Question

Problems for fun:
How many numbers from 1- 200 aren't divisible by 2,3,5, or 7? I'll post a solution when my movie is over.

Answer 1

- the prime numbers

Answer 2

i think i got it

Answer 3

please hold ...

Answer 4

so there are 180 numbers divisible by some combination of the above
so 20 aren't

Answer 5

i guess i should figure out the answer now >.< one sec lol

Answer 6

Half of the numbers are even, hence divisible by 2. Of the remaining numbers, one third are divisible by 3, so we have 133 numbers divisible by 2 or 3.

40 numbers are divisible by 5, of these half are even so we have 20 numbers divisible by 5 and not by 2. 13 numbers of the original 200 are divisible by 3 and 5 (by 15) and of these 7 are odd and 6 even (we know there's one more odd number since 15 is odd) so 7 numbers are divisible by 3 and 5, but not by 2, so of the 20 numbers we had 6 are divisible by 3 and we find that there are 20 - 6 = 14 numbers that are divisible by 5 and not by 3 nor 2. Combining this with the previous result, there are 133 + 13 = 146 numbers that are divisible by 2, 3 or 5.

28 numbers are divisible by 7, half of which are even so we have 14 numbers divisible by 7 and not by 2. 5 numbers of the original 200 are divisible by 5 and 7 (by 45) and of these 3 are odd and 2 are even, so we have 14 - 3 = 11 numbers that are divisible by 7 and not by 2 nor 5. 9 numbers are divisible by 3 and 7 (by 21) 5 of which are odd and one of which is also divisible by 5 (by 105), 11 - 5 + 1 = 7 numbers are divisible by 7 and not by 2, 3 nor 5. Finally, there are 146 + 7 = 153 numbers that are divisible by 2, 3, 5 or 7 between 1 and 200, so there are exactly 47 numbers which are not.

Answer 7

oops i need to go further with my approach

Answer 8

i got 33 i think

Answer 9

the rest

Answer 10

im getting 47

Answer 11

i didnt realize how long this would take to calculate <.< i thought 3 numbers was too short lol.

Answer 12

scanning now~ btw nice job Krebante :)

Answer 13

i caught a mistake in mine thanks you myininaya's pic, i had missed 2, 5, 7 lol

Answer 14

thank's to*

Answer 15

if you cant open my file
there are 100 multiplies of 2
there are 66 multiplies of 3
there are 40 multiplies of 5
there are 28 multiplies of 7
there are 33 multiplies of 2*3
there are 20 multiplies of 2*5
there are 14 multiplies of 2*7
there are 13 multiplies of 3*5
there are 9 multiplies of 3*7
there are 5 multiplies of 5*7
there are 6 multiplies of 2*3*5
there are 4 multiplies of 2*3*7
there are 2 multiplies 2*5*7
there is 1 multiple of 3*5*7
so we 234 from adding the n(multiplies of 2)+n(multiplies of 3)+n(multiplies 5)+n(multiplies of 7)
so we have some overlapping
so need to subtract any intersections
234-n(multiplies of 2*3)-n(multiplies of 2*5)-n(multiplies of 2*7) - n(multiplies of 3*5)-n(multiplies of 3*7)-n(multiplies of 5*7)-n(multiplies of 2*3*5)-n(multiplies of 2*3*7)-n(multiplies of 2*5*7)-n(multiplies of 3*5*7) giving us 167
200-167=33 that are not divisible by 2,3,5,or 7

Answer 16

what did i miss?

Answer 17

ok i will look at yours lol
i didn't realize u posted that

Answer 18

in your method, you have 234 from adding the multiples of the "singles", then you subtract the multiples of the "doubles", now you need to add the multiples of the "triples". so
234-(33+20+14+13+9+5) = 140
Then you add:
140+(6+4+2+1) = 153
So its 200 - 153 = 47

Answer 19

why do we add the triples?

Answer 20

oh wait just in case we took out to much from the doubles i think

Answer 21

my scan was ugly
lol

Answer 22

Even beyond the over counting we are doing in counting the singles, we are also over counting the doubles. for example, 140 is both a multiple of 5*2 and 2*7. So we need the triples to be the yeah you got it :)

Answer 23

yours was alittle bit more pretty
its hard for me to think unless i write down the way i think

Answer 24

you just keep switching the sign from positive to negative (or vice-versa) to take care of over counting. So if i had said, divisible by 2,3,5,7,and11, then it would be
TOTAL - (singles)+(doubles)-(triples)+(quadruples)-(quintuples)

Answer 25

right i get it

Answer 26

i only write nice when i know people are going to look at it lol, otherwise its chicken scratch >.< i can read it while im writing it, but say a day later, its like hieroglyphs lol

Answer 27

nice problem i liked it

Answer 28

its interesting! infact, this can be used to tell you if there are any more prime numbers in that range. since there were 47 numbers left out of the 200, that means there are at most 47 primes left. once i find another one i can reduce that number further. so on and so forth

Answer 29

at first i was going to act like i didn't see your problem but then i was like oh maybe i should look at the opposite of what you said

Answer 30

lol

Answer 31

i was thinking about it because i didn't want to let you down though

Answer 32

im going to try to get my hours corrected
i gotta go to bed
peace

Answer 33

lm glad you tried it out :) regardless of if one gets it right or wrong, it encourages thought.

Answer 34

me too, although im gonna get food lol >.< good night :)

Answer 35

well i was on the right path at least
and i had all the numbers there
i just didn't do the operations right

Answer 36

and at first i did forget some of my intersections

Answer 37

namely the triples

Answer 38

goodnight

Answer 39

night :)