Here is the problem. I just have no clue on how to solve this without spending hours writing them all out.
Find the number of four-digit positive integers divisible by either 3 or 7.
Here is the problem. I just have no clue on how to solve this without spending hours writing them all out.
Find the number of four-digit positive integers divisible by either 3 or 7.
Assume that by “four-digit positive integers” you mean 1000-9999.
The number of integers 1-9999 divisible by 3 is floor(9999/3)=3333.
The number of integers 1-999 divisible by 3 is floor(999/3)=333.
Therefore, the number of integers 1000-9999 divisible by 3 is 3333-333=3000.
The number of integers 1-9999 divisible by 7 is floor(9999/7)=1428.
The number of integers 1-999 divisible by 7 is floor(999/7)=142.
Therefore, the number of integers 1000-9999 divisible by 7 is 1286.
The number of integers 1000-9999 divisible by 21 is 429.
The number of integers 1000-9999 divisible by 3 or 7 is 3000+1286-429=3857.
There are 9000 four digit numbers 3000 of them are divisible by 3 and 1286 of
them are divisible by 7, and 429 are divisible by both (that is by 21).
Therefore 3000+1286-429 are divisible by either 3 or 7 (the 429 are
subtracted because these are the numbers divisible by both 3 and 7 and so
will have been counted in both the 3000 and 1286).
You will need to check that I have the number of four digit numbers
divisible by 3, 7 and 21 correct!
RonL
I am confused on something here. The mathematical "or" statement is true when both statements are true. So shouldn't we be including rather than excluding any 4 digit number that is divisible by both 3 and 7? (ie if it is divisible by 21) I realize that using "English" rules "either/or" statements do mean one or the other, not both. Which should we be using here?
(This kind of question has always bugged me.)
-Dan