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How many 3 digits numbers are there so that each is evenly

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CEO
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How many 3 digits numbers are there so that each is evenly [#permalink] New post 21 Oct 2007, 19:50
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?
VP
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Re: Challenges - Divisibility [#permalink] New post 21 Oct 2007, 20:25
bmwhype2 wrote:
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?


7*15 = 105
7*142 = 994

From 15 to 142 inclusive
Ans = 142-15+1 = 128
Director
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Re: Challenges - Divisibility [#permalink] New post 21 Oct 2007, 20:40
bmwhype2 wrote:
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?


D.

Last term = a + (n-1)d where a is the 1st term and d is the common difference.

1st 3 digit integer divisible by 7 = 105
last 3 digit integer divisible by 7 = 998

998 = 7 + (n-1)7

Solve for n = 128.
Director
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Re: Challenges - Divisibility [#permalink] New post 22 Oct 2007, 05:51
GK_Gmat wrote:
D.

Last term = a + (n-1)d where a is the 1st term and d is the common difference.

1st 3 digit integer divisible by 7 = 105
last 3 digit integer divisible by 7 = 998

998 = 7 + (n-1)7

Solve for n = 128.


I think a = 105, not 7 since we are only looking for 3-digit terms
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Re: Challenges - Divisibility [#permalink] New post 22 Oct 2007, 06:54
bmwhype2 wrote:
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?


here's a faster way-
999-100+1= 900
900/7 = approx. 128
thus the answer is 128, took around 10 seconds.
CEO
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 [#permalink] New post 02 Nov 2007, 09:16
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 [#permalink] New post 02 Nov 2007, 15:02
I dont get this..

please help...

last term=994

first term=105

994=105+(n-1)14 since we are looking for even divisors of 7..

ok..so now i have n approx=64..

i think the answer should 64...howevr if the question asked about all the numbers that divide into 7..then i approx n=128
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Re: Challenges - Divisibility [#permalink] New post 04 Nov 2007, 02:36
bkk145 wrote:
bmwhype2 wrote:
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?


7*15 = 105
7*142 = 994

From 15 to 142 inclusive
Ans = 142-15+1 = 128


how can we know that 994 is the highest number divisible by 7?

working backwards from 999 would take a long time. in addition, i am not sure where u pulled 142 from because there are no divisibility rules of 7 that i know of.
VP
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Re: Challenges - Divisibility [#permalink] New post 04 Nov 2007, 04:59
bmwhype2 wrote:
bkk145 wrote:
bmwhype2 wrote:
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?


7*15 = 105
7*142 = 994

From 15 to 142 inclusive
Ans = 142-15+1 = 128


how can we know that 994 is the highest number divisible by 7?

working backwards from 999 would take a long time. in addition, i am not sure where u pulled 142 from because there are no divisibility rules of 7 that i know of.


I did the division...
999/7 = 142.something
So you know that 142*7 must be the max number less than 999 that is divisible by 7
142*7 = 994
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Re: Challenges - Divisibility [#permalink] New post 18 Nov 2007, 18:26
r019h wrote:
bmwhype2 wrote:
How many 3 digits numbers are there so that each is evenly divisible by 7?

105
111
127
128
142


Do we use AP in this?


here's a faster way-
999-100+1= 900
900/7 = approx. 128
thus the answer is 128, took around 10 seconds.


nice but i think the closeness of the answer choices discourages this.

900/7 = 128.6xxx
hard to gamble with 128 or 127 in the answer choices...
Re: Challenges - Divisibility   [#permalink] 18 Nov 2007, 18:26
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