Why does the summation of the digits of a number divisible by 3 also become divisible by 3?
In: Mathematics
It’s just a natural pattern, which you can see counting upwards all multiples of 3.
* 03 – 06 – 09 … all divisible by 3
* 12 – 15 – 18 … notice that, you have +1 in the tens place, and -1 in the one’s place
* 21 – 24 – 27 … +2 in the tens plance, -2 in the one’s place
* 30 – 33 – 36 – 39 … and now the pattern repeats
This works in the base 10 system because it cycles cleanly over the ten’s place, always maintaining this relationship between the rightmost digits.
Now to further expand this, and you’ll note the hundreds and tens place follow the same (extended) pattern.
* 000 – 003 – 006 – 009
* …
* 030 – 033 – 036 – 039
* …
* 060 – 063 – 066 – 069
* …
* 090 – 093 – 096 – 099
* 102 – 105 – 108
* 111 – 114 – 117 … +1 hundred’s place, +1 ten’s place, -2 one’s place
* 120 – 123 – 126 – 129 … +1 hundred’s place, +2 ten’s place, -3 (or 0) one’s place
Thus, every increase in a value to the digit to the left, is accompanied by an decrease in the value of the one’s place. In this way, they balance out, so whatever’s left if you add them to the one’s place will *always* be a multiple of 3.
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