How many two digit numbers are divisible by either 3 or 5?
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Question

How many two digit numbers are divisible by either 3 or 5?
@satellite73

anonymous · Answer

usually in math "or" means one or the other or both

anonymous · Answer

this is not such a big number that you cannot just count them

anonymous · Answer

if you divide 99 by 3 you get 33, so 33 numbers divisible by 3
if you divide 99 by 5 you get 19.5 so 19 are divisible by 5
but you don't want to count the numbers divisible by both twice, so we have to subtract them off

anonymous · Answer

the numbers divisible by both 3 and 5 are multiples of 15
lets just list them
15
30
45
60
75
90
i count 6 of them

bibby · Answer

|$$|A \cup B| = |A| + |B| - |A \cap B|$$

anonymous · Answer

my guess therefore is 
$$33+19-6$$

anonymous · Answer

well the question asks for two digits number.

anonymous · Answer

oh so it does
guess we have to subtract off some more

anonymous · Answer

30 + 18 - 6 = 42

anonymous · Answer

3, 5, 6, 9

kainui · Answer

there are 90 two digit numbers.

multiples of 3 happen once per every 3 numbers. 1/3
multiples of 5 happen once per every 5 numbers. 1/5
multiples of 3 and 5 happen once per every 15 numbers. 1/15

(90/3)+(90/5)-(90/15)=number of 2-digit numbers divisible by 3 or 5 (inclusive OR)

anonymous · Answer

is that the same as $33+19-6-4$?

calculusxy · Answer

How did u get 18 though? @sourwing is it because it isn't a whole number yet (19.5)?

anonymous · Answer

I counted

calculusxy · Answer

But what do u count?

anonymous · Answer

from 10 to 99 that are divisible by 5

calculusxy · Answer

Is there any other way to count faster from 10-99 that r divisible by 5?

bibby · Answer

the floor of 99/5 - 1 because 5 isn't 2 digits

calculusxy · Answer

Thanks all of you :)

kainui · Answer

They way I've done it completely disregards where the numbers occur. This is possible because 90 is evenly divisible by 3,5, and 15. This means that if you were to "shift" the numbers you're looking at, you will either always gain and lose exactly one number in pairs.

For instance, think of all the numbers divisible by 5 between 1 and 10 inclusive (ten numbers in all)

There's obviously 5 and 10 in there. But suppose we move up 1 to go from 2 to 11 (inclusive)? Still have 5 and 10, so we're good. How about we scoot up to 5 to 14? Aha! Now if we go from 6 to 15 we have exactly gained and lost a divisor of 5, so as long as it's evenly divisible like this, we don't care where they are!