rilarfer wrote: An integer \(n \geq 2\) is said to be tuanis if, when you add the smallest prime divisor of \(n\) and the largest prime divisor of \(n\) (these divisors can be the same), you obtain an odd result. Calculate the sum of all tuanis numbers that are less or equal to \(2023\). $\color{blue}\boxed{\textbf{Answer: 1021086}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $p$ be the smallest prime divisor of $n$, and $q$ be the largest prime divisor of $n$ $$\Rightarrow p+q\equiv 1\pmod{2}$$$$\Rightarrow p=2, q\equiv 1\pmod{2}$$$$\sum^{\le 2023}_{\text{tuanis}}=\sum^{1011}_{i=1} 2i -\sum^{10}_{i=1} 2^i=1011\times 1012-(2^{11}-2)=1021086_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Let $P_k$ be the $k^{\textrm{th}}$ prime divisor of $n$ Since $P_1+P_k\equiv 1 \pmod{2}$, then $P_1=2$ and $P_k\equiv 1 \pmod{2}$. This means that tuanis numbers are even numbers that are not powers of 2. \[2+4+6+8+\cdots+2022=\sum_{i=1}^{1011}2i=1011\cdot 1012=1023132\]\[2+4+8+16+\cdots+1024 = \sum_{i=1}^{10}2^{i}=2^{11}-1=2047\]\[\sum_{tuanis}^{2022}=1011\cdot 1012-(2^{11}-1)=\boxed{1021085}\]

hectorleo123 wrote: rilarfer wrote: $$\sum^{\le 2023}_{\text{tuanis}}=\sum^{1011}_{i=1} 2i -\sum^{10}_{i=1} 2^i=1011\times 1012-(2^{11}-1)=1021085_\blacksquare$$ Are you sure? The tuanis number are even, so their sum is also even.

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rilarfer wrote: hectorleo123 wrote: rilarfer wrote: $$\sum^{\le 2023}_{\text{tuanis}}=\sum^{1011}_{i=1} 2i -\sum^{10}_{i=1} 2^i=1011\times 1012-(2^{11}-1)=1021085_\blacksquare$$ Are you sure? The tuanis number are even, so their sum is also even.

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oh my mistake

First, we assert that tuanis numbers are all the even numbers excluding the powers of two. All prime divisors of odd numbers will be odd, yielding an even sum and the only prime divisor of powers of 2 is 2, yielding again an even sum. In order to verify that all even numbers excluding powers of two, ir suffices to notice that the smallest prime divisor for even numbers is always two and since the numbers are not powers of two, the biggest prime divisor will always be odd, yielding an odd sum. To finish the problem, it is easy to calculate the sum of all even numbers from 1 through 2023 and then subtract all even powers of two. The former sum gives me 1,023,132 (using the formula for the sum of arithmetic sequences) and the latter which I must subtract gives me 2046 (just using properties of powers of 2), giving me a final result of 1,021,086.