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How many two-digit whole numbers yield a remainder of 1 when divided

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How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 19 Jan 2015, 05:25
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How many two-digit 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.

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Re: How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 20 Jan 2015, 03:31
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6
PareshGmat wrote:
anupamadw wrote:
Bunuel wrote:
How many two-digit 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


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 two-digit integer, then n could only be 31, 61, or 91.

Check for more here: positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752.html#p721341

Hope it helps.
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Re: How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 19 Jan 2015, 06:00
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2
Bunuel wrote:
How many two-digit 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 two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 19 Jan 2015, 06:01
1
Bunuel wrote:
How many two-digit 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 two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 20 Jan 2015, 03:13
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Answer = D. Three

LCM of 10 & 6 = 30

Two-digit numbers giving remainder 1 for 30 are

31, 61, 91
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Re: How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 20 Jan 2015, 03:21
anupamadw wrote:
Bunuel wrote:
How many two-digit 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


Can you explain the highlighted calculation? How is that obtained?
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Re: How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 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 two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 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 two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 15 Mar 2015, 22:34
shreygupta3192 wrote:
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


I first found general formula and then applied the restriction.
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Re: How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 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 two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 28 Feb 2018, 22:20
Hi All,

This type of question is rooted in pattern-matching. 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 2-digit numbers that are evenly divisible by 10:

10, 20, 30, ......90

Now, name the 2-digit 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:

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Re: How many two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 29 Mar 2018, 23:06
Bunuel wrote:
How many two-digit 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 two-digit whole numbers yield a remainder of 1 when divided  [#permalink]

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New post 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.
Re: How many two-digit whole numbers yield a remainder of 1 when divided &nbs [#permalink] 08 Apr 2018, 07:01
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