Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 23 May 2013, 15:28

# divisibility by 7

Author Message
TAGS:
Intern
Joined: 11 May 2011
Posts: 24
Followers: 0

Kudos [?]: 7 [0], given: 1

divisibility by 7 [#permalink]  30 Aug 2011, 00:56
00:00

Question Stats:

44% (01:24) correct 55% (01:36) wrong based on 0 sessions
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
A. 128
B. 142
C. 143
D. 141
E. 129

Forum Moderators, Please feel free to remove the post if the question was already posted. I could not find the question hence posted in a new topic.

Regards,
Raghav.V
[Reveal] Spoiler: OA
Senior Manager
Joined: 03 Mar 2010
Posts: 461
Followers: 3

Kudos [?]: 75 [0], given: 21

Re: divisibility by 7 [#permalink]  30 Aug 2011, 03:29
First three digit number which when divided by 7 leaves remainder 5 is 103. here's how.
14*7=98
98+5=103.
next number will be (15*7+5=110). next will be (16*7+5=117)
Sequence 103,110,117,124,......992,999
Last term=first term + (n-1)common difference
999=103+(n-1)7
999-103=(n-1)7
896/7=n-1
128+1=n
n=129.

OA E
_________________

My dad once said to me: Son, nothing succeeds like success.

Manager
Joined: 09 Jun 2011
Posts: 107
Followers: 0

Kudos [?]: 5 [0], given: 0

Re: divisibility by 7 [#permalink]  01 Sep 2011, 20:08
Very Triky Question.

Minimum three digit number is 100 and maximum three digit number is 999.
The first three digit number that leaves remainder 5 when divided by 7 is 103.
14 * 7 = 98 +5 = 103
The second three digit number that leaves remainder 5 when divided by 7 is 110.
15 * 7 = 105 +5 =110
The third three digit number that leaves remainder 5 when divided by 7 is 117
and so on

The last three digit number that leaves remainder 5 when divided by 7 is 999
142 * 7 = 994 + 5 = 999

Therefore, we identify the sequence
103,110,117.....999

use the formula of last term
Last term = first term + (n - 1) * common difference

you will get the answer 129 that is definitely E.
Manager
Joined: 06 Jun 2011
Posts: 162
Followers: 0

Kudos [?]: 12 [0], given: 15

Re: divisibility by 7 [#permalink]  01 Sep 2011, 22:09
1000/7 = 142 is the result with 6 as remainder
100/7= 14 with 2 as remainder

so 15*7=105 is the first three digit # which is divisible by 7 and 994 is the last three digit #. So the total three digit # divisible by 7 = 142-14-1 = 129

and as 994+ 5 =999 is a three digit #

so the # of three digit # which after divided by 7 leave a remainder of 5 are 129
Re: divisibility by 7   [#permalink] 01 Sep 2011, 22:09
Similar topics Replies Last post
Similar
Topics:
are there any rules of divisibility by 7 and 8 1 07 Aug 2006, 08:35
4 Is the sum of integers a and b divisible by 7? 14 22 Oct 2009, 20:04
7 Shortcut: Divisible by 7 ? 5 21 May 2010, 10:17
1 Is N divisible by 7? 8 07 Feb 2012, 15:12
Simple and quick divisibility test for 7 3 06 Feb 2013, 00:27
Display posts from previous: Sort by