# Of the numbers 1,2,3,...,6000, how many are not multiples of 2,3 or 5?

Watch
Announcements
#1
I was looking at Cambridge specimen papers and this came up. There was an old forum about it but no clear answer.
Can anyone confirm?
I got 2000.
What I did:
Just multiples of 2= 3000-1000-600=1400
of 3=2000-1000-400=600
of 5=1200-600-400=200
Then added the repeating multiple of 3 and 2:
(3x2)n, n=1000
Then for the other 2 I found them and took away x amount of repeating multiples
(5x2)=600
(5x3)=400
so 30n=6000, n=200 Therefore there are 800 unique multiples for (5x2) and (5x3)
So 1400+600+200+1000+800=4000
6000-4000=2000
0
1 month ago
#2
(Original post by danm26)
I was looking at Cambridge specimen papers and this came up. There was an old forum about it but no clear answer.
Can anyone confirm?
I got 2000.
What I did:
Just multiples of 2= 3000-1000-600=1400
of 3=2000-1000-400=600
of 5=1200-600-400=200
Then added the repeating multiple of 3 and 2:
(3x2)n, n=1000
Then for the other 2 I found them and took away x amount of repeating multiples
(5x2)=600
(5x3)=400
so 30n=6000, n=200 Therefore there are 800 unique multiples for (5x2) and (5x3)
So 1400+600+200+1000+800=4000
6000-4000=2000
I'd probably treat it like a 3 set venn diagram, so sum each set
* multiples of 2
* multiples of 3
* multiples of 5
Then subtract each of the three combinations of 2 factors (2*3, 2*5, 3*5), and add the combination of three factors (2*3*5). Then subtract from the original.

Gives a different answer. Maybe sketch your approach on a venn diagram as you seem to be trying to sum the mutually exclusive parts.

Edit - an alternative is to note
1/2 are not divislbe by 2
2/3 not divislble by 3
4/5 not divislbe by 5
So
6000*1/2*2/3*4/5 = ...
Last edited by mqb2766; 1 month ago
0
1 month ago
#3
(Original post by danm26)
I was looking at Cambridge specimen papers and this came up. There was an old forum about it but no clear answer.
Another approach is to recognise that the multiples (and non-multiples) will repeat is a sequence of length 2 x 3 x 5 = 30. You could then write a list of the integers from 1 to 30, cross off all the multiples of 2, 3, and 5, then count how many integers are left. then multiply by the number of 30s in 6,000.
0
1 month ago
#4
I'd go about it by halving 6000, then taking two thirds of the sum, and four fifths of that.
Equally, you could find the lowest common multiple of 2, 3 and 5, work out the fraction of non-multiples, and multiply 6000 by that.
0
#5
(Original post by mqb2766)
I'd probably treat it like a 3 set venn diagram, so sum each set
* multiples of 2
* multiples of 3
* multiples of 5
Then subtract each of the three combinations of 2 factors (2*3, 2*5, 3*5), and add the combination of three factors (2*3*5). Then subtract from the original.

Gives a different answer. Maybe sketch your approach on a venn diagram as you seem to be trying to sum the mutually exclusive parts.

Edit - an alternative is to note
1/2 are not divislbe by 2
2/3 not divislble by 3
4/5 not divislbe by 5
So
6000*1/2*2/3*4/5 = ...
Yeah I think this was the ideal method. I got 1600 this time which was what I was typing in my OP until I thought it was wrong. I don't know the answer unfortunately but I reckon this is correct since in the OP I subtracted the common multiples when I should have added them.

Thanks for your help, the Venn diagram really simplified vs my method.
0
1 month ago
#6
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (363)
84.22%
No (68)
15.78%