given the positive integers a , b , c such that . Evaluate the expression

Question by KEITA FC 8C

KEITA FC 8C

19/12/2017 at 12:32

given the positive integers a , b , c such that $2a^a+b^b=3c^c$ . Evaluate the expression $P=2015^{a-b}+2016^{b-c}+2017^{c-a}$

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Kaya Renger Coordinator 19/12/2017 at 20:14

2a^a + b^b = 3c^c

+) With c is an odd number

=> 3c^c is an odd number => 2a^a + b^b is an odd number

<*> With a is an even number => 2a^a is an even number => b is an odd number

=> $\left\{{}\begin{matrix}a\ge2\\b\ge1\\c\ge1\end{matrix}\right.$ =>  $\left\{{}\begin{matrix}2a^a+b^b\ge2.2^2+1^1=7\\3c^c\ge3\end{matrix}\right.$ (removed)

<*> With a is an odd number => 2a^a is an even number => b is an odd number

=> $\left\{{}\begin{matrix}a\ge1\\b\ge1\\c\ge1\end{matrix}\right.$ => $\left\{{}\begin{matrix}2a^a+b^b\ge2.1^1+1^1=3\\3c^c\ge3\end{matrix}\right.$ (satisfy)

+) With c is an even number

=> 3c^c is an even number => 2a^a + b^b is an even number

<*> With a is an even number => 2a^a is an even number => b^b is an even number

=> $\left\{{}\begin{matrix}a\ge2\\b\ge2\\c\ge2\end{matrix}\right.$ =>  $\left\{{}\begin{matrix}2a^a+b^b\ge2.2^2+2^2=12\\3c^c\ge3.2^2=12\end{matrix}\right.$ (satisfy)

<*> With a is an odd number => 2a^a is an even number => b^b is an even number

=> $\left\{{}\begin{matrix}a\ge1\\b\ge2\\c\ge2\end{matrix}\right.$ =>  $\left\{{}\begin{matrix}2a^a+b^b\ge2.1^1+2^2=6\\3c^c\ge3.2^2=12\end{matrix}\right.$ (removed)

From the above analysis, we can see , if a = b = c > 0 then the equation is satisfy .

So P = $2015^{a-b}+2016^{b-c}+2017^{c-a}=2015^0+2016^0+2017^0=1+1+1=3$
Selected by MathYouLike
Faded 22/01/2018 at 12:43

a^a + b^b = 3c^c

+) With c is an odd number

=> 3c^c is an odd number => 2a^a + b^b is an odd number

<*> With a is an even number => 2a^a is an even number => b is an odd number

=> ${\begin{matrix} a \geq 2 \\ b \geq 1 \\ c \geq 1 \end{matrix}$

=>   ${\begin{matrix} 2 a^{a} + b^{b} \geq {2.2}^{2} + 1^{1} = 7 \\ 3 c^{c} \geq 3 \end{matrix}$

(removed)

<*> With a is an odd number => 2a^a is an even number => b is an odd number

=> ${\begin{matrix} a \geq 1 \\ b \geq 1 \\ c \geq 1 \end{matrix}$

=> ${\begin{matrix} 2 a^{a} + b^{b} \geq {2.1}^{1} + 1^{1} = 3 \\ 3 c^{c} \geq 3 \end{matrix}$

(satisfy)

+) With c is an even number

=> 3c^c is an even number => 2a^a + b^b is an even number

<*> With a is an even number => 2a^a is an even number => b^b is an even number

=> ${\begin{matrix} a \geq 2 \\ b \geq 2 \\ c \geq 2 \end{matrix}$

=>   ${\begin{matrix} 2 a^{a} + b^{b} \geq {2.2}^{2} + 2^{2} = 12 \\ 3 c^{c} \geq {3.2}^{2} = 12 \end{matrix}$

(satisfy)

<*> With a is an odd number => 2a^a is an even number => b^b is an even number

=> ${\begin{matrix} a \geq 1 \\ b \geq 2 \\ c \geq 2 \end{matrix}$

=>   ${\begin{matrix} 2 a^{a} + b^{b} \geq {2.1}^{1} + 2^{2} = 6 \\ 3 c^{c} \geq {3.2}^{2} = 12 \end{matrix}$

(removed)

From the above analysis, we can see , if a = b = c > 0 then the equation is satisfy .

So P = $2015^{a - b} + 2016^{b - c} + 2017^{c - a} = 2015^{0} + 2016^{0} + 2017^{0} = 1 + 1 + 1 = 3$
Kaya Renger Coordinator 19/12/2017 at 21:10

Because a,b,c are positive interger numbers so that

a,b,c > 0

If a,b,c are even numbers then $a,b,c\ge2$

If a,b,c are odd numbers then $a,b,c\ge1$

KEITA FC 8C

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