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How many 3 digit positive integers exist that when divided
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30 Aug 2011, 00:56
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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, 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 + (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, 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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Re: divisibility by 7
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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 = 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, 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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Re: How many 3 digit positive integers exist that when divided
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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
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22 Aug 2013, 12:14
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, 17:48
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, 09:55
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, 13:22
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, 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 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, 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
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14 Oct 2016, 12:24
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, 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 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, 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 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, 07:09
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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Re: How many 3 digit positive integers exist that when divided
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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
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