MATH RIDDLE HELP!!!!!!???

One thing to notice is that if you were to add 1 apple to the barrel, then you'd have an exact multiple of 2, 3, 4, 5 and 6.

The lowest common multiple of 2, 3, 4, 5 and 6 is 60
2 x 30 = 60
3 x 20 = 60
4 x 15 = 60
5 x 12 = 60
6 x 10 = 60

This will be true of any multiple of 60 (e.g. 60k, where k is an integer). But we added one apple to do that, so we must subtract it out.

The number of apples in the bin is 60k - 1 and it is a multiple of 7.

Try values of k and see which works:
k = 1 --> 60-1 = 59 (nope)
k = 2 --> 120 - 1 = 119 (yes!)

Double-check:
119 / 2 = 59 remainder 1
119 / 3 = 39 remainder 2
119 / 4 = 29 remainder 3
119 / 5 = 23 remainder 4
119 / 6 = 19 remainder 5
119 / 7 = 47 remainder 0

Answer:
The smallest number of apples is 119 apples.

Obviously there is an odd number of apples in the bin.
Let x = number of apples

So what number is evenly divisible by 7 and not by any of the other numbers?
Number must be odd and not be 21, 35, 63. Let's try 7 x 7 = 49
49/2 = 24 + 1
49/3 = 16 + 1 (b + 2?) does not work.
77/2 = 38 + 1 left over
77/3 = 25 + 2 left over
77/4 = 19 + 1 left over (does not work)
91/2 = 45 + 1 left over
91/3 = 30 + 1 left over (does not work)
119/2 = 59 + 1 left over
119/3 = 39 + 2 left over
119/4 = 29 + 3 left over
119/5 = 23 + 4 left over
119/6 = 19 + 5 left over.
So there are 119 apples in the bin
119/7 = 17.
I don't know of another way of doing this other than trial & error.

Number of apples is congruent to 1 (mod2), 2(mod3), 3(mod4), 4(mod5), 5(mod6) and 0 (mod7). The congruences (mod 4) and (mod 6) show that upon division by 12, there will be 11 left over. The congruences (mod 2) and (mod 3) add no further information, but the congruence (mod 5) shows that upon division by 60, there will be a remainder of 59, or to put it more simply, a remainder of -1. So now the question is, which multiple of 7 is just 1 less than a multiply of 60. A good bet is 119. It's 7*17. It seems to satisfy all the criteria.