[arnabsadhu]

Originally Posted by v-factor

LCM minimum is = 6 (6,6,6,2,2,1)
LCM maximum is = 420 (7,5,4,3,2,2)

M = 414 = 2 * 3^2 * 13
No. of factors is (1+1)*(2+1)*(1+1) = 12

M^2 = 414^2 = 2^2 * 3^4 * 13^2
No. of factors is (2+1)*(4+1)*(2+1) = 45

So answer is 45-12= B)33

Ya I got the same answer too; but by trial & error only. I wonder if ne proper solution exist for such kinda problem. If ne buddy knows that kindly do the favour by posting that.

Regards.
Arnab

[pratyush_sinha]

There exists a method for this.

For maximising the LCM, one has to take the co-primes . In this case 5,6,7 are natural coprimes. The sum is still 18. So to get 5 more break 6 into 3 and 2 and take 4 and 1
So final set 2,2,3,4,5,7==420

To minimise the LCM take numbers those are factors of each other. My first choice would have been 2,4 but taking them I get 1,2,4,4,4,8 which gives the LCM as 8. Then immediately you should think that 6 can be one of the LCMs. Strat trying with 3.6
you get 2,3,3,3,6,6 =6

[MG_1806]

Ans b)33. Method same as others.The key to the problem is to find distinct co-primes so as to maximize the LCM.

[macora]

My answer is also 33.

Getting the smallest and largest LCMs and then solving it.

[thebigshow]

peegeeguy wrote:


If we consider the 6 numbers as 4,4,4,4,4 and 3 (Since the numbers do not have to be distinct), the LCM is 12 which is I believe the minimum possible LCM.



consider the numbers as 5,5,5,5,2,1.......... the lcm in this case is 10

[Aarav]

Originally Posted by thebigshow

peegeeguy wrote:

If we consider the 6 numbers as 4,4,4,4,4 and 3 (Since the numbers do not have to be distinct), the LCM is 12 which is I believe the minimum possible LCM.

consider the numbers as 5,5,5,5,2,1.......... the lcm in this case is 10

You can construct 8 with 8, 4, 4, 4, 2, 1. You can construct 6 with 6, 6, 6, 2, 2, 1. Can't better 6.

The answer is 33 indeed. Well done all those who solved this successfully.

[vivekr]

Hi ppl, am not sure if this is the place to ask this. But, is there a compiled list of last year's QQAD questions somewhere?
I know abt the QQAD 2006 revisited forum, but it'd be helpful if we are able to download the questions with solutions.

[Jaskiran]

Originally Posted by vivekr

Hi ppl, am not sure if this is the place to ask this. But, is there a compiled list of last year's QQAD questions somewhere?
I know abt the QQAD 2006 revisited forum, but it'd be helpful if we are able to download the questions with solutions.

Hey Vivek I think u havent checked d 1st page of dis thread...al u need is dere only...

[pratz]

hi friends,

i think tht the answer to above problem is e)none of the foregoing

though the min and max LCM remains 6 and 420 respectively.

but factors of 420-6=414=2x3x3x23

and for 420^2 = 2x2x3x3x3x3x23x23

So, the answer to the problem can be 12,36...etc and not 22,33,18,26

bye n hv a nice day

[targetABC]

Hi,

Someone with a method of finding the max. and min. LCM other than the methods already discussed?

Thanks.