Prove that there will be atleast one even number in the sum of odd numbered digit and its reverse,if the number is of the form 4n+1.

Question

Prove that there will be at least one even number in the sum of odd numbered digit and its reverse,if the number of digits is of the form $4n+1$.

If the number of digits is of the form $4n+3$ ,there exist number whose sum of the number and its reverse contains all odd numbers.

Answer 1

$$\sum_{i=0}^{n-1}{10^i(d_i+d_{n-i-1})} = \sum_{i=0}^{m-1}{10^i(2x_i+1)}$$ $$d_i \in\left\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\right\}$$ $$x_i \in\left\{0, 1, 2, 3, 4\right\}$$ $$m \in\left\{n, n+1\right\}$$

Let $m=n+b, b\in\left\{0, 1\right\}$

$$\sum_{i=0}^{n-1}{10^i(d_i+d_{n-i-1})} = \sum_{i=0}^{n+b-1}{10^i(2x_i+1)}$$

$$\sum_{i=0}^{n-1}{10^i(d_i+d_{n-i-1}-2x_i-1)} = \begin{cases}0 & b=0\\10^n(2x_n+1) & b = 1\end{cases}$$

$$\sum_{i=0}^{n-1}{10^i(d_i+d_{n-i-1}-2x_i-1)} = 10^n(2x_n+1)b$$

Because the carry of the sum of two numbers is always 1 or 0,

$$\sum_{i=0}^{n-1}{10^i(d_i+d_{n-i-1}-2x_i-1)} = 10^nb$$

Let $y_i=d_i+d_{n-i-1}-2x_i-1$

$$\sum_{i=0}^{n-1}{10^iy_i} = 10^nb$$

By the fact that $2x_i+1$ is the sum of $d_i+d_{n-i-1} \bmod 10$

$$2x_i+1 = d_i+d_{n-i-1}+p_i \bmod 10 = d_i+d_{n-i-1}+p_i-10q_i$$

$$p_i, q_i\in\left\{0, 1\right\}$$

$$y_i = 10q_i-p_i \in\left\{-1, 0, 9, 10\right\}$$

$n$ is odd, so let $n=2k+1$

$$y_i = d_i+d_{2k-i}-2x_i-1$$

For $i=k$

$$y_k = 2(d_k-x_k)-1 \in \left\{-1, 9\right\}$$

For $i=k+t, -k \leq t \leq k$

$$y_{k+t} = d_{k+t}+d_{k-t}-2x_{k+t}-1$$

$$y_{k+t}-y_{k-t} = 2(x_{k+t}-x_{k-t})$$

Consider the range of $x_i$

$$-8 \leq y_{k+t}-y_{k-t} \leq 8$$

Consider all possible values of $y_i$

$$y_{k+t} = y_{k-t}$$

All $10^i$ terms must be canceled out, except for $i=n$, so we have following rules.

1. If $y_i\in\left\{-1,9\right\}$, then $y_{i-1}\in\left\{9, 10\right\}$ and $\exists j<i$ such that $y_j=10$
2. If $y_{i}\in\left\{9, 10\right\}$, then $y_{i+1}\in\left\{-1, 9\right\}$ or $i=n-1$

Now replace $i$ by $k+t$ and $k-t$, and assume $t\geq 0$.

1. If $y_{k+t}=y_{k-t}\in\left\{-1,9\right\}$, then $$y_{k+t-1}=y_{k-t+1}\in\left\{9, 10\right\}$$ $$y_{k-t-1}=y_{k+t+1}\in\left\{9, 10\right\}$$ and $\exists j<k-t$ such that $y_j=10$
2. If $y_{k+t}=y_{k-t}\in\left\{9, 10\right\}$, then $$y_{k-t-1}\in\left\{-1, 9\right\}$$ $$y_{k-t+1}=y_{k+t-1}\in\left\{-1, 9\right\}$$ $y_{k+t+1}\in\left\{-1, 9\right\}$ or $t=k$

If $y_{k+t}=9$ for some $t$, because $y_{2k}=y_0\in\left\{0, 10\right\}$ would never be 9, it is sure that $t\neq k$, then $$y_{k+t-1}=y_{k-t+1}=y_{k+t+1}=y_{k-t-1}=9$$

This will make $y_{i}=9$ for all $i$, which violates $\exists j<k-t$ such that $y_j=10$.

So no $y_i$ could be 9. We can simply the rules.

1. $y_i\in\left\{-1, 0, 10\right\}$
2. If $y_{k+t}=y_{k-t}=-1$, then $$y_{k+t-1}=y_{k-t+1}=10$$ $$y_{k-t-1}=y_{k+t+1}=10$$
3. If $y_{k+t}=y_{k-t}=10$, then $$y_{k-t-1}=-1$$ $$y_{k-t+1}=y_{k+t-1}=-1$$ $y_{k+t+1}=-1$ or $t=k$

We already know that $y_k=-1$, so

If $t$ is odd, then

$$y_{k+t}=10$$

If $t$ is even, then

$$y_{k+t}=-1$$

We know that $y_0 = 10$, because no term will cancel it if $y_0 = -1$, so k is odd.

Let $k = 2v+1$ $$n=2k+1=4v+3$$