(Original post by mik1a)
n = number of students. n < 30.
3 in 4 own footballs, so 3n/4 is an integer.
7 in 8 own football boots, so 7n/8 is also an integer.
there are exactly 3 values on n for which both 3n/4 and 7n/8 are both integers:
n = 24, 16 and 8
Proof? Well, 3/4 = 6/8 and because the second fraction is 7/8, we can say that if there are s values for n where 7n/8 is an integer (and n is an integer), then there must also be s values for n where 6n/8 is an integer (and n is an integer). So we only need to worry about 7n/8.
So if n and 7n/8 are both integers, and n is less than 30, n must be a multiple of 8. If it is, then the division result will be in integer and this multiplied by 7 is also an integer. If it isn't, then the division result will not be an integer, and since 7 is prime, there are no non-integers which multiply by 7 to give integers. There are 3 multiples of 8 less than 30, since 8*4 = 32. They are 8, 16, and 24, whose factors with respect to 8 are 1, 2 and 3.
Hope this help, which it probably doesn't. The fact is there are 3 answers for the first problem.