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The conjecture says that for any a, b belong to the the set of non-negative integers ($a$ and $b$ are not necessarily distinct), taking any natural value of $c$; we have always that $$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base $10$. This conjecture was experimented well.

Example: taking $a=2,b=3,c=5$, we will get $$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base $10$, my question is to prove or disprove this conjecture. Note that the conjecture above was proved before editing . another conjecture that I ask to prove or disprove it is that the sequence of the numbers of the form (10^a+10^2a+1)/3 is the maximally dense sequence in base 10 with the palindromic products property as described above .

The conjecture says that for any a, b belong to the the set of non-negative integers ($a$ and $b$ are not necessarily distinct), taking any natural value of $c$; we have always that $$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base $10$. This conjecture was experimented well.

Example: taking $a=2,b=3,c=5$, we will get $$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base $10$, my question is to prove or disprove this conjecture.

The conjecture says that for any a, b belong to the the set of non-negative integers ($a$ and $b$ are not necessarily distinct), taking any natural value of $c$; we have always that $$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base $10$. This conjecture was experimented well.

Example: taking $a=2,b=3,c=5$, we will get $$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base $10$, my question is to prove or disprove this conjecture. Note that the conjecture above was proved before editing . another conjecture that I ask to prove or disprove it is that the sequence of the numbers of the form (10^a+10^2a+1)/3 is the maximally dense sequence in base 10 with the palindromic products property as described above .

Post Closed as "Not suitable for this site" by Douglas Zare, Gro-Tsen, Greg Martin, Max Alekseyev, Alexey Ustinov

The conjecture says that for any a,b b belong to the the set of non-negative integers (a$a$ and b$b$ are not necessarily distinct)  , taking any natural value of c ;$c$; we have always that ((10^c-1)((10^a+10^2a+1)/3)((10^b+10^2b+1)/3))$$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base 10 $10$. This conjecture was experimented well  . Example

Example: taking a=2,b=3,c=5$a=2,b=3,c=5$, we will get 336733366799999=112344555443211$$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base 10 $10$, my question is to prove or disprove this conjecture  .

The conjecture says that for any a,b belong to the the set of non-negative integers (a and b are not necessarily distinct)  , taking any natural value of c ; we have always that ((10^c-1)((10^a+10^2a+1)/3)((10^b+10^2b+1)/3)) is palindromic in base 10 . This conjecture was experimented well  . Example : taking a=2,b=3,c=5, we will get 336733366799999=112344555443211 which is palindromic in base 10 , my question is to prove or disprove this conjecture  .

The conjecture says that for any a, b belong to the the set of non-negative integers ($a$ and $b$ are not necessarily distinct), taking any natural value of $c$; we have always that $$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base $10$. This conjecture was experimented well.

Example: taking $a=2,b=3,c=5$, we will get $$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base $10$, my question is to prove or disprove this conjecture.

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