Random question

#1

Given a number X which denotes the number of divisors…by using this i have to find a number(smallest) N which is having X divisors?can you help me how to deal with this?

#2

@dare_devil_007 hey one approach is to start from 1 and check number of divisor of number and if it matches X than print it,another approach is like this:

#3

dude you misunderstood the question… the question was given the number of divisors x, find the number N which will be having the x divisors

#4

@dare_devil_007 hey didnt get question ,yhi hai na ki minimum no having no of divisors as X.

#5

lekin mujhe uss link mai code smjh aya h leking usa time complexity zyada h and i want a optimized approach

#6

@dare_devil_007 yhi approach hai jo hm kr skte hai else apko starting se start krke no of divisors check krne pdenge and no of divisors find krne ka efficient way ye rha:

#7

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