Question

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

Answer 1

how many two digit numbers are there?

Answer 2

99

Answer 3

if the number of two digit numbers is x, then 9>x>100

Answer 4

it is hence not 99

Answer 5

49

Answer 6

there are actually 90 two digit numbers, take 99-9

Answer 7

how many numbers from 10 to 99 are divisible by 5 or 3?

Answer 8

It's not 3 AND 5.

Answer 9

The options are
42, 46, 47, 48, 53

Answer 10

So how do I figure this out?

Answer 11

ok, so find the total number of number divisible by 3 OR 5

Answer 12

what is it?

Answer 13

The ones divisible by 5 are 10, 15, 20, 25...95 so there are 18 of those.
The ones divisible by 3 are the ones in which the sum of the digits is a multiple of 3

Answer 14

90/3=30
90/5=18
When I add that I get 48. That answer is not correct according got the answer key.

Answer 15

So that would be 12, 15, 18, 21, 24, 27...

Answer 16

@satellite73

Answer 17

I would start by thinking multiplies of 3 and 5.
\[3n_1 \text{ is all the multiplies of 3} \]
\[5n_2 \text{ is all the multiplies of 5} \]
We want \[12 \le 3n_1 \le 99 \text{ and } 20 \le 5n_2 \le 95 \]
where \[n_1 \text{and} n_2 \text{ are integers }\]
Solve both inequalities

Answer 18

99/3+(100/5-1), did u get that?

Answer 19

0 duhhh

Answer 20

@utopia1234 Why do you have to subtract 1 from it?

Answer 21

I named the first multiple of 3 which is 12 that is a 2 digit
noticed i started the second set at 20 instead of 15 because 15 was already in the first set

Answer 22

33 + 19=52 (number of numbers divisible by 3 or 5 from 1 to 99)

Answer 23

I got something less that than @utopia1234

Answer 24

99/15 round down, u will get 6

Answer 25

52-6 will be the number that that is divisible by 3 or 5 in the first 99 numbers

Answer 26

Yeah! that is what I got.

Answer 27

numbers divisible by 3 and 5 from 1 to 10 includes 3,6,9,5

Answer 28

52-6-4 will give u 42 which is the answer

Answer 29

10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, 99

Answer 30

Count them.

Answer 31

LOL ok heres the method, or the way i was thinking.

Answer 32

Well how about this 12<=3n_1<=99
where n_1 is an integer
solve for n_1
4<=n_1<=33
There are 33-4+1 numbers in this set
Then there is also this set 20<5n_2<95
where n_2 is an integer
solve for n_2
4<=n_2<=19
There are 19-4+1 numbers in this set.
So the answer should be (33-4+1)+(19-4+1)

Answer 33

I'm meaning the inequality above to be a set of integers
not rationals and irrationals just integers