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Does this ques represent Complex or Normal GMAT standard?

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Does this ques represent Complex or Normal GMAT standard?

by raj.may6 » Sun Apr 29, 2012 12:49 pm
THis is question # 68 from OG Quant reviw 2nd Edition (pg 70).


Ques for your quick reference:
When positive integer n is divided by 5, 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 multiple of 35.

When I attempted, I could not make it -- then looked at explanation on pg 106 and what i felt that this is kind of complex question and should not represent Normal GMAT standard on Quant --


What you guys think?
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by tisrar02 » Sun Apr 29, 2012 2:01 pm
Hey raj.may6,

Believe it or not, I got a similar problem like this on the actual test. My belief is that when you are doing quite well, they throw these type of questions at you.
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by tisrar02 » Sun Apr 29, 2012 2:02 pm
Hey raj.may6,

Believe it or not, I got a similar problem like this on the actual test. My belief is that when you are doing quite well, they throw these type of questions at you.
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by jai123oct » Sun Apr 29, 2012 2:15 pm
i think n should be equal to 3 as the no when subdivided according to the conditions given:

let m be the no :

m= 5n+1 (equ .....1)
and
m=7n+ 3 (equ............2)

if u solve this u get lowest value of n = 3 ;

put it in equ 1 u get 1 remainder ; put it again in equ 2 u get 3 as remainder.


now to be divisible by 35 ; the lowest value of k = 32

as K+N is a multiple of 35 and if n = 3 then we get k =32

so 32 is the answer.
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by fyllmax » Sun Apr 29, 2012 3:00 pm
I calculated N=31 and K=4
jai123oct wrote:i think n should be equal to 3 as the no when subdivided according to the conditions given:

let m be the no :

m= 5n+1 (equ .....1)
and
m=7n+ 3 (equ............2)

if u solve this u get lowest value of n = 3 ;

put it in equ 1 u get 1 remainder ; put it again in equ 2 u get 3 as remainder.


now to be divisible by 35 ; the lowest value of k = 32

as K+N is a multiple of 35 and if n = 3 then we get k =32

so 32 is the answer.
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by Stuart@KaplanGMAT » Sun Apr 29, 2012 3:34 pm
raj.may6 wrote:THis is question # 68 from OG Quant reviw 2nd Edition (pg 70).


Ques for your quick reference:
When positive integer n is divided by 5, 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 multiple of 35.

When I attempted, I could not make it -- then looked at explanation on pg 106 and what i felt that this is kind of complex question and should not represent Normal GMAT standard on Quant --


What you guys think?
Like a lot of GMAT challenge questions it may be very hard to solve using number properties but is vastly simplified by picking numbers.

n/5 has rem 1. n could be 1, 6, 11, 16, ... (any number ending in 1 or 6)
n/7 has rem 3. n could be 3, 10, 17, 24, 31...

Since 31 is the first number that matches both criteria, let's set n=31.

We want n+k to be a multiple of 35, so that makes k=35-31=4.

Now, the question asks for the SMALLEST possible value of k. If 4 is the smallest answer, then we're definitely done. If there's an answer smaller than 4, then we'd probably check for the next value of n, which turns out to be 66, also 4 away from a multiple of 35, and be confident that we're done.

Remember - GMAT math isn't about proving theorems or showing your work - it's about getting the job done with as little effort as possible.
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by Stuart@KaplanGMAT » Mon Apr 30, 2012 7:32 pm
raj.may6 wrote: When I attempted, I could not make it -- then looked at explanation on pg 106 and what i felt that this is kind of complex question and should not represent Normal GMAT standard on Quant
One other thing that's important to note... while the OG is a good source of questions, it's generally a horrible source of explanations (especially in math). The OG math explanations tend to be the "how to get 10/10 on your grade 10 math test" solution, rather than the "how to answer this question as quickly and painlessly as possible (and in under 2 minutes) solution.

There are almost always more efficient solutions than those shown in the Guide.
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