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If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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18 Nov 2012, 13:06
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If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15
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Last edited by Bunuel on 19 Nov 2012, 04:01, edited 1 time in total.
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Re: Fraction and Inequality [#permalink]
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monsoon1 wrote: 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? You actually dont need to test numbers. It could be purely algebric approach coupled with some logical deductions. we have\(c= ab/(a+b)\) or \(c = \frac{1}{(1/a+1/b)}\) If you notice this expression and remember that c has to be integer, that would mean Denominator has to be 1 (Since numerator is already 1). There is only one such possiblity of a and b that could give you 1/a+1/b =1 So you dont need to test any number. Hope it helps. Lets kudos
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Re: Fraction and Inequality [#permalink]
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monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 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.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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14 Nov 2013, 00:22
JepicPhail wrote: 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... 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.)
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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jlgdr wrote: monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 Hey all, This one was a bit tricky indeed. Is there some other way we can notice quickly which number satisfies the ab/a+b constraint giving 'c' as an integer value Cheers! J From F.S 1, we know that for a to be positive, c<b. The given equation is valid for b=2,c=1 and also for b=3,c=2. Insufficient. Now, back to your question. We know that\(\frac{a+b}{2}\geq{\sqrt{ab}}\) Also, from the question stem, we know that \(\frac{a+b}{ab} =\frac{1}{c}\) Thus, \((a+b) = \frac{ab}{c}\). Replacing this in the first equation, we get \(\frac{ab}{2c}\geq{\sqrt{ab}}\) Or,\(c\leq{\frac{\sqrt{ab}}{2}} \to c\leq{\frac{\sqrt{15}}{2}} \to c<{2}\). Thus, the only positive integer less than 2 is One and thus c=1.Sufficient.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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07 Oct 2013, 13:19
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Nilohit wrote: 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. 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



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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sharkr wrote: VeritasPrepKarishma wrote: 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.
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)
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If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 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.
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If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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KARISHMA315 wrote: monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 Is there any algebraic approach to be sure that B is sufficient I am pretty sure I have an algebraic/number theoretic derivation of the general form of the solutions, but it's ugly and really not how you want to go about this. Still, for completeness and since you asked... \(\frac{1}{c} = \frac{a+b}{ab}\) \(c = \frac{ab}{a+b}\) Now note that, since c is a positive integer, a+b must divide ab. This is only possible if a and b share a common factor greater than 1, based on the principle that "multiple + nonmultiple = nonmultiple"; unless a and b share some factor greater than 1, a+b will not share any factor greater than 1 with a and, by the same token, a+b will not share any factor greater than 1 with b. So a and b have a greatest common factor larger than 1, and we will call this factor g. We will define a = gp and b =gq. Now rewrite the expression as \(c = \frac{gpgq}{gp+gq}\) \(c = \frac{g^2pq}{g(p+q)}\) \(c = \frac{gpq}{p+q}\) Now recall that p and q by definition are positive integers that have no common factors greater than 1. So p+q shares no factors besides 1 with either p or q. This means that c will be an integer if and only if g is divisible by p+q. We can rewrite one final time by taking g = k(p+q). This gives, at last, \(a = k(p+q)p\) and \(b = k(p+q)q\) so \(ab = k^2(p+q)^2pq\) It is now easy to verify that \(ab \leq 15\) is sufficient, since p = q = k = 1 gives \(1^2 * (1+1)^2 * 1 * 1 = 4\), but if p = q = 1 and k = 2 then a = 4, b = 4, and the product is 16, and if k = 1, either p or q = 1, and the other of q or p = 2, then a = 3, b = 6 (or vice versa), and the product is 18. And this should work for any positive integers p and q that share no common factors greater than 1 and for any positive integer k, in which case the result will be c = kpq. Try it! For instance: p = 3 q = 8 k = 4 gives a = 132 b = 352 and, sure enough, c will be an integer: c = 96 Another set: p = 7 q = 11 k = 77 a = 9702 b = 15246 c = 5929 Now, aren't you glad you asked for algebra?



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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14 Apr 2017, 05:53
monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 We are given that a, b, and c are positive integers such that 1/a + 1/b = 1/c. We can multiply the entire equation by abc and we have: bc + ac = ab c(b + a) = ab c = ab/(b + a) Statement One Alone:b ≤ 4 Knowing only that b is less than or equal to 4 is not enough information to answer the question. For example, if b = 4 and a = 4, then c = (4 x 4)/(4 + 4) = 2. However, if b = 2 and a = 2, then c = (2 x 2)/(2 + 2) = 1. Statement one alone is not sufficient. Statement Two Alone:ab ≤ 15 We can try all pairs of possible positive integer values of a and b such that ab ≤ 15 and see which pairs yield a positive integer value of c (keep in mind that c = ab/(b + a)). Furthermore, for a pair of values of a and b (e.g., a = 5 and b = 3), even though we can switch the two values and say b = 5 and a = 3, it won’t change the result of c, since multiplication and addition are commutative (i.e., ab = ba and a + b = b + a). That is, if we consider a = 5 and b = 3, we don’t need to consider a = 3 and b = 5. Thus, let’s only consider all of the cases in which a ≥ b. Lastly, we don’t have to consider b = 1 because if b = 1, then a(1)/(1 + a) = a/(1 + a) will never be an integer. Keeping in mind what we have mentioned, the following are all of the cases we need to consider: 1) a = 5, b = 3: c = ab/(b + a) = 15/8 → not an integer 2) a = 5, b = 2: c = ab/(b + a) = 10/7 → not an integer 3) a = 4, b = 3: c = ab/(b + a) = 12/7 → not an integer 4) a = 4, b = 2: c = ab/(b + a) = 8/6 → not an integer 5) a = 3, b = 3: c = ab/(b + a) = 9/6 → not an integer 6) a = 3, b = 2: c = ab/(b + a) = 6/5 → not an integer 7) a = 2, b = 2: c = ab/(b + a) = 4/4 = 1→ IS an integer Thus, c must be equal to 1. Answer: B
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Re: Fraction and Inequality [#permalink]
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19 Nov 2012, 17:38
Vips0000 wrote: monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 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?



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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07 Oct 2013, 12:38
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.



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Re: Fraction and Inequality [#permalink]
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07 Oct 2013, 12:59
I can't see how you did the following: Vips0000 wrote: monsoon1 wrote: we have\(c= ab/(a+b)\) or \(c = \frac{1}{(1/a+1/b)}\)
Could you explain? thanks. Edit: Think I got it now. Reciprocal of both sides of the original equation? or is there a different way?



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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27 Oct 2013, 11:42
monsoon1 wrote: If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c? (1) b ≤ 4 (2) ab ≤ 15 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
Last edited by jlgdr on 29 Mar 2014, 07:09, edited 1 time in total.



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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13 Nov 2013, 23:07
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...



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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13 Nov 2013, 23:22
JepicPhail wrote: 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?
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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13 Nov 2013, 23:44
mau5 wrote: JepicPhail wrote: 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.



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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20 Feb 2014, 20:20
Quote: 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. 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??



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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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20 Feb 2014, 21:11
Rohan_Kanungo wrote: Quote: 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. 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.
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