Prove that the (recursive) sum of the digits of any number multiplied by 9 is 9

Asked Feb 04 '20 at 12:08

Active Feb 04 '20 at 12:15

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Prove that $$\sum_{digits}...\sum_{digits}9a = 9, a∈ℕ$$

Examples

9x2 = 18 -> 1+8 = 9

9x21 = 189 -> 1+8+9 = 18 -> 1+8 = 9

9x543 = 4887 -> 4+8+8+7 = 27 -> 2+7 = 9 

9x87,523 = 787,707 -> 7+8+7+7+0+7 = 36 -> 3+6 = 9 

9x98,745,632,584 = 888,710,693,256 -> 8+8+8+7+1+0+6+9+3+2+5+6 = 63 -> 6+3 = 9

edited Feb 04 '20 at 12:15

Martin Sleziak

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asked Feb 04 '20 at 12:08

owninggreendragsdude

I don't know how to mathematically express "sum of the digits", but I gave it a shot – owninggreendragsdude Feb 04 '20 at 12:10
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See "casting out nines" https://en.wikipedia.org/wiki/Casting_out_nines Congratulations on finding this! Just because it has been done before, does not diminish your accomplishment. – GEdgar Feb 04 '20 at 12:13
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The question you're asking can be rephrased as "prove that the digital root of any (positive) multiple of $9$ is $9$". – Ben Grossmann Feb 04 '20 at 12:13
Much more concise and precise @Omnomnomnom. – owninggreendragsdude Feb 04 '20 at 12:21
@owninggreendragsdude perhaps, but the real goal there was to give you the appropriate term to google. – Ben Grossmann Feb 04 '20 at 12:23

Prove that the (recursive) sum of the digits of any number multiplied by 9 is 9

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