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 and 55r=c11aβ. Equating the sums of the numerator and denominator:
a+5c4c2cβ=11a+c=10a=5a.β
Since gcd(a,c)=1, we must have a=2 and c=5. Thus, r=252β.
Case 2: b=11c, where gcd(c,5)=1
Then r=11caβ and 55r=c5aβ. Equating the sums:
a+11c10c5cβ=5a+c=4a=2a.β
Again, gcd(a,c)=1 implies a=5 and c=2. Thus, r=225β.
Case 3: b=55c
Then r=55caβ and 55r=caβ. Equating the sums:
a+55c54ccβ=a+c=0=0,β
which is invalid. So there is no solution in this case.
Combining the valid cases, the set of possible values for r is S={252β,225β}. The sum of these values is:
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