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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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17 00:00

Difficulty:   75% (hard)

Question Stats: 60% (02:15) correct 40% (02:31) wrong based on 290 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: 348
Schools: Simon '16 (M\$)
Re: divisibility by 7  [#permalink]

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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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Manager  Joined: 09 Jun 2011
Posts: 74
Re: divisibility by 7  [#permalink]

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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: 72
Re: divisibility by 7  [#permalink]

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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
Senior Manager  Joined: 10 Jul 2013
Posts: 289
How many 3 digit positive integers exist that when divided  [#permalink]

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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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Asif vai.....
Math Expert V
Joined: 02 Sep 2009
Posts: 58378
Re: How many 3 digit positive integers exist that when divided  [#permalink]

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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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Posts: 40
Location: India
Concentration: Marketing, Operations
Schools: Schulich '16 (A)
GMAT 1: 690 Q48 V36 WE: Operations (Telecommunications)
Re: How many 3 digit positive integers exist that when divided  [#permalink]

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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: 184
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 Re: divisibility by 7  [#permalink]

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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 V
Joined: 02 Sep 2009
Posts: 58378
Re: divisibility by 7  [#permalink]

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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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Manager  Joined: 26 Sep 2013
Posts: 184
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 Re: divisibility by 7  [#permalink]

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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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Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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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.
VP  P
Joined: 07 Dec 2014
Posts: 1222
How many 3 digit positive integers exist that when divided  [#permalink]

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

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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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Joined: 24 Jul 2016
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Re: How many 3 digit positive integers exist that when divided  [#permalink]

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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  B
Joined: 04 Feb 2013
Posts: 14
Re: How many 3 digit positive integers exist that when divided  [#permalink]

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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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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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Software Tester currently in USA ( ) Re: How many 3 digit positive integers exist that when divided   [#permalink] 05 Oct 2019, 16:04
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