5. You should check this for yourself.
We can also see that for any positive integer k, k*LCM-1 will also share the same property. For example, for k=3, 60*3-1=179 is also a solution to conditions 1-5. Also, check this for yourself.
We conclude therefore, to satisfy condition 6, we only need to find (by trial and error), a value of k such that 7 divides (60*k-1), in which case (60k-1) will be the required number.
I will leave it to you to complete the problem. Please post any time if you have questions.