Not so easy math

#1

bitmasking question
please provide some hint to deal with time complexity

#2

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.

#3

after getting all mask how i will get to answer

#4

How many numbers from 1 to 100 are div by 3 ?
Thats equal to 100/3.
Use the formula N/a to get the number of numbers till N which are divisible by a.

#5

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