Mathematics — CBSE Class 10 board question

Step 1: Prime factorise each number.

$48 = 2^4 \times 3$
$60 = 2^2 \times 3 \times 5$
$65 = 5 \times 13$

Step 2: Find LCM.

LCM $(48, 60, 65) = 2^4 \times 3 \times 5 \times 13 = 16 \times 3 \times 5 \times 13 = 3120$

Step 3: Find the greatest multiple of 3120 less than 10,000.

$10000 \div 3120 = 3.205…$

So the greatest multiple $= 3 \times 3120 = \mathbf{9360}$

The greatest number less than 10,000 exactly divisible by 48, 60 and 65 is 9360.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

---

• A number divisible by all three given numbers must be a multiple of their LCM.
• Use prime factorisation (Fundamental Theorem of Arithmetic) to find LCM: take the highest power of every prime factor appearing.
• Then divide 10,000 by the LCM, take the floor of the quotient, and multiply back.
• The key step students miss is verifying it is less than 10,000 (not equal), so use $\lfloor 10000/\text{LCM} \rfloor \times \text{LCM}$ — here that gives 9360, not 10,000 itself.