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M21-35

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Bunuel
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M21-35 [#permalink] New post   16 Sep 2014, 01:15
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If \(x\) is a positive integer, what is the remainder when \(x\) is divided by 3?


(1) The remainder when \(x\) is divided by 4 is 1

(2) The remainder when \(x\) is divided by 7 is 6
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Bunuel
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Re M21-35 [#permalink] New post   16 Sep 2014, 01:15
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Official Solution:


If \(x\) is a positive integer, what is the remainder when \(x\) is divided by 3?

(1) The remainder when \(x\) is divided by 4 is 1.

Given: \(x=4q+1\), so \(x\) could be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... This information is not sufficient to determine the remainder when \(x\) is divided by 3.

(2) The remainder when \(x\) is divided by 7 is 6.

Given: \(x=7p+6\), so \(x\) could be: 6, 13, 20, 27, 34, 41, ... This information is not sufficient to determine the remainder when \(x\) is divided by 3.

(1)+(2) If \(x=13\) then the remainder is 1 when divided by 3, but if \(x=41\) then the remainder is 2 when divided by 3.Not sufficient.


Answer: E
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samuraijack256
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Re: M21-35 [#permalink] New post   03 Jul 2015, 22:40
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Bunuel wrote:
Official Solution:


(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.


Answer: E


Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).

In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?
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chetan2u
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Re: M21-35 [#permalink] New post   03 Jul 2015, 23:03
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samuraijack256 wrote:
Bunuel wrote:
Official Solution:


(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.


Answer: E


Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).

In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?


Hi,
The faster method in these Qs is to find the first common multiple and, thereafter, add the LCM of two numbers..

In this case when we have got the first number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41..
the next 41+28=69 and so on...

so 13 gives remainder 1
41 gives remainder 2
69 gives remainder 0... and so on
hope it helped..
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Re: M21-35 [#permalink] New post   03 Jul 2015, 23:10
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samuraijack256 wrote:
Bunuel wrote:
Official Solution:


(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.


Answer: E


Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).



In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?


another way would be to equate two eq from two statements..

\(x=4q+1\), \(x=7p+6\)...
\(4q+1=7p+6\)....
\(4q=7p+5\)....

now substitute p as 1,3,5... as p has to be odd to make the eq even ..
where we getan integer value of q... those are the numbers we rae looking for..
for ex
p=1,q=3... so number=4q+1=13..
p=3, q is not an int..
p=5, q=10 number =4q+1=41.. and so on
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Re: M21-35 [#permalink] New post   28 Aug 2017, 09:34
I think this is a high-quality question and I agree with explanation.
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Re: M21-35 [#permalink] New post   29 Oct 2018, 17:33
Hi Bunuel- Can you please confirm if my logic is right while doing this exercise?

n = 4k + 1 -> Insuf

n = 7k + 6 -> Insuf

n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).

As 28k is not divide by 3, the Reminder will change.

Thanks in advance!
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Re: M21-35 [#permalink] New post   29 Oct 2018, 22:09
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efr wrote:
Hi Bunuel- Can you please confirm if my logic is right while doing this exercise?

n = 4k + 1 -> Insuf

n = 7k + 6 -> Insuf

n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).

As 28k is not divide by 3, the Reminder will change.

Thanks in advance!


Correct with small but crucial correction: quotient in all three cases (highlighted) will not be the same so you should not sue one variable for them (use p, q, k instead).
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Bunuel
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Re: M21-35 [#permalink] New post   24 May 2023, 02:03
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M21-35 [#permalink]
24 May 2023, 02:03
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