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🧠 Rule:
The HCF (Highest Common Factor) of numbers must always divide the LCM (Least Common Multiple).

We are given:

• LCM of three different numbers = 120

• We are to find which number cannot be their HCF.

✅ Step 1: List the factors of 120
Factors of 120:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

So, any of these could be a potential HCF — but with a condition:

✅ Step 2: The HCF must divide all the numbers, and it must also be less than or equal to each of the numbers.

Now we need to pick from a list of options (not given directly), but let’s assume the question gives options like:

A) 1
B) 2
C) 3
D) 5
E) 30

Let’s analyze each:

• 1: Can be HCF of any set of different numbers. ✅

• 2, 3, 5: All are factors of 120 and valid possible HCFs. ✅

• 30: Also divides 120, so technically it could be an HCF, but let’s test further.

⛔ Test for 30:
If HCF = 30, then all three numbers must be multiples of 30. So examples: 30, 60, 90…

But:

• LCM of 30 and 60 is 60

• LCM of 30, 60, 90 = 90

→ But LCM of any such set is always at least 90 or more — and certainly will be a multiple of 30.

Now try 30, 60, and 120
→ LCM(30, 60, 120) = 120
HCF = 30
Valid ✅

So 30 can be the HCF.

But let’s try HCF = 40

• 40 is a factor of 120
  But can we find three different numbers whose HCF is 40 and LCM is 120?

Try multiples of 40: 40, 80, 120
→ LCM(40, 80, 120) = 240 → ❌

Try: 40, 60, 120
→ HCF = 20 → ❌

Try: 40, 80, 160 → LCM = 160 → ❌

You’ll find that HCF = 40 cannot give LCM = 120.

✅ So the correct answer is:

🚫 Final Answer: 40 cannot be the HCF of three numbers whose LCM is 120.