Find the greatest number of $6$ digits exactly divisible by $24, 15$ and $36$.
Step-by-Step Solution
Key Concept: Find $\text{LCM}(24, 15, 36) = 360$. Divide largest 6-digit number $999999$ by $360$, and subtract the remainder.
Stepwise Solution:
Prime factorisation: $24 = 2^3 \times 3, 15 = 3 \times 5, 36 = 2^2 \times 3^2$.
$\text{LCM}(24, 15, 36) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$. [1.0 Mark]
Largest 6-digit number is $999999$.
$999999 / 360 = 2777$ with a remainder of $279$. [0.5 Mark]
Required number $= 999999 - 279 = 999720$. [0.5 Mark]
Marking Scheme:
• Calculating $\text{LCM}(24, 15, 36) = 360$: 1.0 Mark
• Dividing 999999 by 360 to find remainder 279: 0.5 Mark
• Subtracting remainder to get 999720: 0.5 Mark
Correct Answer: