Reciprocals!
Number Theory
Level
pending
Find three distinct positive integers with the least possible sum such that the sum of there reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer.
Submit your answer as sum of those numbers!
The answer is 11.
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1 solution
Let the 3 numbers taken be x, y, and z. Also, consider 3 numbers a,b,c such that the 3 below equations are satisfied: 1/x + 1/y = a/z ----->(1)
1/y + 1/z = b/x ----->(2)
1/z + 1/x = c/y ----->(3) =====> y/z + y/x = c -----> (3) (2) - (3) ===> 1/y - 1/x = b/x - c/y
===> (1+c)/y = (1+b)/x
===> y/x = (1+c)/(1+b) ----->(4)
Similarly, y/z = (1+c)/(1+a)-------->(5)
Substituting (4) and (5) in (3) : (1+c)[ 1/(a+1) + 1/(b+1)] = c Which on Simplification gives: a + b + c = abc - 2 But x+ y + z carries minimum value. Hence, x= 1 + b, y= 1 +c and z= 1+ a. So x + y + z carries minimum value when a + b + c does which happens when abc-2 does. If you just take a few seconds to look at the possibilities, we quickly find out that this value of a + b + c is 8 which occurs at (a, b, c )= (1, 2, 5) and their permutations. Hence, the minimum such possible value o x + y + z is 11 So our final answer is 11 Hope this helps