If a, b, and c are positive integers such that 1/a + 1/b = 1

given is c= ab/(a+b)
thus, since c is integer, ab/a+b must be an integer.

statement 1: b ≤ 4
No information can be drawn. Not sufficient
statement 2: ab ≤ 15
Only possible value of ab, such that ab/a+b is integer could be when a=2,b=2. Thus, c=1.
Sufficient.

Ans B it is.

How did you find that only these values would satisfy?
Did you test several numbers?

The question also doesn't give us the clue whether the numbers are the same or different.So, we have to test many numbers.right?

Can you please show the steps or any other way to get to the correct answer?

I have a question here. If the solution given by Vips0000 is correct and to me it seems perfect, then neither of the statements is actually needed to solve the question there can be only one combination possible from the question stem itself; a=b=2 and consequently c=1. In such a scenario shouldn't the answer be D, since both statements independently will lead us to the answer.

I can't see how you did the following:





Could you explain? thanks.


Edit: Think I got it now. Reciprocal of both sides of the original equation? or is there a different way?

Well, lets say the ab<or= 15 restriction was not there. Then if a=10 and b=10 then ab/(a+b) would equal 100/20...which is an integer.

The only possible values when ab<or=15 are 2 and 2.


but for the b<or=4 statement, you could still have a case where say b=4 and a=12, and ab/a+b=48/16=3. Then you would have 2 (or more) possible values for a and b if you also include the possible value set of "a=2 and b=2".


Vips0000 reasoning isn't actually perfect. In the case of \(\frac{1}{1/a+1/b}\) the denominator does not have to be "1". \(\frac{1}{1/a+1/b}\) could equal \(\frac{1}{1/4+1/12}\) which reduces to \(\frac{1}{4/12}\) then \(1*\frac{12}{4}\) which equals 3

Updating with new solution by jlgdr

OK so we have that ab / a+b = C is an integer. Therefore let's hit the first statement.

Statement 1 says that b<=4. We have two choices here (actually 3). Let's begin with (2,2) C would equal 2. Now if we pick (4,4), C is again two so same answer. But if we pick (6,6) then C= 3. So not sufficient.

From statement 2 we know that ab<=15. Hence ab has to be 2,2 since both a,b are positive integers and of course C = 2.

Therefore B stands

Gimme kudos
Cheers
J

Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...

I don't understand how you can get the answer from the First fact statement.

Also, how can you get that c=1, without fact statement 2?

Oh, I see now why you need 15. \(sqrt(15)\) is less than 4, so C is less than or equal to something like 1/2, 2/2, 3/2, etc... and since only integer here is 1, C equals 1.

Actually, this is not correct. c needn't be 1 in every case.

Take a = 2, b = 2. In this case c = 1

Take a = 4, b = 4. In this case,
1/4 + 1/4 = 1/2
c = 2

Take a = 3, b = 6. In this case,
1/3 + 1/6 = 1/2
c = 2

Take a = 15, b = 30. In this case
1/15 + 1/30 = 1/10
etc

Basically, you have to look for values such that when the numerators add up, the sum is divisible by the denominator. a and b cannot be 1 since we need the sum to be less than 1.

Statement 1: b <= 4
This gives you different values of c. c could be 1 or 2. Not sufficient.

Statement 2: ab <= 15
a and b could be 2 each. There is no other set of values. Try the small pairs (2, 4), (3, 3).
Hence (B) alone is sufficient.
(Note the algebraic solution provided by mau5 for statement 2.)

What if a = 1 and b =1, it would still satisfy all the conditions i.e. ab<15 and c would be an integer only. Doesnt this gives 2 solutions for statement 2 as well??

If a = 1, b = 1,

1 + 1 = 1/c
c = 1/2
c is not an integer in this case.

What are the numerators, the sum, and the denominator referring to?

Thanks in advance.

1/a + 1/b = 1/c

You have to look for values of a and b (denominators) such that when you add 1/a and 1/b, the numerator you get is divisible by the denominator of the fraction (which you get after adding)

e.g. if a = 3, b = 6

1/a + 1/b = 1/3 + 1/6 = (2 + 1)/6 = 3/6
The numerator is 3 and the denominator is 6. The numerator is divisible by the denominator such that you get 1 in the numerator after cancelling.

1/a + 1/b = 1/3 + 1/6 = 1/2

When you get 1 in the numerator, the denominator of the resulting fraction is the value of c (i.e. 2 above)

Thanks for the reply.
For clarification, don't you mean the denominator (ie. 6) is divisible by the numerator (ie. 3)?
3 / 6 = 0*6 + 3 whereas 6/3 = 3*2 (no remainder).

Source : Manhattan Advanced Quant Question No. 3

OFFICIAL SOLUTION

Since a, b, and c are positive integers, 1/a, 1/b, and 1/c are each less than or equal to 1. Also, 1/a and 1/b must both be less than 1/c, implying that a and b must both be greater than c. Furthermore, either 1/a or 1/b must be no less than ½ of 1/c, because if both fractions are less than ½ of 1/c, the sum will be less than 1/c , which implies that either a or b must be less than or equal to 2c.

These implications, along with the integer constraints and the given equation, greatly reduce the number of possible values for a, b, and c. We should make a comprehensive list for the first few c values. A good approach is to work backwards from the target value of c (= 1, 2, 3, etc.) and try to find integer values of a and b that fit the equation. There are only a few possibilities in each case. One pair that always works is making both a and b equal to 2c. Also, if we make a equal to c + 1, then there is always an integer value for b (which winds up equaling ac or c(c + 1), as we can show by a little algebra).

This question can be rephrased to “Which (a, b) pairs listed above are valid, and what is the resulting value of c?”

(1) INSUFFICIENT: If b ≤ 4, then valid (a, b) pairs are (2, 2) and (6, 3) and (4, 4) and (12, 4). This implies that c could be 1, 2, or 3.

(2) SUFFICIENT: If ab ≤ 15, then the only valid (a, b) pair is (2, 2) and c must be 1.

By the way, fractions of the form 1/integer are known as Egyptian fractions, because they were used first in ancient Egypt.

The correct answer is B.

Is there any algebric approach to be sure that B is sufficient