integers

Yikes....

I get 38. I must have missed something...

Between 200 and 300 inclusive there are a total of 101 numbers

Multiples of 2

2*100 = 200
2*150=300

Total of 150-100+1 = 51 numbers

Multiples of 3

3*67=201
3*100=300

100-67+1 = 34

Multiples of 5

5*40
5*60


21 numbers multiples of 5

Total: 106


But multiples of 6 would have been double counted for 2 and 3.

So multiples of 6 are 6*34
6*50

17 such numbers

Multiples of 10 would have been double counted for 2 and 5

10*20
10*30

11 such numbers


Multiples of 15 would have been double counted for 3 and 5

15*14
15*20

7 such numbers

Multiples of 30 would have been triple counted for 2,3 and 5

30*7
30*10

4 such numbers


using 3 sets formula

101 = 106 -17-11-7-2(4) + None

None = 38

Divisible by 2 = 1 + (300-200)/2 = 51

Divisible by 3 = 1 + (300 - 201)/3 = 34

Divisible by 5 = 1 + (300-200)/5 = 21

Divisible by 2 & 3 = 1 + (300 -204)/6 = 17

Divisible by 3 & 5 = 1 + (300 -210)/15 = 7

Divisible by 2 & 5 = 1 + (300 -200)/10 = 11

Divisible by 2 & 3 & 5 = 1 + (300 -210)/30 = 4

Divisible by 2 or 3 or 5 = 51 + 34 + 21 -17 -7 -11 + 4
= 110 - 35 = 75

Not divisible by 2,3,5 = 101 - 75 = 26

Thanks
raama

Realized my mistake.

The answer should be 26.....


All the multiples of 2 cannot be part of the answer so 101-51 = 50

So 50 is the maximum we can get. From this 50 we have to subtract all unique multiples of 3 and 5

3*67 = 201
3*100 = 300

34/2 = 17 odd multiples

5*40
5*60

10 odd multiples


3 and 5 double counted 7 times but only 3 odd multiples

Unique multiples of 3 and 5 27-3 = 24

So 50-24 = 26

I think the OA's may be incorrect.....

Regards,
CR