Question

Tricky Puzzle :
A integer contains 110 digits such that any 10 consecutive digits have no repeating digits. Find the sum of all its digits. ​

Answer 1

$\large\underline{\large\underline{\text{Clues}}}$

The digits must satisfy the two conditions.

Each digit consists of a different one from 0, 1, …, 8, 9 in a pair of 10 numbers.

Every ten digits are equal.

$\large\underline{\large\underline{\text{Explanation}}}$

Let's find the sum of a pair consisting of 10 different digits.

$\cdots\longrightarrow0+1+\cdots+8+9=\dfrac{1}{2}\times10\times9$

$\cdots\longrightarrow0+1+\cdots+8+9=45.$

Every ten digits repeat 11 times, so the sum is,

$\cdots\longrightarrow45\times11=495.$

So, the answer is,

$\cdots\longrightarrow\boxed{495.}$

$\large\underline{\large\underline{\text{Learn more}}}$

Where $a,d,l$ are the first term, common difference, last term, the arithmetic terms are,

$\cdots\longrightarrow a,a+d,a+2d,\cdots,l-2d,l-d,l$

Let $S_{n}$ express the sum of $n$ consecutive numbers,

$\begin{aligned}S_{n}&=a+(a+d)+(a+2d)+\cdots+(l-2d)+(l-d)+l\\S_{n}&=l+(l-d)+(l-2d)+\cdots+(a+2d)+(a+d)+a\end{aligned}$

$\cdots\longrightarrow 2S_{n}=n(a+l)$

$\cdots\longrightarrow\boxed{S_{n}=\dfrac{1}{2}n(a+l).}$

Where $a,n,l$ are the first term, number of terms, last term, the arithmetic series is,

$\cdots\longrightarrow S_{n}=\dfrac{1}{2}n(a+l)$

$\cdots\longrightarrow S_{n}=\dfrac{1}{2}\{a+a+(n-1)d\}$

$\cdots\longrightarrow\boxed{S_{n}=\dfrac{1}{2}\{2a+(n-1)d\}.}$