Not so easy math

#1

How to do this question with bitmasking?

#2

Hi Anant
You need to use inclusion-exclusion principle mainly.
The same way you do it for questions like find number of numbers div by 2&3.
You find number of numbers div by 2+ number of numbers div by 3 - number of numbers div by 6.
Same way!
Bitmasking:
00 div by none
01 div by 2 only
10 div by 3 only
11 div by both 2 and 3

I hope you can relate.
Similarly in the question, you shall have mask with number of bits equal to number of prime numbers less than 20.

#3

I didn’t get it completely. Can you please explain in detail.

#4

if you have a,b,c,d,e as your prime numbers and you want to make combinations, that lets take abc, lets now take bde…
You can make all combinations using a mask.
Numbers from 0 to 31 (in binary form) would help denoting which one to take.

for eg.
10110 = take a,c,d
01111 = take b,c,d,e
00000 = take none
11111 = take a,b,c,d,e

in this way bitmasking can be used.

#5

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