"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+1 is a multiple of 35?"
A. 3
B. 4
C. 12
D. 32
E. 35
Solution B
The explanation in the guide is not very clear to me and seems to not be the optimal way to get the answer.
What I did instinctively under time pressure was: n=5k+1 and n=7k+3.
5k+1=7k+3 and k=-1
Therefore, substituting, n=-4. Since 0 is a multiple of every number; for k=4, n+k=0, which is also the right answer.
I'm really curious. Is what I did a correct other way to get the rigth answer or is it completely wrong and it just happens to lead to the right answer?
Help with PS#68 from Quant Review 2nd edition.
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- outreach
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is the Q properly worded
there are two unrelated variable k and n
there are two unrelated variable k and n
jeremy8 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+1 is a multiple of 35?"
A. 3
B. 4
C. 12
D. 32
E. 35
Solution B
The explanation in the guide is not very clear to me and seems to not be the optimal way to get the answer.
What I did instinctively under time pressure was: n=5k+1 and n=7k+3.
5k+1=7k+3 and k=-1
Therefore, substituting, n=-4. Since 0 is a multiple of every number; for k=4, n+k=0, which is also the right answer.
I'm really curious. Is what I did a correct other way to get the rigth answer or is it completely wrong and it just happens to lead to the right answer?
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- selango
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I think the question must be n+k is a multiple of 35.
n is different when divided by 5 and 7.So they need to be expressed by 2 diff variables.
n=5a+1
n=6,11,16,21,26,31......
n=7b+3
n=10,17,24,31,38..............
n+k is a multiple of 35.So n+k=35q
Try substituting value of k from options.Let start with the least one.
k=3,n=35q-3
n=32,67,102
But n does not satisfy both the equations.
k=4,n=35q-4
n=31,66....31,66 is available in both the equations and satisfy.If u check,all values satisfy both the equations.
Hence k=4
Pick B
n is different when divided by 5 and 7.So they need to be expressed by 2 diff variables.
n=5a+1
n=6,11,16,21,26,31......
n=7b+3
n=10,17,24,31,38..............
n+k is a multiple of 35.So n+k=35q
Try substituting value of k from options.Let start with the least one.
k=3,n=35q-3
n=32,67,102
But n does not satisfy both the equations.
k=4,n=35q-4
n=31,66....31,66 is available in both the equations and satisfy.If u check,all values satisfy both the equations.
Hence k=4
Pick B
--Anand--
- selango
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Whenever the problem saying that n when divided by different numbers,try to express n by 2 different variables.jeremy8 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+1 is a multiple of 35?"
A. 3
B. 4
C. 12
D. 32
E. 35
Solution B
The explanation in the guide is not very clear to me and seems to not be the optimal way to get the answer.
What I did instinctively under time pressure was: n=5k+1 and n=7k+3.
5k+1=7k+3 and k=-1
Therefore, substituting, n=-4. Since 0 is a multiple of every number; for k=4, n+k=0, which is also the right answer.
I'm really curious. Is what I did a correct other way to get the rigth answer or is it completely wrong and it just happens to lead to the right answer?
From ur example,if we take 5k+1=7k+3;k=-1 and n=-4
Is n+k=-5 is a muliple of 35.No
Also n=-4 divided by 5 and 7 leaves remainder 1 and 3.No
Always check whether the given equation is satisified.
--Anand--
Anand, thanks for your answer, it makes sense now.selango wrote:I think the question must be n+k is a multiple of 35.
n is different when divided by 5 and 7.So they need to be expressed by 2 diff variables.
n=5a+1
n=6,11,16,21,26,31......
n=7b+3
n=10,17,24,31,38..............
n+k is a multiple of 35.So n+k=35q
Try substituting value of k from options.Let start with the least one.
k=3,n=35q-3
n=32,67,102
But n does not satisfy both the equations.
k=4,n=35q-4
n=31,66....31,66 is available in both the equations and satisfy.If u check,all values satisfy both the equations.
Hence k=4
Pick B