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How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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27 Oct 2010, 11:39
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How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine? (A) 37 (B) 38 (C) 39 (D) 40 (E) 46
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Re: Number system 1 [#permalink]
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27 Oct 2010, 17:27
zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(1) 37 (2) 38 (3) 39 (4) 40 (5) 46 There are total of \(\frac{200100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) (check this: totallybasic94862.html?hilit=multiple%20range ); There are total of \(\frac{198108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\); There are total of \(\frac{196112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\); As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+71=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\); So there are \(5112=39\) even numbers which are not divisible neither by 7 nor by 9. Answer: C.
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Re: Number system 1 [#permalink]
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Updated on: 27 Oct 2010, 19:46
total 51 even integers
out of which 7 will be devided by 7 and 6 will be devided by 9 (for 7..find intervals of 14 as we are taking even numbers 112,126...196
(for 9..find intervals of 18 as we are taking even numbers 108,126...198)
126 is common
so 5176+1 = 39 will not be divided by 7 or 9
Originally posted by krushna on 27 Oct 2010, 13:02.
Last edited by krushna on 27 Oct 2010, 19:46, edited 1 time in total.



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Re: Number system 1 [#permalink]
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28 Oct 2010, 08:48
Thanks Bunuel. It is clear but I am not quite clear on why it should be 18 or 14 because of even numbers. What happens if it is odd numbers instead of even numbers?



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Re: Number system 1 [#permalink]
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28 Oct 2010, 08:59



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Re: Number system 1 [#permalink]
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28 Oct 2010, 17:04
Bunuel wrote: vgan4 wrote: Thanks Bunuel. It is clear but I am not quite clear on why it should be 18 or 14 because of even numbers. What happens if it is odd numbers instead of even numbers? I'm not sure understood your question. Stem says: "How many even integers n, ..." So even multiples of 9 are multiples of 2*9=18. I must have been sleeping when I asked that question. Sorry. It is clear now.



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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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08 Dec 2013, 23:16
zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(A) 37 (B) 38 (C) 39 (D) 40 (E) 46 Even From 16 x 7 all the way up to 28 x 7 comes up to 7 such numbers From 12 x 9 all the way up to 22 x 9 which comes up to 6 such numbers From 2 x 63  1 such number Total number of even numbers divisible by either 7 or 9: 7 + 6  1 = 12 From 100 to 200 there are 51 even numbers as the series start from 100. Number neither divisible: 51  12 = 39 Odd From 15 x 7 all the way up to 27 x 7 which count up to 7 numbers From 13 x 9 all the way up to 21 x 9 which count up to 5 numbers From 63 x 3  1 number Total = 7 + 5  1 = 11 From 100 to 200 there are 50 odd numbers Total number of odd numbers not divisible: 50  11 = 39 Pushpinder Gill
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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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10 Dec 2013, 08:00
I marked B (5176=38) because I counted 126 twice. Will try to improve...
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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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21 Aug 2014, 09:11
Doubt:
if we change the range from 100 <= n <= 200 to 200 <= n <= 300 and use the same line of reasoning  it leads to erroneous results as illustrated below...
number of even integers between 200 and 300 = (300  200)/2 + 1 = 51 integers;
even multiples of 7 : (294  203)/14 + 1 = 7.5 : clearly, this is not possible.
and works for even multiples of 9 : (297207)/18 + 1 = 6.
Bunuel (or anyone), could you please explain.
Thanks!



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How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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22 Aug 2014, 04:52



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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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19 Nov 2014, 23:25
The best way is to write all even integers in a row, eliminate all multiples of 7 and 9 and count numbers left. You can do that in 10 minutes. Good way to raise your self esteem))



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How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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29 Jun 2015, 10:43
I know this is a pretty old thread but I'm starting to study to GMAT and come up with an idea to solve that problema
Basically count the number of multiples for 7 and 9.
in every 70 numbers are 10 multiples of 7 > 105, 112, 119, ... and every 30 numbers are 3 multiples of 7, so thus far we have 13 multiples of 7, from which 6 are even
in every 90 numbers are 10 multiples of 9 > 108, 117, 126, ... and every 10 numbers are 1 multiple of 9, so thus far we have 11 multiples of 9, from which 6 are even
(look at the first multiple to determine the number of even multiples)
so we have 51 even numbers  6  6 = 39



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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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22 Aug 2015, 03:20
Bunuel wrote: zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(1) 37 (2) 38 (3) 39 (4) 40 (5) 46 There are total of \(\frac{200100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) (check this: totallybasic94862.html?hilit=multiple%20range ); There are total of \(\frac{198108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\); There are total of \(\frac{196112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\); As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+71=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\); So there are \(5112=39\) even numbers which are not divisible neither by 7 nor by 9. Answer: C. Bunuel here's what i did: #muliples of 9 in the set = 8 #muliples of 7 in the set = 1 #muliples of LCM(9,7)=63 in the set = 3 #even multiples in set =51 answer is 51831 = 5112 = 39 good?



