Assume, $\Phi : A \to A, A=\{0,1,2,\cdots \}$ is a function, which is defined as,

\[\Phi(x) = \begin{cases}

0 \quad \text{if } x \text{ is a prime}\\

\Phi(x - 1) \quad \text{if } x \text{ is not a prime} \end{cases} \]

Find \[ \sum_{x=2}^{2010} \Phi(x)\]

(Corrected)

## Dhaka Higher Secondary 2010/1 (Secondary 2010/8)

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### Re: Dhaka Higher Secondary 2010/1

is this BDMO bot user like gamebot in AOPS ,

the problem was corrected from $x=2$ to 2010 not $x=0$

the aswer is 0 for every $x$ will be $f(x)=0$ ,even if we think $2$ is only prime number

the problem was corrected from $x=2$ to 2010 not $x=0$

the aswer is 0 for every $x$ will be $f(x)=0$ ,even if we think $2$ is only prime number

### Re: Dhaka Higher Secondary 2010/1 (Secondary 2010/8)

I have no idea about summations, Do not even know what does the symbols and variables mean there( I beg someone teach me what are they, physics give me a lot of trouble for this), but comin to the function, for any number n, if it is prime then $\phi (n)$ is 0, and if it is not prime then it will continue to go $\phi (n-1)$ .. until it gets prime, and then it gets 0 as well!!! (I hope I got the question and concept right) so ultimately for any n, $\phi (n)$ =0!!! now please clear e the summation thing so that I can atleast try to post a solution.......

### Re: Dhaka Higher Secondary 2010/1 (Secondary 2010/8)

The answer is 0..