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The formula to find the digital root/ sum is:

digital root of n = 1 + ( (n - 1) % 9 )

Can someone explain me the intuition behind this formula? Why does this result give the sum of digits?

user3426358
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2 Answers2

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Numbers in base $10$ are "arranged" into boxes of $10$ entries. For example, $21$ looks like $$ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \cdot\\ \blacksquare \square \square \square \square \square \square \square \square \square $$

and $123$ looks like:

$$ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \cdot\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\ \cdot\\ \blacksquare \blacksquare \blacksquare \square \square \square \square \square \square \square $$

When you take the modulo $9$, you remove all the "full" packages such as the $100$ and only $1$ remains. (Remove the first $9$ columns, and $10$ boxes remain. $10 \equiv 1\mod 9$). This builds up, so $200$ loses for each $100$ box $99$ boxes, and $2$ boxes remain.

Hope this helps to visualize the operation.

OFRBG
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  • When you take the modulo 9, you remove all the "full" packages such as the 100 I can not see a full 100 package from your visualization. According to your visualization a full package contains 10 elements, not 100. – some1 here Apr 14 '21 at 16:44
  • @some1here modulo operations are associative e.g. (20 mod 9) = (10 mod 9) + (10 mod 9). In the case of 100, we split it into (10 mod 9) ten times, which gives us 10 times 1. That remaining 10 boxes get modded again, and we get the final 1. – OFRBG Apr 28 '21 at 15:16
  • @some1here More formally, $(x \times 10 \mod 9 ≡ x)$, which you can expand for any times you can factor out 10 out of $x$. – OFRBG Apr 28 '21 at 15:26
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From here:

The digital root is the value modulo 9 because $10\equiv 1{\pmod {9}}$, and thus $10^{k}\equiv 1^{k}\equiv 1{\pmod {9}}$, so regardless of position, the value mod $9$ is the same – $ a\cdot 100\equiv a\cdot 10\equiv a{\pmod {9}}$ – which is why digits can be meaningfully added. Hope it helps.