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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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19 Jan 2015, 05:25
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Question Stats:

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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

Project PS Butler : Question #53

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

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20 Jan 2015, 03:31
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7
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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19 Jan 2015, 06:00
1
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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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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20 Jan 2015, 03:13
2
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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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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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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15 Mar 2015, 21:47
hi Bunuel
dont 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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15 Mar 2015, 22:34
shreygupta3192 wrote:
hi Bunuel
dont 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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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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Re: How many two-digit 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 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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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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08 Apr 2018, 07:01
1
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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10 Feb 2019, 21:15
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

Project PS Butler : Question #53

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We have to take the LCM and never the product

N1 = 10 k + 1
N2 = 6k + 1

N3 = 30 k + 1

k = 1,2,3

Only 3, two digit values can be yield.

D
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Re: How many two-digit whole numbers yield a remainder of 1 when divided   [#permalink] 10 Feb 2019, 21:15
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