CAT 2007: Quantitative Questions a Day 51 to 100 - The Discussions

arnabsadhu

Quote:

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

Quote:

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

Quote:

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.