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

[sastry751]

ans is 33 max 420 min is 6 so M is 414 and then 414=2*3*3*23 subtracted these factors from 414squares got the answer

[dhruv.dingliwal]

For max LCM,
I took 3,4,5,6,7 and any two nos. that sum upto 6
I m getting 420 as LCM for that case.
Please come up with ur thoughts guys..

[getneonow]

Q)51
Max LCM for 23 by 6 numbers we get if we start with unique numbers whose GCD is 1. so lets take 2,3,5,7..the sum till now is 17..so to make 23 we have to look for two numbers whose sum is 6 and which contribute to the LCM..the contribution wud b max if we take the numbers as 2,4 . so numbers are 2,2,3,4,5,7 whose LCM is 420

Least LCM is for numbers 1,2,2,6,6,6 i.e LCM as 6

so M= 420-6 =414 divisors divide M but not M^2
=>divisors of 414^2 but not 414

414=2^1 * 3^2 *23^1 .so number of divisors =(1+1)(2+1)(1+1)=12
414*414 = 2^2 * 3^4 * 23^2. number of divisors=(2+1)(4+1)(2+1)=45
so 45-12=33 is the answer

[getneonow]

oops.. a small err i have committed..

for 414 = 2^1 * 3^2 * 23^1 ..number of divisors = 12
for 414*414 = 2^2 * 3^4 * 23^2.. number of divisors = 45

so 45-12 =33 (b) is the answer .

Cheers
Neo

[marijuana_user]

QQAD 51 ::

For minimum LCM Choose the numbers in such a way that contri is min.

One can choose 4 4 4 4 4 3 ----> LCM=12

or trying further 1 2 2 6 6 6 thus LCm---> 6
hence min LCM is 6

MAx LCM can be 420 for the combination 2 4 2 3 5 7

Difference = 420-6 = 414= M
Number of divisors of M=12(including 1 and 414)
M^2= 171396 = 2^2*3^4*23^2
number of divisors= 45(including all till 414 and greater than that)

hence answer 45-12=33

regards!

[Jaskiran]

very well explained by Neo...ans is 33(b)

[apocalypse_then]

peegeekay = pgk = praveen gopalakrishnan? this post is slightly impertinent to the issue at hand....sorry

[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

[vivekr]

Me too getting (b) 33. same approach as getneonow

[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 by trail & error. But I wonder if there exist ne proper solution method for this kinda problem. If nebody knows, Kindly do the favour by posting ASAP.

Regards
Arnab