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06 Apr 2017, 13:20
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KARISHMA315
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 + 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 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?
14 Apr 2017, 05:53
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monsoon1
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
04 Jan 2018, 10:03
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mau5
Whoa! That's some really interesting point you are making. Never heard of this... that the average of two average of two numbers is always greather than the square root of the product of two numbers! Does this property hold always?
