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

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CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
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How many 3 digits numbers are there so that each is evenly [#permalink]  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
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 133 [0], given: 0

Re: Challenges - Divisibility [#permalink]  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
Joined: 09 Aug 2006
Posts: 766
Followers: 1

Kudos [?]: 68 [0], given: 0

Re: Challenges - Divisibility [#permalink]  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
Joined: 20 Aug 2007
Posts: 852
Location: Chicago
Schools: Chicago Booth 2011
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Re: Challenges - Divisibility [#permalink]  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
Senior Manager
Joined: 04 Jun 2007
Posts: 374
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Re: Challenges - Divisibility [#permalink]  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
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 349 [0], given: 4

Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 180 [0], given: 2

I dont get this..

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
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 349 [0], given: 4

Re: Challenges - Divisibility [#permalink]  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
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 133 [0], given: 0

Re: Challenges - Divisibility [#permalink]  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
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 349 [0], given: 4

Re: Challenges - Divisibility [#permalink]  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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