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Let $D(a)$ denotes sum of digits of $a$ in decimal. Examples $D(49)=4+9=13$

Let $P(a)$ denotes parity of $a$. Example $P(2)=0$ as even and $P(3)=1$ as odd.

Questions: show that there is no $a$ such that $$P(D(a))= P(D(a+1))=P(D(a+2)).$$

Sequence: Parity of sum of digits of n A179081

Sequence $\{P(D(n))\}=0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1,...$ for $n=0,1,2,...$

let $D(a, b)$ denotes sum of digits of $a$ in base $b$. In general, we can ask, is above claim true for all bases $b$?

We can understand the pattern for $b=odd$ base, note $D(odd,b)=odd$ and $D(even, b)=even$ thus $P(D(a+1,b))=P(D(a,b)+1)$

Pruthviraj
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    Well, argue that, if the units place digit in $n$ is not $9$, then $D(n+1)=D(n)+1$. – lulu Jun 29 '21 at 15:01

1 Answers1

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Hint:

We will divide the numbers into two cases- Those ending with $9$ and those which are not.

Let, us select a number of the form (ending with $9$) $$k9$$ where $k\in\mathbb{N}$. Clearly, the next number is of the form $$(k+1)0$$ Now, $k$ be either odd or even. Note, $D(k9)=k+9$ is of the opposite parity of $k$ (Why?). And for the second number, $D((k+1)0)=k+1+0=k+1$ which is of opposite parity of $k$. So, if a number ends with $9$, the number and its successive number are of same parity $(0, 0)$ or $(1, 1)$.

Now, select a number of the form (not ending with $9$) $$kn$$ where, $k\in\mathbb{N}$ and $n\in\{1,2,...,8\}$. The next number is of the form $$k(n+1)$$ Clearly, $D(kn)=k+n$ and $D(k(n+1))=k+n+1$ which are of opposite parity.

Now, see if the result you want follows.

Soham
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