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19 votes
19 votes
Given positive integers x and y such that x doesn't equal y and $1/x+1/y=1/18$ what is the smallest possible value for x+y?

User Diatrevolo
by
2.4k points

2 Answers

20 votes
20 votes

Solution:

We are given that

1/x + 1/y = 1/12.

Multiplying both sides by 12xy, we get 12y + 12x = xy, which we can write as

xy - 12x - 12y = 0.

We can then apply Simon's Favorite Factoring Trick, by adding 12*12 = 144 to both sides:

xy - 12x - 12y + 144 = 144.

The left-hand side factors:

(x - 12)(y - 12) = 144.

We want to minimize x + y, which is equivalent to minimizing the sum of the two factors x - 12 and y - 12.

Since the product of x - 12 and y - 12 is a constant, we can minimize their sum by making them as close to each other as possible. Normally, we could set x - 12 = y - 12 = 12, but the problem states that x does not equal y. The next two factors of 144 (whose product is 144) that are close to 12 as possible are 9 and 16. Hence, we can set x - 12 = 9 and y - 12 = 16, to get x = 21 and y = 28, so the minimum value of x + y is 21 + 28 = 49.

User Qamar
by
2.9k points
15 votes
15 votes

Answer:

Hello,


\boxed{Answer: 75}

Explanation:

x,y integers, x,y >0 ,x≠y


(1)/(x) +(1)/(y) =(1)/(18)\\\\(x+y)/(xy) =(1)/(18)\\\\x+y=(xy)/(18)\\\\x=(18y)/(y-18)\\x=(18y-324 +324)/(y-18)\\x=18+(324)/(y-18)\\


(1)/(x) +(1)/(y) =(1)/(18)\\\\(x+y)/(xy) =(1)/(18)\\\\x+y=(xy)/(18)\\\\x=(18y)/(y-18)\\x=(18y-324 +324)/(y-18)\\x=18+(324)/(y-18)\\\begin{array}y-18&y&x&x+y\\---&---&---&---\\1&19&18+324=342&362\\2&20&18+162=180&200\\3&21&126&147\\4&22&99&121\\6&24&108&132\\12&30&45&75\\18&36&36&72\\\end{array}\\\\\\\boxed{Answer: 75}\\\\

User Hokam
by
3.5k points
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