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Let $x_1$ be a given positive integer. A sequence $\{x_n\}_ {n\geq 1}$ of positive integers is such that $x_n$, for $n \geq 2$, is obtained from $x_ {n-1}$ by adding some nonzero digit of $x_ {n-1}$.

Prove that

a) the sequence contains an even term;

b) the sequence contains infinitely many even terms

Now i did not understand the question and what it is asking to prove , what is meaning of "$x_n$, for $n \geq 2$, is obtained from $x_ {n-1}$ by adding some nonzero digit of $x_ {n-1}$"

So can somebody pls give me some example ?

thankyou

Ishan
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    What is it you don't understand, is it "digit"? I guess "digit" means "decimal digit", so it $x_{n-1}=1024$, then $x_n$ is either $1024+1=1025$ or $1024+2=1026$ or $1024+4=1028$. – bof Nov 18 '20 at 09:36
  • If you start with $x_1=727$, then I would choose the digit $2$ of $727$, add it to $727$ to obtain $x_2=727+2=729$. Now I choose the digit $2$ of $x_1=729$ add it to $729$ to obtain $x_3=729+2=731$. Now I can choose either $7$ or $3$ or $1$ and add it to $731$ to obtain $x_4$ which can take the values $738$ or $734$ or $732$ – GraduateStudent Nov 18 '20 at 09:38
  • got it thanks.. – Ishan Nov 18 '20 at 10:24

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Hint-Notation- $a_1 a_2 a_3 \dots a_n$ is an $n$ digit number, that is $23$ is a $2$ digit number with $a_1=2$ and $a_2=3, 92938$ is a $5$ digit number with $a_1=9,a_2=2,a_3=9,a_4$ $=3$ and $a_5=8$

If you start with $x_1=a_1a_2a_3 \dots a_k$, then if $a_k$ is even then we are done, else show that after few steps you will get $x_n=b_1b_2 \dots b_k$ where $b_1,b_2,b_3 \dots b_{n-1} $ are all odd. Then if $b_n $ is $odd$, then $x_{n+1}$ is even else $x_n$ is even

Repeated application of the above procedure will give you infinitely many even numbers

GraduateStudent
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