The sum of digits of a $5$-digit number $abcde$ is equal to $10$, such that every digit is different and $a \neq 0 $ and $e \neq 0 $. If we add this number to it's reverse number $edcba$ we get a number with digits that are all same. How many $5$-digit numbers $abcde$ exist such that it fits the rules above?


Since $abcde+edcba=xxxxx$ then $a+e=b+d=2c=x$

Also we have $10=a+b+c+d+e=(a+e)+(b+d)+c=\frac 52x$

So $x=\frac {20}5=4$ and $c=\frac 12x=2$

Since $0+1+2+3+4=10$ then since $a,b,c,d,e$ are different, they can only take these values.

Considering $c=2$ is already taken we are left with possibilities $(0+4)$ and $(1+3)$ for $(a+e)$ and $(b+d)$

But $a,e\neq 0$ so $\{a,e\}=\{1,3\}$ and consequently $\{b,d\}=\{0,4\}$.

To resume, the possible numbers are



$$0+1+2+3+4=10$$ Then $\{a,b,c,d,e\}=\{0,1,2,3,4\}$ and $a\not =0$ and $e\not= 0$

  • 1
    $\begingroup$ Using brute force I got a=1, e=3, b=4, d=0, c=2, and all its swaps. Is this it or is there more, and is there a way to not brute force it? @Roman83 $\endgroup$ – Ayy Lmao Nov 7 '17 at 12:59

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