Latest cocubes Aptitude Question SOLUTION: the numbers between (200,300) which are not divisible by 2,3,4,5

1) There are 101 integers (300 - 200 + 1 = 101). Since the set begins with an even and ends with an even, there are 51 evens.

2) Question says integers are not divisible by 2, leaving all of the odds (101 - 51 = 50 integers).

3) Question says integers are not divisible by 5, removing all the integers ending in 5 (already took out those ending in 0). Take out 10 integers (2?5, ? = 0 to 9), leaving us with 40 integers.

4) We have to remove the remaining numbers that are multiples of 3. Those are 201, 207, 213, 219, 231, 237, 243, 249, 261, 267, 273, 279, 291, and 297...a total of 14 numbers. 26 numbers left!

(1/2)*(2/3)*(4/5)*300 - (1/2)*(2/3)*(4/5)*200=26

It is a set problem.
There 101 numbers between them.
51 of them are divisible by 2 (first 200, last 300)
34 of them are divisible by 3 (first 201, last 300)
21 of them are divisible by 5 (first 200, last 300)
11 of them are divisible by both 2 and 5 (first 200, last 300) (its divisibility by ten)
17 of them are divisible by both 2 and 3 (first 204, last 300) (its divisibility by six)
7 of them are divisible by both 3 and 5 (first 210, last 300) (its divisibility by fifteen)
4 of them are divisible by all (210,240,270,300)
So,
4 of them are divisible by all
3 of them are divisible by 3 and 5 but not by 2
13 of them are divisible by 2 and 3 but not 5
7 of them are divisible by 2 and 5 but not by 3
This part is tricky:
For 2; (51 numbers total); 4 divisible by all, 13 of them are divisible by 2 and 3 but not 5, 7 of them are divisible by 2 and 5 but not by 3 => that makes 27 are divisible by only 2 (51-4-13-7)
For 3; (34 numbers total); 4 divisible by all, 13 of them are divisible by 2 and 3 but not 5, 3 of them are divisible by 3 and 5 but not by 2 => that makes 14 are divisible by only 3 (34-4-13-3)
For 5; (21 numbers total); 4 divisible by all, 3 of them are divisible by 3 and 5 but not 2, 7 of them are divisible by 2 and 5 but not by 3 => that makes 7 are divisible by only 5 (21-4-3-7)
Totaling them;
4 + 3 + 13 + 7 + 27 + 14 + 7 = 75 are divisible by either 2, 3 or 5.
So 26 are not divisible by any.

it means they are askng prime numbers and numbers divisible 7 203,209,211,217,219 etc so there are 26

26 numbers that are not divisible by 2,3,4,5

1. Sample set 201-299=99
2. Removing multiples of 2, equals 49
3. Removing odd multiples of 3, equals 17
4. Removing all prime multiles of 5, (41,43,47,49,53,55,59) equals 7

99-49-17-7=26

i think there are 25 numbers that are not divisible by 2,3,4,5.

total = 99;
multiples of 2 = 49;
multiples of 3 = 33;
multiples of 5 = 19;
Added More;
multiples of 6(2 x 3) = 16;
multiples of 10(2 x 5) = 9;
multiples of 15(3 x 5) = 6;
Deleted more;
multiples of 30(2 x 3 x 5) = 3;
Answer = 99 - 49 - 33 - 19 + 16 + 9 + 6 - 3;
Answer = 26;

203,205,209 qre not divisible by 2,3,4,5

105*95/100