Introduction to real valued functions

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$Suppose a, b and c are positive integers with a < b < c such that \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 . The value of (a + b + c) is$

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Solution

$\begin{array}{l} We must have \frac{1}{a} < 1, so a > 1 \\ Since \frac{1}{a} > \frac{1}{b} > \frac{1}{c} . \\ \Rightarrow \frac{1}{a} > \frac{1}{3} \\ \Rightarrow a < 3 \Rightarrow a = 2 \\ \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{2} where 2 < b < c \\ Similarly \frac{1}{b} > \frac{1}{4} \\ so b < 4 \Rightarrow b = 3 \end{array}$
$\begin{array}{l} Now c = 6 satisfies the equation \\ \Rightarrow a + b + c = 11 \end{array}$

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Introduction to real valued functions

Remember concepts with our Masterclasses.

Suppose a,b and c are positive integers with a<b<c such that 1a+1b+1c=1. The value of (a+b+c) is

We must have 1a<1, so a>1 Since 1a>1b>1c . ⇒1a>13⇒a<3⇒a=2⇒1b+1c=12 where 2<b<c Similarly 1b>14 so b<4⇒b=3 Now c=6 satisfies the equation ⇒a+b+c=11

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$Suppose a, b and c are positive integers with a < b < c such that \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 . The value of (a + b + c) is$