Thread: Hcf/lcm

How do I find the smallest 4-digit number which is divisible by 18, 24 & 32 using HCF/LCM?

Thanks,

Ron

Originally Posted by rn5a

How do I find the smallest 4-digit number which is divisible by 18, 24 & 32 using HCF/LCM?

Thanks,

Ron

Calculate HCF and LCM.

LCM will be a three digit number. It can be converted to four digits by multiplying by 10. You can minimize the number, dividing it HCF if possible.

Originally Posted by sa-ri-ga-ma

Calculate HCF and LCM.

LCM will be a three digit number. It can be converted to four digits by multiplying by 10. You can minimize the number, dividing it HCF if possible.

I don't understand why you multiply by 10.

Let the lcm be called L, it is indeed 3 digits, so take . We never need the gcd (aka hcf).

Originally Posted by undefined

I don't understand why you multiply by 10.

Let the lcm be called L, it is indeed 3 digits, so take . We never need the gcd (aka hcf).

Hear the LCM has three digits which is divisible by all the three given numbers. But the problem requires smallest 4 digit number divisible by the given numbers. That is why I multiplied by 10 and to make it still smaller, I divided it by HCF

Originally Posted by sa-ri-ga-ma

Hear the LCM has three digits which is divisible by all the three given numbers. But the problem requires smallest 4 digit number divisible by the given numbers. That is why I multiplied by 10 and to make it still smaller, I divided it by HCF

Doesn't that give 1440 instead of 1152? Still don't really understand your reasoning. My reasoning is: Disregarding the "smallest 4 digit number" part, the entire solution set is given by k*L where L is the lcm and k is any integer. So find the smallest k such that k*L is 4 digits.

You are right. Thank you.