Problem:
Let S be the set of all positive rational numbers r such that when the two numbers r and 55r are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of S can be expressed in the form qpβ, where p and q are relatively prime positive integers. Find p+q.
Solution:
Let r=baβ, where gcd(a,b)=1. Then,
55r=b55aβ.
Assume gcd(55,b)=1. Then the sum of the numerator and denominator of r is a+b, while that of 55r is 55a+b. These sums are unequal unless a=0, which contradicts gcd(a,b)=1. Hence,
gcd(55,b)ξ =1.
Since 55=5β
11, we consider three cases depending on the factors of b.
Case 1: b=5c, where gcd(c,11)=1
Then,
r=5caβ,55r=c11aβ.
Equating the sums:
a+5c=11a+cβ4c=10aβ2c=5a.
Since gcd(a,c)=1, we must have a=2, c=5. Thus,
r=252β.
Case 2: b=11c, where gcd(c,5)=1
Then,
r=11caβ,55r=c5aβ.
Equating the sums:
a+11c=5a+cβ10c=4aβ5c=2a.
Again, gcd(a,c)=1 implies a=5, c=2. Thus,
r=225β.
Case 3: b=55c
Then,
r=55caβ,55r=caβ.
Equating the sums:
a+55c=a+cβ54c=0βc=0,
which is invalid. So there is no solution in this case.
Combining valid cases:
S={252β,225β},252β+225β=55044+125β=550169β.
Therefore, the final answer is:
169+550=719β.
The problems on this page are the property of the MAA's American Mathematics Competitions