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remainder problem

by Gurpinder » Mon Aug 23, 2010 12:27 pm
when positive integer N is divided by 5, the remainder is 1. When N is divided by 7, the remainder is 3. What is the smallest positive integer K such that K+N is a multiply of 35?

3,4,12,32,35


my approach was

n=5x+1
n=7x+3
5x+1=7x+3 --> solve to get n = -1.

and i am stuck here?

I thought the equations that I have above are suppose to work for these remainder problems! Whats going on!

Thanks to all!
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by Stuart@KaplanGMAT » Mon Aug 23, 2010 1:40 pm
Gurpinder wrote:when positive integer N is divided by 5, the remainder is 1. When N is divided by 7, the remainder is 3. What is the smallest positive integer K such that K+N is a multiply of 35?

3,4,12,32,35


my approach was

n=5x+1
n=7x+3
5x+1=7x+3 --> solve to get n = -1.

and i am stuck here?

I thought the equations that I have above are suppose to work for these remainder problems! Whats going on!

Thanks to all!
Almost certainly the quickest way to solve this problem is by brute force.

We have to rules to satisfy:

1) n/5 has remainder 1. Since n must be positive, we could have n = 1, 6, 11, 16, (any number ending in 1 or 6).

2) n/7 has remainder 3. So, n could be 3, 10, 17, 24, 31, ...

The first overlapping number is 31, so to get a multiple of 35 we simply add 4.

Now, there's one answer lower than 4, so since the question asks for the "smallest possible" value of k, we have to think about whether 3 could also work.

We don't have to think very long - since by the first rule n will always end in 1 or 6, and every multiple of 35 ends in 5 or 0, there's no way that adding 3 to n will ever produce a multiple of 35: choose "4".

In fact, based on the answer choices, we didn't even need to go past the very first step - once we determined that n was going to end in 1 or 6, we know that, to generate a number ending in 0 or 5, we have to add a number ending in either 4 or 9. Only "4" fits the bill!

Here's an important takeaway for all problem solving questions: keep your eye on the answer choices, they provide invaluable assistance in choosing the most efficient approach on the GMAT.
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by kmittal82 » Mon Aug 23, 2010 1:47 pm
Gurpinder wrote:when positive integer N is divided by 5, the remainder is 1. When N is divided by 7, the remainder is 3. What is the smallest positive integer K such that K+N is a multiply of 35?

3,4,12,32,35


my approach was

n=5x+1
n=7x+3
5x+1=7x+3 --> solve to get n = -1.

and i am stuck here?

I thought the equations that I have above are suppose to work for these remainder problems! Whats going on!

Thanks to all!
Hi Gurpinder,

Where you went wrong is that you used the same quotient when dividing by 5 and by 7
Number = Divisor*Quotient + Remainder

n = 5x + 1
n = 7y + 3

You used 'x' in both cases, which is not always true.
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by Gurpinder » Tue Aug 24, 2010 6:12 am
kmittal82 wrote:
Gurpinder wrote:when positive integer N is divided by 5, the remainder is 1. When N is divided by 7, the remainder is 3. What is the smallest positive integer K such that K+N is a multiply of 35?

3,4,12,32,35


my approach was

n=5x+1
n=7x+3
5x+1=7x+3 --> solve to get n = -1.

and i am stuck here?

I thought the equations that I have above are suppose to work for these remainder problems! Whats going on!

Thanks to all!
Hi Gurpinder,

Where you went wrong is that you used the same quotient when dividing by 5 and by 7
Number = Divisor*Quotient + Remainder

n = 5x + 1
n = 7y + 3

You used 'x' in both cases, which is not always true.
Hey Mittal,

If you dont mind, can you please show me how you would solve this one using the equations you gave.

Thanks,
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by Gurpinder » Tue Aug 24, 2010 6:15 am
Stuart Kovinsky wrote:
We have to rules to satisfy:

1) n/5 has remainder 1. Since n must be positive, we could have n = 1, 6, 11, 16, (any number ending in 1 or 6).

2) n/7 has remainder 3. So, n could be 3, 10, 17, 24, 31, ...

The first overlapping number is 31, so to get a multiple of 35 we simply add 4.
Hey Stuart,

So is the equation method wrong? Am I always suppose to tackle this one by just getting numbers that satisfy each condition.

Also, can you please just explain the LCM method that you used to get 31. Why are we looking for LCM. Are we doing the same thing as if in the equation that I had I was putting the variable for 1 equal to the other one. So by getting the LCM, we are satisfying the condition for x/5 AND x/7?

