Oranges are packed into boxes of two types: box type A holds 10 oranges each,…
2024
Oranges are packed into boxes of two types: box type A holds 10 oranges each, and box type B holds 25 oranges each. A carton contains a total of 1000 oranges, packed using any number (including zero) of type A boxes and any number (including zero) of type B boxes. In how many different combinations of the number of type A and type B boxes can these 1000 oranges be packed?
- A.
19
- B.
18
- C.
21
- D.
20
Attempted by 6 students.
Show answer & explanation
Correct answer: C
Concept: This is a linear Diophantine equation of the form ax + by = c. When gcd(a, b) divides c, the non-negative integer solutions form a family: one variable steps forward by b/gcd(a, b) at a time, and each such step pairs with exactly one non-negative integer value of the other variable.
Setting up: Let A be the number of type A boxes (10 oranges each) and B be the number of type B boxes (25 oranges each). The total gives 10A + 25B = 1000.
Divide the equation by the common factor 5: 2A + 5B = 200.
Solve for B: B = (200 - 2A) / 5. For B to be a non-negative integer, (200 - 2A) must be divisible by 5 and non-negative.
200 is already divisible by 5, so 2A must also be divisible by 5. Since gcd(2, 5) = 1, this forces A itself to be a multiple of 5.
A must also satisfy 0 <= A <= 100 (otherwise B would go negative). So A can take the values 0, 5, 10, 15, ..., 100.
Counting the terms of this arithmetic sequence: (100 - 0)/5 + 1 = 21 values of A, each pairing with exactly one non-negative integer value of B.
Cross-check: Check both ends of the range: A = 0 gives B = (200 - 0)/5 = 40, and 0(10) + 40(25) = 1000. A = 100 gives B = (200 - 200)/5 = 0, and 100(10) + 0(25) = 1000. Both boundary cases hold, confirming every one of the 21 values of A is a valid combination.
Result: There are 21 different combinations of type A and type B boxes that pack exactly 1000 oranges.