# Thread: Ques from GRE Nova maths-Number theory

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## Ques from GRE Nova maths-Number theory

Ques :
If each of the non-zero numbers a,b,c is divisble by 3 ,then abc must be divisbile by which of the following numbers?

A 8
C 81
D 121
E 159

Reasoning: if each number-a,b,c is divisble by 3,then
say a= 3 x p ,b=3 x q, c=3 x r
abc = 3x3x3x(pqr) divisible by 27.

How come??
Say number is 333 -(not divisible by 27 but each digit divisbile by 3)
I dont understand if in question 'abc' meant to be 'a x b x c' or 'abc'.

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i think in question 'abc' means the product of three numbers a,b,c.

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I think there is something wrong with the question. It should have stated that a is not equal to b and b not equal to c. If they meant product then it should have been mentioned clearly.

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Yes,This is a small error in thw ques ans answer as well in Nova GRE book.I hope not to see such ambiguities on the real exam.

For abc -anyone would be clearly considering it as number formed by digits -abc ...
With that ,the reasoning given doest fit well.
If 'abc' ='a x b x c' ,then it fits but Q should have written a.b.c atleast to represent product.

I have seen many such small errorsi n Novas book (specially in answers-explanation ) which dumbfounds me and wastes so much of my time to come out of confusion.

5. Good post? |
Originally Posted by Deep79
Ques :
If each of the non-zero numbers a,b,c is divisble by 3 ,then abc must be divisbile by which of the following numbers?

A 8
C 81
D 121
E 159

Reasoning: if each number-a,b,c is divisble by 3,then
say a= 3 x p ,b=3 x q, c=3 x r
abc = 3x3x3x(pqr) divisible by 27.

How come??
Say number is 333 -(not divisible by 27 but each digit divisbile by 3)
I dont understand if in question 'abc' meant to be 'a x b x c' or 'abc'.

abc here refers to a*b*c and I don't think there is any problem with the question.

we can write the equation as follows :

a/3 = 0------------1.

b/3 = 0------------2.

c/3 = 0------------3.

Now if we multiply the three equations (number 1 * number 2 * number 3) we get

a/3*b/3*c/3 = 0*0*0

or we can write it as : abc/27 = 0

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assume a=3; b=6; c=9 then abc= 3*6*9=162 ; hence you can write it as (3*1)*(3*2)*(3*3) = 27*6; there you will get 27 as the answer

7. Good post? |
assume a=3; b=6; c=9 then abc= 3*6*9=162 ; hence you can write it as (3*1)*(3*2)*(3*3) = 27*6; there you will get 27 as the answer
Yes this is also a good method which you can adopt to solve this kind of problems but at time calculations can be cumbersome , there you need to approach this problem using some other logic.

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