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Suppose: If all 9 numbers is smaller or equal than 1

We see: 

The sum of them is small or equal than 9

So at least one of them must be bigger than 1 to have the sum 10 :)

We have:

f(5) - f(4) = (a.5³ + b.5² + c.5 + d) - (a.4³ + b.4² + c.4 + d) 

= (125a + 25b + 5c + d) - (64a + 16b + 4c + d) 

= 61a + 9b + c 

f(7) - f(2) = (a.7³ + b.7² + c.7 + d) - (a.2³ + b.2² + c.2 + d) 

= (343a + 49b + 7c + d) - (8a + 4b + 2c + d) 

= 335a + 45b + 5c \(⋮5\) (1) 

Because a > 0  

=> 61a + 9b + c < 335 + 45b + 5c 

=> f(7) - f(2) > 2012 (2) 

(1), (2) => f(7) - f(2) isn't a prime number 

She has the same number of apples and oranges

\(\Rightarrow x=4y\)

Or \(\dfrac{x}{y}=\dfrac{4}{1}\)

We have: \(36,5,m,n\in Z^+\)

\(\Rightarrow36^m,5^n\in Z^+\)

\(\Rightarrow36^m\ge36^1=36;5^n\ge5^1=5\)

\(\Rightarrow36^m-5^n\ge36-5=31\)

\(\Rightarrow\left(36^m-5^n\right)_{Min}=31\Leftrightarrow m=n=1\)

Sorry Dao Trong Luan for the wrong title. Can you do it again?

\(\left|36^m-5^n\right|\).

If it's the last title, we have m = 1 and n big [unsolved]

We have:

\(xy=1\Leftrightarrow y=\dfrac{1}{x}\)

So \(\left|x+y\right|=\left|\dfrac{1}{x}+x\right|\)

Other way:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\left(a,b\ne0\right)\)

So \(x+\dfrac{1}{x}\ge2\)

And \(\left|x+\dfrac{1}{x}\right|\ge2\)

Or \(\left|x+y\right|\ge2\)

\(\left|x+y\right|_{Min}=2\Leftrightarrow x=y=1\)