Thanks,
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by kmittal82 » Tue Aug 24, 2010 6:24 am
Gurpinder wrote:
kmittal82 wrote:
Gurpinder wrote:when positive integer N is divided by 5, the remainder is 1. When N is divided by 7, the remainder is 3. What is the smallest positive integer K such that K+N is a multiply of 35?

3,4,12,32,35


my approach was

n=5x+1
n=7x+3
5x+1=7x+3 --> solve to get n = -1.

and i am stuck here?

I thought the equations that I have above are suppose to work for these remainder problems! Whats going on!

Thanks to all!
Hi Gurpinder,

Where you went wrong is that you used the same quotient when dividing by 5 and by 7
Number = Divisor*Quotient + Remainder

n = 5x + 1
n = 7y + 3

You used 'x' in both cases, which is not always true.
Hey Mittal,

If you dont mind, can you please show me how you would solve this one using the equations you gave.

Thanks,
Hi Gurpinder,

Its challenging to actually "solve" the question using the equations above (due to quite a few variables). The easiest approach would be brute force method as Stuart mentioned.
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by Gurpinder » Tue Aug 24, 2010 7:42 am
kmittal82 wrote:
Gurpinder wrote:
kmittal82 wrote:
Gurpinder wrote:when positive integer N is divided by 5, the remainder is 1. When N is divided by 7, the remainder is 3. What is the smallest positive integer K such that K+N is a multiply of 35?

3,4,12,32,35


my approach was

n=5x+1
n=7x+3
5x+1=7x+3 --> solve to get n = -1.

and i am stuck here?

I thought the equations that I have above are suppose to work for these remainder problems! Whats going on!

Thanks to all!
Hi Gurpinder,

Where you went wrong is that you used the same quotient when dividing by 5 and by 7
Number = Divisor*Quotient + Remainder

n = 5x + 1
n = 7y + 3

You used 'x' in both cases, which is not always true.
Hey Mittal,

If you dont mind, can you please show me how you would solve this one using the equations you gave.

Thanks,
Hi Gurpinder,

Its challenging to actually "solve" the question using the equations above (due to quite a few variables). The easiest approach would be brute force method as Stuart mentioned.
Alright! Thanks! So just out of curiosity, because I want to be able to take something away from this question, should I always use brute force?

Since these equations cannot be solved since there are 2 different variables, when would be an idle situation to use equations?

Can you, if possible, explain the purpose of LCM in the brute force method.

Sorry I am asking so many questions, its just that I never came across this problem in my MGMAT books so I want to completely learn it now than stare at the screen on test date.

Thanks,
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- Alfred A. Montapert, Philosopher.
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by Stuart@KaplanGMAT » Tue Aug 24, 2010 9:12 am
Gurpinder wrote:
Stuart Kovinsky wrote:
We have to rules to satisfy:

1) n/5 has remainder 1. Since n must be positive, we could have n = 1, 6, 11, 16, (any number ending in 1 or 6).

2) n/7 has remainder 3. So, n could be 3, 10, 17, 24, 31, ...

The first overlapping number is 31, so to get a multiple of 35 we simply add 4.
Hey Stuart,

So is the equation method wrong? Am I always suppose to tackle this one by just getting numbers that satisfy each condition.

Also, can you please just explain the LCM method that you used to get 31. Why are we looking for LCM. Are we doing the same thing as if in the equation that I had I was putting the variable for 1 equal to the other one. So by getting the LCM, we are satisfying the condition for x/5 AND x/7?

Thanks,
Hi,

I didn't use the LCM - 31 isn't the LCM of anything. "LCM" stands for lowest common multiple, which we definitely didn't calculate here (the LCM of 5 and 7 is 35, but that doesn't help in this question).

The only time we use LCM on remainder questions is when our two rules have the same remainder. For example, if the question had been:
If n is a positive integer, n/5 has remainder 1 and n/7 has remainder 1, which of the following could be the value of n?
Then we would think:

Since 5 and 7 don't have any common factors, the LCM of 5 and 7 is 35. So, n/35 will also have a remainder of 1 and, since n is positive, n must belong to the set {1, 36, 71, ...}

To your first question, the equation method isn't "wrong", it's just generally slower on these types of questions on the GMAT. If you were doing a problem set and the question were worth 10 marks, the equation method is what would get you 10/10. On the GMAT, however, we get no marks for our elegant solutions, so we prefer fast and dirty to long and elegant.
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