There are a number of chocolates in a bag. If they were to be equally divided…
2026
There are a number of chocolates in a bag. If they were to be equally divided among 14 children, there are 10 chocolates left. If they were to be equally divided among 15 children, there are 8 chocolates left. Obviously, this can be satisfied if any multiple of 210 chocolates are added to the bag. What is the remainder when the minimum feasible number of chocolates in the bag is divided by 9?
- A.
2
- B.
5
- C.
4
- D.
6
Attempted by 629 students.
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Correct answer: A
Let the number of chocolates be N.
Condition 1: When divided among 14 children, 10 are left, so N ≡ 10 (mod 14). Write N = 10 + 14a.
Condition 2: When divided among 15 children, 8 are left, so substitute into N ≡ 8 (mod 15):
10 + 14a ≡ 8 (mod 15)
14a ≡ -2 ≡ 13 (mod 15)
Since 14 ≡ -1 (mod 15), we get -a ≡ 13 (mod 15), so a ≡ 2 (mod 15).
Minimum value: take a = 2, so N = 10 + 14×2 = 38.
Required remainder: 38 = 4×9 + 2, so the remainder is 2.
The note about adding multiples of 210 gives other valid numbers, but the question asks for the minimum feasible number.