M05-36

Question

If  a ,  b , and  c  are distinct positive integers, is  \frac{(\frac{a}{b})}{c}  an integer?

(1)  \frac{a}{c} = 3 
 
(2)  a = b + c

Bunuel · Accepted Answer

Official Solution:

If  a ,  b , and  c  are distinct positive integers, is  \frac{(\frac{a}{b})}{c}  an integer? 
 
Firstly, simplify the expression  \frac{(\frac{a}{b})}{c}  to  \frac{a}{bc} .
 
(1)  \frac{a}{c} = 3 . 
 
Substitute the value of  \frac{a}{c}  into the simplified expression to get  \frac{a}{bc}=\frac{3}{b} . If  b  equals 1 or 3, then the result is an integer. However, if  b  represents any other positive integer, the outcome is not an integer. Not sufficient.
 
(2)  a = b + c . 
 
Substituting  a  into the equation gives  \frac{a}{bc}=\frac{b+c}{bc}=\frac{b}{bc}+\frac{c}{bc}=\frac{1}{c}+\frac{1}{b} . For this expression to become an integer, either  b=c=1  or  b=c=2  would need to be true. However, we know that  b  and  c  are distinct positive integers, so neither case is possible. Therefore,  \frac{1}{c}+\frac{1}{b}  is not an integer. Sufficient.

Answer: B

fearose · Answer

what is b=-1 and c=1 i n that case also 1/b+ 1/c= 0 (which is integer) fulfills the condition. So answer should be E

roneileonel · Answer

b= -1 is not possible since the question says "If a, b, and c are distinct positive integers".

thembaseeker · Answer

Simplifying expression  \frac{(\frac{a}{b})}{c}  gives a/bc

Now

Statement 1 : a/c=3

therefore a/bc=3/b, No data related to b is given.

Insufficient

Statement 2 : a=b+c

a/bc=(b+c)/bc=1/b+1/c......(i)

As b and c are distinct. Therefore expression will never be an integer.

Sufficient.

Answer : B

rishi02 · Answer

I think this is a  high-quality  question and I agree with explanation.

vrgmat · Answer

Bunuel ,

How did you arrive at this? - "Now, this expression to be an integer (1) either b=c=1 must be true or b=c=2 must be true". Why cant b and c be two distinct positive integers such that their decimal add upto 1? How do you concretely say b=c=1 or b=c=2. How do you rule out other possibilities.

I thought ( even though it may not be possible ), we may have numbers like 1/b = 0.6 & 1/c = 0.4 which add upto 1. I thought we may have such decimals which add upto 1. How do you rule out every such possibility?

asricharan · Answer

I think this is a  high-quality  question and I agree with explanation.

praveen8047 · Answer

Even I was thinking on the same lines but later on, I tried to figure it out on my own.
Addressing your query:
Two positive fractions will add up to 1, only when they are in this format : 1/x + (x-1)/x i.e. 1/3 + 2/3 or 1/10 + 9/10. 
And x & x-1 are co-prime i.e. there is no common factors between them hence they can never be simplified further.
So, if one of the denominator is Integer, other denominator has to be non-integer to add up to 1.
Or, if both the denominators are integer, sum will never be equal to 1.
Hence, for Statement (2), equation : a/bc= 1/c + 1/b will never be Integer if both c & b will be integer.

Charanbas · Answer

I think this is a  high-quality  question and I agree with explanation.

NischalSR · Answer

Hello  Bunuel , thanks for the explanation.

May I know the reasoning behind this statement - "Now, this expression to be an integer (1) either b=c=1 must be true or b=c=2 must be true". 
Can't b and c be two distinct positive integers such that their decimal add up to an integer? I tried and could find any such pair, but maybe there is a pair which gives an integer on addition.

Can we say for sure that such a case is never possible?

Bunuel · Answer

If b and c are more than 2, then both 1/b and 1/c are less than 1/2. Therefore, 1/b + 1/c = (less than 1/2) + (less than 1/2) = less than 1.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

M05-36

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