alfat wrote:find out total number of integer between 100 & 200 that are divisible by3 & not todivisible by 5 nor 7!!
Okay. The total number of such integers will be given by:
div3 - div15 - div21 + div105
I'll explain. Numbers that are divisible by both 3 and 5 are divisible by 15, and vice-versa. Numbers that are divisible by both 3 and 7 are divisible by 21, and vice-versa. Numbers that are divisible by 3, 5 and 7 are divisible by 3 * 5 * 7 = 105, and vice-versa.
Therefore, to find the "total number of integers between 100 & 200 that are divisible by 3 & not divisible by 5 nor 7" we need to:
1. Find the number of integers between 100 and 200 that are divisible by 3;
2. Subtract from this the total number of integers between 100 and 200 that are divisible by 15;
3. Subtract from this the total number of integers between 100 and 200 that are divisible by 21;
4. Now, we have effectively subtracted the total number of integers between 100 and 200 that are divisible by 105 TWICE. This is bad. We don't want to double-discount! Therefore, we must add that figure back in.
So... what now? Well, the smallest multiple of 3 in the range we're talking about is 102. The largest is 198. To find the number of multiples of 3 in between and including these two, we take:
(198 - 102)/3 + 1 = 96/3 + 1 = 32 + 1 = 33
The smallest multiple of 15 in the range we're talking about is 105. The largest is 195. To find the number of multiples of 15 in between and including these two, we take:
(195 - 105)/15 + 1 = 90/15 + 1 = 6 + 1 = 7
The smallest multiple of 21 in the range we're talking about is 105. The largest is 189. To find the number of multiples of 21 in between and including these two, we take:
(189 - 105)/21 + 1 = 84/21 + 1 = 4 + 1 = 5
And there is only one multiple of 105 on this range, and that is 105, since the next one up will be 210. So, we have:
33 - 7 - 5 + 1 = 22
