For All Your Quant Queries
with 7 comments
Written by Implex
July 30, 2009 at 1:03 pm
Posted in Algebra, CAT 2009, Mock Cat, Mock Quant, Number Thoery, Online CAT, Problem of the week, Problems
Tagged with cat, IIM, Number Theory, Online CAT
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August 1, 2009 at 5:16 pm
can u plz explain the method ?
August 3, 2009 at 11:51 am
I used hit and trial, guessed it would be the quickest approach to solve it.
Start with 1,2,3,4 then 1,2,3,6 (which fetches the answer)
August 4, 2009 at 7:10 pm
hit N trial method N u will get 126.
August 10, 2009 at 10:09 am
If the number is even then 1 and 2 are gonna be its two divisors.The remaining 2 divisors could be p and 2p for sure then(taking p as a prime number).We can check p for 3,5,7 and 3 will give the answer by meeting all the conditions.
A nkur Dudeja
August 18, 2009 at 1:56 pm
Rahul pls put the approach to this question and answer to it..
August 20, 2009 at 7:50 pm
Let the least number be N, 1 is its least divisor.Let 2nd,3rd and 4th least divisors be x,y and z respectively. We consider the following values of divisor a and the corresponding values of a^3, from x,y and z exactly 1 or all 3 are odd.(N is even) a=1:a^3=1 a=2:a^3=8 a=3:a^3=27 a=4:a^3=64 a=5:a^3=125 a=6:a^3=216 For x, y and z=(2,3,4),2*N=100(i.e.N=50). But 3 is not a divisor of 50. For x,y,z=(2,3,6),2*N=252(i.e.N=126) and the 1,2,3,6 are four least distinct divisor of 126.The required number is 126. The sum of digits is 9.
August 15, 2013 at 5:01 pm
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