Question

Set X consists of the numbers between 1 and 200 that are divisible by 6. Set Y consist of the numbers between 1 and 200 that are divisible by 8. How many numbers are in both set X and set Y?

Answer 1

@oldrin.bataku

Answer 2

200/6 for set X round down

Answer 3

200/8 for Y ...also round down

Answer 4

Oh I think I misunderstood the question

Answer 5

lololol haha it's fine

Answer 6

every 3rd multiple of 8 will be in the set, so 24, 48...

Answer 7

OOPS!

Answer 8

24n<200 n<200/24 n< 25/3 round down

Answer 9

Given that \(6\) and \(8\) divide our number, we know that \(\operatorname{lcm}(6,8)=24\) should divide our number, correct?

Answer 10

yes

Answer 11

\(\operatorname{lcm}\) denotes their least common multiple. So we see that the elements in the intersection are specifically \(24k\) for integer \(k\). For how many \(k\) is \(1\le24k\le200\)?

Answer 12

$$1\le24k\le200\\\frac1{24}\le k\le8\frac26$$

Answer 13

i.e. how many integers are there in \([1,8]\)? Well, surely just \(8\): \(1,2,3,4,5,6,7,8\)

Answer 14

The reasoning behind the above is that \(6=2\times3,8=2\times2\times2\) -- so surely our number must be divisible by \(2\) at least three times and at least \(3\) once; this means \(2\times2\times2\times3=24\) (i.e. their least common multiple) must evenly divide our number. How many multiples of \(24\) are there in \([1,200]\)? that's what we did above!