Adding extra digit $1$ or $7$ to a multidigit number changes remainder mod $3$

Question

The problem where this remark is made is pasted below. Suppose we have number $$k=a_n\cdots a_1a_0,$$ and suppose $$k\equiv 0 \pmod{3}.$$ Then why would $$k'=a_n\cdots a_1a_01 \equiv 1 \pmod{3}?$$

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Answer 1

This is because, as $10\equiv 1\mod 3$, any number is congruent mod. $3$ to the sum of its digits, i.e. $$k\bmod 3\equiv a_n+\dots+a_1+a_0\mod 3,$$ so if $k\equiv 0\mod 3$, then $$k'= a_n \dotsm a_1 a_01=10k+1\equiv 1\cdot 0+1=1\mod 3.$$

Answer 2

Note that $$k' = 10k + 1 \implies k' \equiv k+1\pmod{3}.$$

Answer 3

If $k=3m$, then $k'=10k+1=3\cdot 10m+1$.

Answer 4

There is a much more general and much more fascinating theorem from which the result from your post follows:

Theorem 1: A number is congruent mod 3 to its sum of digits.

It is quite clear that the result you are looking for follows from this. Another thing that follows from Theorem 1, and which you have probably heard of already at some point in your life, is

Theorem 2: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

If, like a younger version of myself, you have heard of Theorem 2 but did not know how to prove it, this is your lucky day! Using the other two answers you can easily cook up a proof of Theorem 1 (using induction) and then Theorem 2 follows.

PS: Another approach to proving Theorem 1 is to derive it from its big cousin:

Theorem 3: A number is congruent mod 9 to its sum of digits

Which itself can be proven along the lines sketched in the other two answers.