Seating Rows and Certificate Bundles: HCF, LCM and a Bad Shortcut
At a district science fair, three schools send 84, 126 and 210 students. The organisers must seat them in rows. Every row must contain the same number of students, all from one school, no student may be left out, and the row size must be as large as possible. At the registration desk, a stack of participation certificates causes another problem. Whether the clerk counts them in bundles of 8, 12 or 18, five certificates remain each time, and the total lies between 520 and 600. A volunteer says the row problem is solved by multiplying 84 × 126 × 210, and that the certificate total must be a multiple of 8 × 12 × 18. The gates open in an hour, and the organisers need numbers they can defend, not guesses.
1. Find the largest possible row size and the number of rows needed for each school. State whether the volunteer's multiplication rule is correct.
Read the worked answer
Write each roll in prime factors:
- 84 = 2² × 3 × 7
- 126 = 2 × 3² × 7
- 210 = 2 × 3 × 5 × 7
The largest row size that divides all three rolls is the HCF, taking the lowest power of each common prime:
HCF = 2 × 3 × 7 = 42.
So each row has 42 students. The rows needed are 84 ÷ 42 = 2, 126 ÷ 42 = 3 and 210 ÷ 42 = 5, giving 10 rows altogether.
The volunteer is wrong. The product 84 × 126 × 210 = 2,222,640 is far larger than any school's roll, so it cannot be a row size. A product equals the LCM only when the numbers are pairwise coprime; here the numbers share factors 2, 3 and 7. The actual LCM is 2² × 3² × 5 × 7 = 1260, still not the product. The correct operation for 'largest equal group' is HCF, not multiplication.
2. Find the number of certificates. Explain why the volunteer's second claim fails and why the stated range matters.
Read the worked answer
A number leaving remainder 5 on division by 8, 12 and 18 has the form N = LCM(8, 12, 18) × k + 5.
Prime factors: 8 = 2³, 12 = 2² × 3, 18 = 2 × 3². Taking the highest powers gives LCM = 2³ × 3² = 72.
So N = 72k + 5. Test multiples near the range:
- k = 7: 72 × 7 + 5 = 509, below 520
- k = 8: 72 × 8 + 5 = 581, inside 520 to 600
- k = 9: 72 × 9 + 5 = 653, above 600
Hence N = 581 certificates.
The volunteer's product 8 × 12 × 18 = 1728 is not the least common multiple. A multiple of 1728 plus 5 would be at least 1733, outside the range, and 581 does not satisfy the bundle condition for 1728. The range is essential: without it, 5, 77, 149, 221, … all work. Even with a looser range such as 500 to 600, both 509 and 581 would satisfy the remainder condition, so the tighter bound is what makes the answer unique.