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# How many 3 digit positive integers exist that when divided

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Intern
Joined: 11 May 2011
Posts: 18
How many 3 digit positive integers exist that when divided  [#permalink]

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29 Aug 2011, 23:56
2
12
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:10) correct 38% (02:31) wrong based on 244 sessions

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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
Senior Manager
Joined: 03 Mar 2010
Posts: 372
Schools: Simon '16 (M\$)

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30 Aug 2011, 02:29
5
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
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##### General Discussion
Manager
Joined: 09 Jun 2011
Posts: 83

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01 Sep 2011, 19:08
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: 93

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01 Sep 2011, 21:09
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
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Joined: 10 Jul 2013
Posts: 313
How many 3 digit positive integers exist that when divided  [#permalink]

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22 Aug 2013, 10:56
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
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Asif vai.....

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Joined: 02 Sep 2009
Posts: 53066
Re: How many 3 digit positive integers exist that when divided  [#permalink]

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22 Aug 2013, 11:00
Asifpirlo wrote:
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

Merging similar topics. Please refer to the solutions above.
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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22 Aug 2013, 11:14
1
Used sheer calculations.

3 digit no. divided by 7 leaves remainder 5
xyz=7a+5
Started with a =13, xyz=096
for a=14, xyz=105
therefore 1st value of a is 14.
now for highest value of a, started with a =140 => xyz=985
for a=141 => xyz=992
for a=142 => xyz=999
therefore highest value of a is 142
that makes the no. of values for which xyz remains 3 digit no. are - 142-14+1 = 129

What is the simplest way to solve this?
Manager
Joined: 26 Sep 2013
Posts: 191
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41

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14 Oct 2013, 16:48
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

Can someone explain what this means, and where I can read up on this concept? I have never run into it
Math Expert
Joined: 02 Sep 2009
Posts: 53066

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15 Oct 2013, 08:55
1
1
AccipiterQ wrote:
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

Can someone explain what this means, and where I can read up on this concept? I have never run into it

For arithmetic progression if the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by: $$a_ n=a_1+d(n-1)$$.

For more check here: math-number-theory-88376.html

Hope it helps.
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Joined: 26 Sep 2013
Posts: 191
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41

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15 Oct 2013, 12:22
Bunuel wrote:
AccipiterQ wrote:
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

Can someone explain what this means, and where I can read up on this concept? I have never run into it

For arithmetic progression if the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by: $$a_ n=a_1+d(n-1)$$.

For more check here: math-number-theory-88376.html

Hope it helps.

Thank you, are there other problems like this one to practice on?
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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14 Oct 2016, 05:36
Raghava747 wrote:
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

ouch...I ignored the 103 number...
my approach differ from others...
i started with finding the multiples of 5...
lowest one 105 or 15*7
greatest one 994 or 142*7
142-15 = 127.
for each multiple of 7, we can add 5, and get a number that when divided by 7, will yield a remainder of 5.
since 994 = 999-5, 999 works too.
127+1 = 128.
now..if we take into consideration 103, then we have 129.
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Joined: 07 Dec 2014
Posts: 1154
How many 3 digit positive integers exist that when divided  [#permalink]

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14 Oct 2016, 10:45
Raghava747 wrote:
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

let x+1=number of 3 digit integers leaving remainder of 5
lowest integer=103
highest integer=999
103+7x=999
x=128
x+1=129
E.
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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14 Oct 2016, 11:24
1
Raghava747 wrote:
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

Define the SET

S = { 103 , 110 , 117.......992 , 999 }

No of digits will be $$\frac{( 999 - 103 )}{7} +1$$ = 129

Hence correct answer will be (A) 129
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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14 Oct 2016, 11:44
ProfX wrote:
Raghava747 wrote:
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
I got B.142

Sent from my iPhone using GMAT Club Forum mobile app

Answer must be (E) 129, please post your approach to identify the part where you went wrong..
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Abhishek....

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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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14 Oct 2016, 11:59
Abhishek009 wrote:
ProfX wrote:
Raghava747 wrote:
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
I got B.142

Sent from my iPhone using GMAT Club Forum mobile app

Answer must be (E) 129, please post your approach to identify the part where you went wrong..
I already found it. I just divided 1000 by 7 thinking all multiples of 7 plus 5 will give me the answer (also subtracted one at the end I think). However, we only need the 3 digit numbers. I always make mistakes like this

Sent from my iPhone using GMAT Club Forum mobile app
Intern
Joined: 04 Feb 2013
Posts: 14
Re: How many 3 digit positive integers exist that when divided  [#permalink]

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04 Jan 2018, 06:09
Bunuel wrote:
AccipiterQ wrote:
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

Can someone explain what this means, and where I can read up on this concept? I have never run into it

For arithmetic progression if the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by: $$a_ n=a_1+d(n-1)$$.

For more check here: http://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.

Is this a feasible solution?: a 3 digit number can either leave a remainder of 0 / 1 / 2 / 3 / 4 / 5 or 6 when divided by 7. So the answer will be 1/6th of 900 (3 digit numbers) = 128 + 1 (for 999) = 129
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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11 Feb 2019, 23:57
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Re: How many 3 digit positive integers exist that when divided   [#permalink] 11 Feb 2019, 23:57
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