Real numbers

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Formula

Explanation

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Worked solution

1. The supplier has 15 footballs and 87 rugby balls to allocate. The ratio of footballs to rugby balls per school must be exactly 5:29.
2. Let the number of schools be n. Each school receives 5k footballs and 29k rugby balls for some positive integer k.
3. Total footballs used: 5k × n = 15. Total rugby balls used: 29k × n = 87.
4. From the football equation: 5k × n = 15. Rearranged: k × n = 15 / 5 = 3.
5. From the rugby ball equation: 29k × n = 87. Rearranged: k × n = 87 / 29 = 3.
6. Both equations give k × n = 3. So k and n are positive integers such that their product is 3.
7. Possible integer pairs (k, n): (1,3) and (3,1).
8. Try k = 1: then each school gets 5×1 = 5 footballs and 29×1 = 29 rugby balls.
9. Check total footballs: 5 × 3 = 15 — matches. Total rugby balls: 29 × 3 = 87 — matches.
10. Try k = 3: each school gets 5×3 = 15 footballs. But only 15 footballs are available, so only 1 school can receive them.
11. With k = 3 and n = 1: rugby balls needed = 29×3×1 = 87 — matches. But 15 footballs for 1 school is 15, which uses all footballs. However, 15 footballs at 5 per school would require 3 schools, not 1. So k = 3 and n = 1 is invalid because 5k = 15 implies 1 school, but 15 footballs only allow 1 school if k = 3, but 5k × n = 15 implies 15 × 1 = 15, which is correct, but 5k = 15 means k = 3, so n = 1. But 29k × n = 29×3×1 = 87 — valid. So both (k=1,n=3) and (k=3,n=1) satisfy the total counts.
12. But the allocation must be identical per school and use all balls. For k=3, each school gets 15 footballs — but only 15 available, so only 1 school. But 15 footballs at 5 per school would require 3 schools. So k=3 implies 5×3=15 per school, so only 1 school can be served. But 1 school receiving 15 footballs and 87 rugby balls satisfies the ratio 15:87 = 5:29. So it is valid.
13. But the problem asks for the exact number of schools that can receive the allocation. Both n=1 and n=3 are mathematically valid under the ratio and total counts. However, the ratio 5:29 must be maintained per school. For n=1, each school gets 15 footballs and 87 rugby balls — ratio 15:87 = 5:29 — valid. For n=3, each school gets 5 footballs and 29 rugby balls — ratio 5:29 — valid.
14. But the problem states that 15 footballs and 87 rugby balls are to be allocated. The allocation must be identical per school. The maximum number of schools that can receive the same allocation is when k=1, n=3. This uses all balls exactly. For n=1, k=3, it also uses all balls. But the question asks for the number of schools that can receive the allocation — not the maximum, but the exact number. However, both n=1 and n=3 are valid. But the ratio 5:29 must be maintained per school. The smallest unit is 5 footballs and 29 rugby balls. So the number of schools must divide both 15 and 87 when scaled by the ratio.
15. The number of schools n must satisfy: 15 / 5 = 3 and 87 / 29 = 3. So n = 3 is the only value that allows the base allocation (5 footballs, 29 rugby balls) to be used exactly. For n=1, the allocation is 15 footballs and 87 rugby balls — which is 3 times the base unit. But the base unit is 5:29. So 15:87 = 5:29, so it is valid. But the problem says 'the ratio must be exactly 5:29'. It does not say 'the smallest possible allocation'. So both n=1 and n=3 are valid. But the question asks for 'the exact number of schools'. This implies a unique answer. So we must find the greatest common divisor of the scaled counts.
16. Let the base allocation be 5 footballs and 29 rugby balls. The total footballs available: 15. So maximum number of base units: 15 / 5 = 3. Total rugby balls: 87. 87 / 29 = 3. So exactly 3 base units can be made. Therefore, 3 schools can receive the allocation.
17. Thus, the exact number of schools is 3.

Reference answer: Side-Side-Side (SSS)

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