find the number of 4 digit positive intergers divisible by either 3 or 7.?

Answer 1

Isn't it something like 10000/3 + 10000/7 - 10000/21

The last part is because numbers divisible by 3 and 7 will be counted twice. Be sure to round each calculation down.

I agree with rahidz2003 that 4 digits would be 1000 to 9999, unless you write your numbers with leading zeros ( never mind).

So you would need to subtract the divisible numbers in the first 1000 or 999/3 + 999/7 - 999/21 from the original calculation. However I get a total result of 3857 for the answer

Answer 2

The first number > 4 digits is 1002 for 3 (3*343
The first number > 4 digits is 1001 for 7 (7*143)
The last number > 4 digits for 3 is 9999 (3*3333)
The last number > 4 digits for 7 is 9996 (7*1428)

For 3 do the following it is (9999-1002)/3 = 8997/3= 2999
For 7 do the following it is (9996-1001)7=8995/7=1285

2999+1285 = 4284

Answer 3

Hmmm...4 digit integers are from 1000 to 9999 (9000 of them)

Divisible by 3 (1002, 1005, ... ,9999) : 3000 of those
Divisible by 7 (1001, 1008, ... ,9996) : 1286 of those
Divisible by both 3 and 7 (1008, ... ,9996) : 429 of those

So, 9000 - (3000 + 1286) - 429 =

4285 numbers.

I think.