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How many 3 digit positive integers exist that when divided
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29 Aug 2011, 23:56
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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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Re: divisibility by 7
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30 Aug 2011, 02:29
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 + (n1)common difference 999=103+(n1)7 999103=(n1)7 896/7=n1 128+1=n n=129. OA E
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Re: divisibility by 7
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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.



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Re: divisibility by 7
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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 = 142141 = 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
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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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Re: How many 3 digit positive integers exist that when divided
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Re: How many 3 digit positive integers exist that when divided
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22 Aug 2013, 11:14
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  14214+1 = 129
What is the simplest way to solve this?



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Re: divisibility by 7
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14 Oct 2013, 16:48
jamifahad 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 + (n1)common difference 999=103+(n1)7 999103=(n1)7 896/7=n1 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



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Re: divisibility by 7
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15 Oct 2013, 08:55
AccipiterQ wrote: jamifahad 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 + (n1)common difference 999=103+(n1)7 999103=(n1)7 896/7=n1 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: \(a_ n=a_1+d(n1)\). For more check here: mathnumbertheory88376.htmlHope it helps.
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Re: divisibility by 7
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15 Oct 2013, 12:22
Bunuel wrote: AccipiterQ wrote: jamifahad 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 + (n1)common difference 999=103+(n1)7 999103=(n1)7 896/7=n1 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: \(a_ n=a_1+d(n1)\). For more check here: mathnumbertheory88376.htmlHope 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
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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 14215 = 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 = 9995, 999 works too. 127+1 = 128. now..if we take into consideration 103, then we have 129.



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How many 3 digit positive integers exist that when divided
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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
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14 Oct 2016, 11:24
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
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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 appAnswer must be (E) 129, please post your approach to identify the part where you went wrong..
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Re: How many 3 digit positive integers exist that when divided
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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 appAnswer 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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Re: How many 3 digit positive integers exist that when divided
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04 Jan 2018, 06:09
Bunuel wrote: AccipiterQ wrote: jamifahad 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 + (n1)common difference 999=103+(n1)7 999103=(n1)7 896/7=n1 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: \(a_ n=a_1+d(n1)\). For more check here: http://gmatclub.com/forum/mathnumbertheory88376.htmlHope 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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