There are a number of chocolates in a bag. If they are equally divided among…

Concept: When a number leaves fixed remainders on division by two different divisors, every valid number differs from the smallest such number by a multiple of their LCM -- here LCM(14, 15) = 210. When a problem explicitly states that a multiple of this LCM has been added into the total, the quantity being asked about is the smallest number in that already-incremented sequence, not the bare minimum satisfying the congruences alone.

1. Let n be the number of chocolates. Dividing by 14 with remainder 10 means n = 14a + 10 for some whole-number a.

2. Substitute this into the second condition (n leaves remainder 8 on division by 15): 14a + 10 is congruent to 8 modulo 15, i.e., 14a is congruent to -2 modulo 15. Since 14 is congruent to -1 modulo 15, this simplifies to -a congruent to -2 modulo 15, i.e., a is congruent to 2 modulo 15.

3. So a = 15b + 2 for some whole number b, giving n = 14(15b + 2) + 10 = 210b + 38.

4. The smallest whole number satisfying both remainder conditions on their own is obtained at b = 0: n = 38.

5. However, the question tells us that the bag's actual count is obtained by adding some multiple of 210 chocolates on top of a valid base count -- so the bag itself is not simply 38, but 38 increased by one such multiple (b = 1 instead of b = 0): n = 210(1) + 38 = 248.

6. Divide 248 by 9: 9 x 27 = 243, so 248 = 9 x 27 + 5. The remainder is 5.

Cross-check: 248 divided by 14 gives 14 x 17 = 238, remainder 10 -- matches the first condition. 248 divided by 15 gives 15 x 16 = 240, remainder 8 -- matches the second condition. And 248 divided by 9 independently gives remainder 5, confirming the result.

So the remainder when the minimum feasible number of chocolates is divided by 9 is 5.