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Bunuel
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Kiara229
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Hey , can you elaborate this method a bit?

Thanks.

Kiara229
IMO D

I solved the problem by doing the prime factorization of 42 - 2,3,7 and used all combinations of products with the integer 3. Therefore, I came to 4 multiples of 3 that can divide 42 evenly. (3,2*3,3*7,2*3*7)
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Hi Sameer1,

The method Kiara229 used is a slick shortcut, so let me unpack the logic behind it.

Start with the prime factorization: 42 = 2 × 3 × 7.

The key idea: a divisor of 42 that is a multiple of 3 must contain that 3 as one of its building blocks. If it didn't have a 3 in it, it couldn't be a multiple of 3. So lock the 3 in place first.

Now the only freedom you have is what to attach to that 3, using the leftover primes {2, 7}. For each leftover prime you make one simple yes/no choice - include it or not:

- 3 alone → 3
- 3 × 2 → 6
- 3 × 7 → 21
- 3 × 2 × 7 → 42

That's 2 choices for the 2 (in or out) × 2 choices for the 7 (in or out) = 4 divisors. Every one of them divides 42 and is a multiple of 3, which gives you answer D.

This is exactly the same list Aravind04 got by writing out all 8 divisors and picking the ones divisible by 3 - the prime-factor method just skips straight to them.

Quick one to lock it in: how many divisors of 30 are multiples of 3?

- 30 = 2 × 3 × 5. Fix the 3, then choose freely from the leftovers {2, 5}: 2 × 2 = 4 of them (3, 6, 15, 30).

Same move every time: nail down the required prime, then count the on/off combinations of what's left.

Answer: D

Sameer1
Hey , can you elaborate this method a bit?

Thanks.


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