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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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22 Aug 2015, 04:00
zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(A) 37 (B) 38 (C) 39 (D) 40 (E) 46 There are 14 numbers from 105 to 196 divisible by 7 there are 11 numbers from 108 to 198 divisible by 9 Out of these 25 numbers,126 and 189 are divisible by both 7 and 9 these two numbers fall in both the sets Therefore, 252=23 numbers are divisible by 7 and 9 There are 101 numbers from 100 to 200 10123=78 numbers are divisible neither by 7 nor by 9 Out of 78, 39 are even numbers.



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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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05 Nov 2016, 22:10
zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(A) 37 (B) 38 (C) 39 (D) 40 (E) 46 Please check the solution as attached. Answer: Option C
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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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06 Nov 2016, 04:49
Bunuel wrote: zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(1) 37 (2) 38 (3) 39 (4) 40 (5) 46 There are total of \(\frac{200100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) There are total of \(\frac{198108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\); There are total of \(\frac{196112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\); As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+71=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\); So there are \(5112=39\) even numbers which are not divisible neither by 7 nor by 9. Answer: C. Hi Bunuel, A very stupid question  instead even integers, how do you get the number of odd integers between 100 and 200? Thanks! Uve



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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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06 Nov 2016, 04:59
uvemdesalinas wrote: Bunuel wrote: zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(1) 37 (2) 38 (3) 39 (4) 40 (5) 46 There are total of \(\frac{200100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) There are total of \(\frac{198108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\); There are total of \(\frac{196112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\); As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+71=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\); So there are \(5112=39\) even numbers which are not divisible neither by 7 nor by 9. Answer: C. Hi Bunuel, A very stupid question  instead even integers, how do you get the number of odd integers between 100 and 200? Thanks! Uve __________________ Total  Even = Odd.
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Re: How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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06 Nov 2016, 05:11
Bunuel wrote: zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(1) 37 (2) 38 (3) 39 (4) 40 (5) 46 There are total of \(\frac{200100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) There are total of \(\frac{198108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\); There are total of \(\frac{196112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\); As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+71=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\); So there are \(5112=39\) even numbers which are not divisible neither by 7 nor by 9. Answer: C. Thanks Bunuel, But Total = 200  100= 100 integers or Total = 200  100 + 1 = 101 ?



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How many even integers n, where 100 <= n <= 200, are divisib [#permalink]
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06 Nov 2016, 05:14
uvemdesalinas wrote: Bunuel wrote: zerotoinfinite2006 wrote: How many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine?
(1) 37 (2) 38 (3) 39 (4) 40 (5) 46 There are total of \(\frac{200100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) There are total of \(\frac{198108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\); There are total of \(\frac{196112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\); As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+71=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\); So there are \(5112=39\) even numbers which are not divisible neither by 7 nor by 9. Answer: C. Hi Bunuel, A very stupid question  instead even integers, how do you get the number of odd integers between 100 and 200? Thanks! Uve Here is the solution for odd numbers Total Integers from 100 to 200 inclusive = 101 (51 even and 50 odd) Total Odd Integers from 100 to 200 = 50Total No. that are multiple of 7 from 1200 = 200/7 = 28 (i.e. 14 even and 14 odd) Total No. that are multiple of 7 from 1100 = 100/7 = 14 (i.e. 7 even and 7 odd) Total No. that are ODD multiple of 7 from 100200 = 147 = 7Total No. that are multiple of 9 from 1200 = 200/9 = 22 (i.e. 11 even and 11 odd) Total No. that are multiple of 9 from 1100 = 100/9 = 11 (i.e. 5 even and 6 odd) Total No. that are ODD multiple of 9 from 100200 = 116 = 5Total No. that are odd multiple of 7 and 9 will be multiple of 63 and count of such numbers from 100200 = 1 So Total No. that are odd multiple of 7 and/or 9 from 100200 = 7+51 = 11 So Total No. that are NOT odd multiple of 7 and/or 9 from 100200 = 5011 = 39 Answer: Option C
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