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Manhattan GMAT-Hey-Im a little confused here

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by dddanny2006 » Mon Nov 11, 2013 3:49 am
Answer the questions with a Yes,No or Cannot be determined.
1.If a is divided by 7 or by 18,an integer results.Is (a/42) an integer?

2.If j is divisible by 12 and 10,is j divisible by 24?

These problems are from Manhattan Number Properties guide.What I fail to understand is the difference between the 2 problems.They have been solved by the PrimeBox method.Why in the 2nd answer we see three boxes drawn?,in comparison to the one thats drawn in the 1st problem solution?I have attached the way they have drawn their boxes.

Thanks
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by ganeshrkamath » Mon Nov 11, 2013 6:53 am
dddanny2006 wrote:Answer the questions with a Yes,No or Cannot be determined.
1.If a is divided by 7 or by 18,an integer results.Is (a/42) an integer?

2.If j is divisible by 12 and 10,is j divisible by 24?

These problems are from Manhattan Number Properties guide.What I fail to understand is the difference between the 2 problems.They have been solved by the PrimeBox method.Why in the 2nd answer we see three boxes drawn?,in comparison to the one thats drawn in the 1st problem solution?I have attached the way they have drawn their boxes.

Thanks
I don't know the PrimeBox method, but here's my solution:

1. a is a multiple of LCM(7,18)
a = k * 126 _____________(k is an integer)
a/42 = k*126/42 = 3k (also an integer)
Yes.

2. j is a multiple of LCM(12,10)
j = m * 60 ______________(m is an integer)
j/24 = m*60/24 = 5m/2
This may or may not result in an integer.
Cannot be determined.

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by Abhishek009 » Mon Nov 11, 2013 7:05 am
dddanny2006 wrote:Answer the questions with a Yes,No or Cannot be determined.
1.If a is divided by 7 or by 18,an integer results.Is (a/42) an integer?

2.If j is divisible by 12 and 10,is j divisible by 24?

These problems are from Manhattan Number Properties guide.What I fail to understand is the difference between the 2 problems.They have been solved by the PrimeBox method.Why in the 2nd answer we see three boxes drawn?,in comparison to the one thats drawn in the 1st problem solution?I have attached the way they have drawn their boxes.

Thanks
Really in love with these kind of problems , post more if U have such..

1.If a is divided by 7 or by 18,an integer results.Is (a/42) an integer?

The green highlighted part simply means that the number A is completely divisible by both 7 and 18

The question states that the number A is divisible by 7 as well as 18

We can write A as -

A = 7 M * 18 N

Now the question here ask - Is (a/42) an integer?

Find out the factors of the denominator -

42 = 2*3*7


The part (a/42) can be re stated as -

A = 7 M * 18 N / 42 = 2*3*7

Which in every possible case will yield an integer...


2.If j is divisible by 12 and 10,is j divisible by 24?

Going to solve this one with the help of LCM

Since J is divisible by 12 and 10 , the least number possible is LCM ( 12 and 10 ) 60

So the number series of this pattern will be -

60 , 120 , 180 .......

Test each one individually

60/24 = Not completely divisible

120/24 = 5 Completely divisible

180 / 24 = Not completely divisible

So , we can't predict this one...
Abhishek
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by dddanny2006 » Mon Nov 11, 2013 7:48 am
Thanks your your inputs guys.I'd prefer if someone explains the box method to me.
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by Mathsbuddy » Tue Nov 12, 2013 9:18 am
I don't know what the box method is, but maybe this will simplify the algebra:

Question 1)

a is a multiple of both 7 and 18, therefore a is a multiple of 7 x 18 = 126 (LCM)
As 126 is a multiple of 42 (because 126/42 = 3) then it is TRUE.

Question 2)

j is a multiple of both 12 and 10, therefore j is a multiple of 60 (LCM, divisible by both 12 and 10)
60/24 = 2.5 which is not an integer
Odd multiples of 2.5 produce a non-integer
Even multiples of 2.5 produce an integer
Therefore the answer is TRUE 50% of the time and NOT TRUE 50% of the time
Conclusion: Cannot be determined without more information.
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by ceilidh.erickson » Wed Nov 13, 2013 3:00 pm
The primary difference between these two questions is that 7 and 18 do not SHARE FACTORS, but 12 and 10 do.

"Prime boxes" are handy ways of visualizing the prime factors of a given number - the component pieces. If a number is divisible by 18, for example, we can say that it has a 2, 3, and 3 in its "prime box." It might also have other factors, but those are the ones we know about. So we can say that it has AT LEAST a 2 and two 3's.

1) a is a multiple of both 7 and 18
Translate this as: "a has in its prime box at least one 7, and at least one 2 and two 3's." Since 42 is composed of one 2, one 3, and one 7, then a definitely has all of the component pieces of 42, so it's divisible.

2) j is a multiple of both 12 and 10
Translate this as: "j has in its prime box at least two 2's and one 3 (the factors of 12), and at least one 2 and one 5 (the factors of 10)."
If we say that j has "at least two 2's" and "at least one 2", can we then conclude that it has three 2's? No, because one of those 2's might be overlapping. All we can conclude is that it has at least two 2's in its prime box.

Since 24 has three factors of 2, it might go into j, but it might not. Inconclusive.
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Harvard Graduate School of Education
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