Proof in mirror numbers / multiples of 3

Question

Happy 2019,

What's the proof for: any integer number subtracted by its mirror number is always a multiple of 3?

This is, abfc – cfba will always be a multiple of 3 (abfc, integer number).

Thanks

This has been solved here on this site. Extra for you: and $abfc+cfba$ will always be a multiple of $11$, see here. — Dietrich Burde, Jan 06 '19 at 18:09

score 0 · Answer 1 · answered Jan 06 '19 at 18:07

0

I suggest that you prove and use the followinng lemma: if $n$ is an integer and $s$ is the sum of its digits, then $n-s$ is divisible by $9$.

answered Jan 06 '19 at 18:07

Aphelli

score 0 · Answer 2 · answered Jan 06 '19 at 18:09

0

Hint: You may want to look at the following expression

$$(1000a+100b+10f+c)-(1000c+100f+10b+a)$$

Can you simplify the above expression and reach a conclusion? Note that the same idea applies regardless of how many digits the number has.

answered Jan 06 '19 at 18:09

KM101

score 0 · Accepted Answer · answered Jan 06 '19 at 18:10

You can write any general number $N_1=abfc$ as

$$=1000a + 100b +10f + c$$

Similarly you can write its mirror $N_2$ as

$$=1000c + 100f +10b + a$$

Subtracting $N_1$ and $N_2$

$$N_1 - N_2= 999a +90b -90f - 999c = 3(333a + 30b - 30f - 333c)$$

As you can see this number is clearly divisible by $3$ irrespective of the values of $a,b,c,f$

3 Answers3