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# How many 3 digit positive integers exist that when divided

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Intern
Joined: 11 May 2011
Posts: 15
How many 3 digit positive integers exist that when divided  [#permalink]

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30 Aug 2011, 00:56
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17
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Difficulty:

75% (hard)

Question Stats:

61% (02:16) correct 39% (02:31) wrong based on 293 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
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Joined: 03 Mar 2010
Posts: 347
Schools: Simon '16 (M\$)

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30 Aug 2011, 03:29
1
6
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
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Joined: 09 Jun 2011
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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.
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Joined: 06 Jun 2011
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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
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How many 3 digit positive integers exist that when divided  [#permalink]

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22 Aug 2013, 11:56
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: 58427
Re: How many 3 digit positive integers exist that when divided  [#permalink]

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22 Aug 2013, 12:00
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, 12:14
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?
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14 Oct 2013, 17:48
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: 58427

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15 Oct 2013, 09:55
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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15 Oct 2013, 13:22
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, 06:36
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
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How many 3 digit positive integers exist that when divided  [#permalink]

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14 Oct 2016, 11:45
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, 12:24
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, 12:44
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, 12:59
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
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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, 07:09
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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05 Oct 2019, 16:04
Quote:
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

I guess yes it is.
But you have to be really comfortable with what you are doing. Chances are less that such an approach will strike you during the exam but if you're confident then it should work.

Also if you don't add that 1 you're still boomed.

Thank you!
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Re: How many 3 digit positive integers exist that when divided   [#permalink] 05 Oct 2019, 16:04
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