Explaining why adding the digits of a multiple of $9$ (repeatedly if necessary) yields $9$

Question

I'm new to this forum. I think my math abilities are above average but nowhere near the top. I would like to hear an explanation of why this happens.

it's a mystery to me why this happens I'm sure some here can explain it.

I haven't got a clue.

here we go:

$9\cdot 5=45$ add up the digits - $4+5=9$

$9\cdot 8=72$ add up the digits - $7+2=9$

$9\cdot 7=63$ add up the digits - $6+3=9$

$9\cdot 17=153$ add up the digits - $1+5+3 =9$

$9\cdot 77=693 $ add up the digits - $6+9+3$

and get $18$ then add up those digits $1+8=9$

$9\cdot 4,796=43,164$ add up the digits $4+3+1+6+4$

$9\cdot 104,675 = 942,075$ add up the digits $9+4+2+0+7+5$ and get $27$ then add up those digits $2+7=9$

$9\cdot5,327,894 = 47,951,046 $add up the digits $4+7+9+5+1+0+4+6$ and get $36$ then add up those digits $3+6=9$

Related (duplicate?): "Don't understand casting out nines". A site search should reveal other relevant entries. — Blue, Jul 25 '21 at 13:35
Does this answer your question? Divisibility by $9$ There are many other duplicates. Search the site for "divisibility by 9". — Ethan Bolker, Jul 25 '21 at 14:00

score 1 · Answer 1 · answered Jul 25 '21 at 13:36

this one is pretty intuitive to answer. Let's write out the tens/hundreds in front of the nine, so that: $1 \cdot 9 = 9 \rightarrow 00...009$

$2 \cdot 9 = 9+9 \rightarrow 00...018$

$3 \cdot 9 = 9+9+9 \rightarrow 00...027$

$4 \cdot 9 = 9+9+9+9 \rightarrow 00...036$

as you can see, when you add a 9, the "tens" increase and the "ones" decrease. Meaning that the sum of both always stays constant.

When you go from $9 \cdot 9 = 00...081$ to $10 \cdot 9 = 00...090$ the counter "resets" and you start all over, just one decimal place to the left.

I hope that's a fairly intuitive explanation!

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