sadullaevd wrote:
Can you plz explain how you came to the answer? Why didnt you try to look at even number? Im stuck, because i thought we would need some more numbers to test.
There is a way to find out no of factors of a number say n
If it can be expressed as n = x^a*y^b*z*c.....
Where x,y,z...each are prime numbers and a,b,c,d ... etc are integers
No of factors of that number n = (a+1)(b+1) .....
I will demonstrate it by an example 12 = 2^2*3^1
No of factors of 12 = (2+1)(1+1) = 6
And let us try to list them - 1,2,3,4,6 and 12
That is the background you need to attempt this question
Now option 2 says - 2n is divisible by twice as many positive integers as n
Now imagine when n is odd, if you multiply by 2 the new number has a new prime factor 2 and it will have 2 (1+1) times more factors and if n is even, this factor is not 2 (b becomes b+1) and the factor is b+1/b
I will again demonstrate by example
12 =2^2*3^1 ; no of factors = 2*3 =6
24 = 12*2 = 2^3*3^1 , no of factors = 4*2 =8
(I hope you notice number of factors did not double)
Now let us take an odd number 15
15 = 3*5 ; no of factors = 2*2 =4
30 = 2*3*5 ; no of factors = 2*2*2 = 8
Hope you understood, I am glad that you asked the question
Always borrow money from a pessimist, he doesn't expect to be paid back.
