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If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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23 Mar 2017, 18:08

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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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06 Apr 2017, 12:20

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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...

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 + non-multiple = non-multiple"; 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 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.

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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