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Take a number say 987654. Sum it's digits

9 + 8 + 7 + 6 + 5 + 4 = 39 3 + 9 = 12 1 + 2 = 3 

i.e. keep doing this till you get a single digit answer.

Now I take the same number & do it in other different ways, I still end up with the same answer.

987 + 654 = 1641 16 + 4 + 1 = 21 2 + 1 = 3 

Or

98765 + 4 = 98769 9876 + 9 = 9885 988 + 5 = 993 99 + 3 = 102 1 + 0 + 2 = 3 

How come I always get the same answer (3 in this case). This is not special for 987654. It's for any number you take.

What's the reason or theory behind this?

(PS - I am not sure what's the right tag for this question. Please correct if necessary).

  • 0
    OK, I see your point.2012-10-24

2 Answers 2

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What you have stumbled upon is known as taking the digital root, which is also sometimes referred to as a part of "Vedic mathematics." You might also be interested in reading about "Casting out nines". These sources should help point you in the right direction.

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In short: in the end of this 'sum of digits' procedure, what is left coincides with the remainder modulo $9$. (More precisely, if $9|n$ then $9$ will be the result, unless $n=0$.)

The main point is $10\equiv 1 \pmod{9}$, hence $10^k\equiv 1 \pmod{9}$ for all $k\in\Bbb N$. So for any number with decimal digits $N=\overline{a_1a_2...a_n}$ $N=\sum_{k=1}^n 10^{n-k}\cdot a_k \equiv \sum_k a_k \pmod{9} $ So, meanwhile the process, the remainder mod $9$ always stays the same.

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    Technically we can say that reducing modulo 9 is a ring homomorphism - it respects the basic operations of multiplication and addition. It doesn't matter whether you do your arithmetic before you do the reduction, or afterwards, or part before and part after, you get the same eventual result (mod 9). This works so well for decimal numbers mod 9, because $10 \equiv 1 \mod 9$, so that the integer 10 acts like 1 when reduced mod 9, ie as the identity element for multiplication. (if our number base were 12 we'd be working mod 11 ...)2012-10-24