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How many integers between 100 and 150, inclusive, cannot be

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How many integers between 100 and 150, inclusive, cannot be  [#permalink]

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New post 03 Nov 2008, 09:24
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How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5

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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 10:40
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 150-100+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

51-11-13=27
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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 10:53
amitdgr wrote:
How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5



lets see.. 150=102+(n-1)3, 150-99=51/3= n=17

for 5, lets see 150=100+(n-1)5, 150-95=5n 55=5n or n=11

so 17+10=27 as 150 is already a multiple of 3 and 5..
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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 11:26
fresinha12 wrote:
amitdgr wrote:
How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5



lets see.. 150=102+(n-1)3, 150-99=51/3= n=17

for 5, lets see 150=100+(n-1)5, 150-95=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 ?
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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 13:15
fresinha12 wrote:
amitdgr wrote:
How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5



lets see.. 150=102+(n-1)3, 150-99=51/3= n=17

for 5, lets see 150=100+(n-1)5, 150-95=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+11-4= 24.
Total number of integers = 51. So not divisible by 3 or 5 = 51 -24 = 27.
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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 14:21
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

27-3 = 24.........can be evenely devided thus 51-24 = 27 can not
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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 20:26
1
nth term = a + (n-1)d
a = initial term
d = difference between each terms
n = number of terms = 150-100+1 = 51
nth term = 150

150 = 105 + (n-1)*15 (numbers that are divisible by 3*5) -------> n= 4
150 = 102 + (n-1)*3 (numbers that are divisible by 3) -------> n = 17
150 = 100 + (n-1)*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
51-17-11+4 = 27
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Re: PS : Number Prop  [#permalink]

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New post 03 Nov 2008, 22:09
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 ... 34-7 = answer of 27
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Re: PS : Number Prop  [#permalink]

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New post 04 Nov 2008, 04:48
alternative approach,

A B n=(B-A)+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.

===> 51-17-11+4 = 27

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