Author 
Message 
VP
Joined: 30 Jun 2008
Posts: 1026

How many integers between 100 and 150, inclusive, cannot be [#permalink]
Show Tags
03 Nov 2008, 08:24
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum. How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5
_________________
"You have to find it. No one else can find it for you."  Bjorn Borg



Manager
Joined: 12 Oct 2008
Posts: 104

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 09:40
I think it is 27. It might look time consuming, but it took me a little over 1,5 min, what is okay
no. of integers 150100+1=51 divisible by 5 are all which end with 5 or 0. We have 11 of such integers in the range.
divisible by 3 are the ones, which sum of the digits is multiple of 3: 102, 108, 111, 114, 117, 123, 126, 129, 132, 138, 141, 144, 147. 13 Integers
511113=27



Current Student
Joined: 28 Dec 2004
Posts: 3342
Location: New York City
Schools: Wharton'11 HBS'12

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 09:53
amitdgr wrote: How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5 lets see.. 150=102+(n1)3, 15099=51/3= n=17 for 5, lets see 150=100+(n1)5, 15095=5n 55=5n or n=11 so 17+10=27 as 150 is already a multiple of 3 and 5..



VP
Joined: 30 Jun 2008
Posts: 1026

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 10:26
fresinha12 wrote: amitdgr wrote: How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5 lets see.. 150=102+(n1)3, 15099=51/3= n=17 for 5, lets see 150=100+(n1)5, 15095=5n 55=5n or n=11 so 17+10=27 as 150 is already a multiple of 3 and 5.. why did we do 17 + 10 when we had n=11 for multiples of 5 ? because we have already counted 150 as a multiple of 3 ?
_________________
"You have to find it. No one else can find it for you."  Bjorn Borg



Intern
Joined: 29 Oct 2008
Posts: 2

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 12:15
fresinha12 wrote: amitdgr wrote: How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5 lets see.. 150=102+(n1)3, 15099=51/3= n=17 for 5, lets see 150=100+(n1)5, 15095=5n 55=5n or n=11 so 17+10=27 as 150 is already a multiple of 3 and 5.. If 17 are divisible by 3 and 11 by 5, so 17 + 11 1 (for 150, as you mention) = 27, but this is the numbers divisible by 3 or 5. The question asks for numbers NOT divisible by 3 or 5. You have the final answer right in this case. But you just subtracted 1 for 150, whereas 105, 120, 135, 150 are all divisible by 3 and 5. So following your calculations.. Divisible by 3 = 17, divisible by 5 = 11, divisible by both = 4. So divisible by 3 or 5 = 17+114= 24. Total number of integers = 51. So not divisible by 3 or 5 = 51 24 = 27.



Retired Moderator
Joined: 05 Jul 2006
Posts: 1747

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 13:21
amitdgr wrote: How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5 multiples of 3 = 51/3 = 17 multiples of 5 = 51/5 = 10.xxx total = 27 however there is 51/15 =3..x ie: 3 multiples in common 273 = 24.........can be evenely devided thus 5124 = 27 can not



VP
Joined: 30 Jun 2008
Posts: 1026

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 19:26
1
This post received KUDOS
nth term = a + (n1)d a = initial term d = difference between each terms n = number of terms = 150100+1 = 51 nth term = 150 150 = 105 + (n1)*15 (numbers that are divisible by 3*5) > n= 4 150 = 102 + (n1)*3 (numbers that are divisible by 3) > n = 17 150 = 100 + (n1)*5 (numbers that are divisible by 5) > n = 11 since we are looking for numbers that cannot be evenly divided by 3 nor 5, then: total no of nos  nos divisible by 3  nos divisible by 5 + nos divisible by 15 511711+4 = 27
_________________
"You have to find it. No one else can find it for you."  Bjorn Borg



Manager
Joined: 23 Aug 2008
Posts: 63

Re: PS : Number Prop [#permalink]
Show Tags
03 Nov 2008, 21:09
Nice answers guys, wish my algebra was that reliable. To be honest I didn't trust my formula, so I wrote out the pattern and added up. I'm not sure what would be faster for most poeple in the exam, but I find this way a more reliable and much quicker for me personally. FYI.
For those divisible by 3, pattern was N,N,Y (Y=divisible, N=not) 100,101,102 ... 148,149,150; therefore out of 51 numbers, 51/3*2=34 are not divisible by 3
We have eliminated those divisible by 3, so now out of those not divisible by 3, eliminate those which are divisible by 5. Numbers divisible by 5, omitting those already eliminated by bieng divisible by 3: 100,110,115,125,130,140,145 = 7 numbers [in reality I wrote down all of them from 100 to 150 and crossed off the ones with digits adding up to 3]
Last part ... 347 = answer of 27



Manager
Joined: 15 Oct 2008
Posts: 54

Re: PS : Number Prop [#permalink]
Show Tags
04 Nov 2008, 03:48
alternative approach,
A B n=(BA)+1 100/3 150/3 34 50 17 100/5 150/5 20 30 11 100/15 150/15 7 10 4 (# that overlap of 3,5)
Then from 100  150 has 51 no.
===> 511711+4 = 27




Re: PS : Number Prop
[#permalink]
04 Nov 2008, 03:48






