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How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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19 Jan 2015, 05:25
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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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19 Jan 2015, 06:00
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Bunuel wrote: How many twodigit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?
A. None B. One C. Two D. Three E. Four
Kudos for a correct solution. The possible number N can be written as follow: N = Multiple of LCM(6,10) + 1st such number N = 30x + 1 Possible values = 1, 31, 61, 91 Answer : 3 such 2 digit number. D.



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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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19 Jan 2015, 06:01
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Bunuel wrote: How many twodigit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?
A. None B. One C. Two D. Three E. Four
Kudos for a correct solution. n=10p+1 > Number could be 11 21 31 41 51 n=6q+1 > Number could be 7 13 19 25 31 n= 30q+31 so n could be 31,61,91 IMO D



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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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20 Jan 2015, 03:13
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Answer = D. Three LCM of 10 & 6 = 30 Twodigit numbers giving remainder 1 for 30 are 31, 61, 91
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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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20 Jan 2015, 03:21
anupamadw wrote: Bunuel wrote: How many twodigit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?
A. None B. One C. Two D. Three E. Four
Kudos for a correct solution. n=10p+1 > Number could be 11 21 31 41 51 n=6q+1 > Number could be 7 13 19 25 31 n= 30q+31so n could be 31,61,91 IMO D Can you explain the highlighted calculation? How is that obtained?
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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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20 Jan 2015, 03:31
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PareshGmat wrote: anupamadw wrote: Bunuel wrote: How many twodigit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?
A. None B. One C. Two D. Three E. Four
Kudos for a correct solution. n=10p+1 > Number could be 11 21 31 41 51 n=6q+1 > Number could be 7 13 19 25 31 n= 30q+31so n could be 31,61,91 IMO D Can you explain the highlighted calculation? How is that obtained? Positive integer n is divided by 10, the remainder is 1 > \(n=10q+1\), where \(q\) is the quotient > 1, 11, 21, 31, 41, ... Positive integer n is divided by 6, the remainder is 1 > \(n=6p+1\), where \(p\) is the quotient > 1, 7, 13, 19, ... There is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is divisor and \(r\) is a remainder) based on above two statements:Divisor \(x\) would be the least common multiple of above two divisors 10 and 6, hence \(x=30\). Remainder \(r\) would be the first common integer in above two patterns, hence \(r=1\). Therefore general formula based on both statements is \(n=30m+1\). Thus n could be 1, 31, 61, 91, ... Since n is a twodigit integer, then n could only be 31, 61, or 91. Check for more here: positiveintegernleavesaremainderof4afterdivisionby93752.html#p721341Hope it helps.
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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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20 Jan 2015, 04:44
Find the least common factor and multiples of the number +1 Least common factor of 10 and 6 is 30 (two digits multiples of 30 are 30,60,90.. Add +1 to the numbers) so totally 3 numbers are possible



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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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15 Mar 2015, 21:47
hi Bunuel don`t you think with respect to your answer.. since q is the quotient..how can you put q=0 and get 1 as common from both equations I mean if you put q=0,then n=1 but n is a two digit number so the first common value needs to be 31 i.e N(two digit)=30m+31.. thanks



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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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15 Mar 2015, 22:34



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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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16 Aug 2017, 09:26
31,61 & 91 are the only three two digit numbers that when divided by 10 and 6 each leaves a remainder of 1. Option D.



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How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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28 Feb 2018, 22:20
Hi All, This type of question is rooted in patternmatching. Once you find the patterns behind this question, it won't be hard to solve. Instead of trying to do every step all at once, I suggest that you break the prompt into "pieces": First, name the 2digit numbers that are evenly divisible by 10: 10, 20, 30, ......90 Now, name the 2digit numbers that have a remainder of 1 when divided by 10: 11, 21, 31,.....91 Now that we've established the numbers that fit the first 2 "restrictions" in the prompt, let's factor in numbers that are ALSO divisible by 6: 30, 60, 90 And ALSO have a remainder of 1 when divided by 6: 31, 61, 91 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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29 Mar 2018, 23:06
Bunuel wrote: How many twodigit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?
A. None B. One C. Two D. Three E. Four
Kudos for a correct solution. Two digit numbers divided my 10 yielding remainder 1 = 11,21,31,41,51,61,71,81,91 Two digit numbers divided my 6 yielding remainder 1 = 31,61,91 Three common numbers. Hence (D)
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Re: How many twodigit whole numbers yield a remainder of 1 when divided [#permalink]
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08 Apr 2018, 07:01
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The possible number N can be written as follow: N = Multiple of LCM(6,10) + 1st such number N = 30x + 1 Possible values = 1, 31, 61, 91 Answer : 3 such 2 digit number. D.




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