Another Problem with factors and multiples

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AWR83: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Wed Oct 22, 2008 10:17 pm

by AWR83 » Sat Jan 03, 2009 5:14 pm

would someone plz explain how to get the answer of the following:

How many of the 1st 200 positive integers are divisible by 3 or 4 without any remainder?
2
3
100
108
116

Thanks.

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Source: Beat The GMAT — Problem Solving | Sat Jan 03, 2009 5:14 pm

logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

Re: Another Problem with factors and multiples

by logitech » Sat Jan 03, 2009 5:21 pm

Hmm,

3 or 4 = divisible 3+divisible 4 - divisible with 12 ( think like a set )

200/3 = 66 numbers

200/4 = 50 numbers

200/12 = 16 numbers

60+56 - 16 = 100

Am I right ?

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Sat Jan 03, 2009 6:10 pm

3*1(3) to 3*66(198)

So total of 66-1+1 = 66 multiples of 3 (1)

4*1(4) to 4*50(200)

50-1+1 = 50 multiples of 4 (2)

We would have double counted the common multiples of 3 and 4 i.e multiples of 12 which we need to subtract after adding (1) and (2)

12*1 to 12*16

So 16-1+1 = 16 multiples of 12

66+50-16 = 100

Choose C)

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