In Hiring Worker why does AM>=GM fail?
e.g.
Let the numbers be x1,x2…xn.
Let x1x2…xn =p(given)
X1+x2+x3+…xn=s .
Using am & gm inequality
(X1+x2+…+xn)/n>=(x1x2…xn)^1/n
S/n>=p^(1/n).
S>=n*p^(1/n).
In Hiring Worker why does AM>=GM fail?
e.g.
Let the numbers be x1,x2…xn.
Let x1x2…xn =p(given)
X1+x2+x3+…xn=s .
Using am & gm inequality
(X1+x2+…+xn)/n>=(x1x2…xn)^1/n
S/n>=p^(1/n).
S>=n*p^(1/n).
Problem can be reframed as :
Find K numbers such that their LCM is x and their sum is minimum .
Brute-force/dp can be used to find it .(As there are only 8 distinct primes)
Me too thinking about the same.
8 distinct primes how?
The maximum size of x is 1e6 so and multiplying primes more than the first 8 primes itself leads to a larger value than 1e6 , so maximum number of primes will be 8.
I understood this but couldn’t code
I guess my solution could have worked if I had atleast thought it this way
can we reframe it like this too ?
"Find K numbers such that their product is X and their sum is minimum "
Say, you have to find 3 numbers whose LCM is 18 and sum is minimum .
But you, say product is 18 and sum is minimum ,
you find 3 numbers : 3 * 6 * 1 (say) whose product is 18 and sum is 10 . But, that is wrong because LCM is (3,6,1) = 6 ! = 18
That’s where I was wrong , thankyou for clarifying it !