1. ## Hcf/lcm

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

Thanks,

Ron

2. 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.

3. 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 $L\cdot\lceil\frac{1000}{L}\rceil$. We never need the gcd (aka hcf).

4. 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 $L\cdot\lceil\frac{1000}{L}\rceil$. 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

5. 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.

6. You are right. Thank you.