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.

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- Sep 28th 2006, 01:31 PMceasar_19134Need Help with numbers
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.__ - Sep 28th 2006, 01:58 PMPlato
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. - Sep 28th 2006, 02:01 PMCaptainBlack
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 - Sep 28th 2006, 02:36 PMtopsquark
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 - Sep 28th 2006, 02:58 PMPlato
If |A| stands for the number of elements in set A, then |AuB|=|A|+|B|-|A^B|.

We subtract off those elements that we have counted twice.

Both of us did subtract the number of multiples of 21. - Sep 28th 2006, 03:08 PMtopsquark