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Hello guys/girls I was bored and I just played around with math. I am stuck and it's about raised numbers. (9, 6 and 3)

So this is how you calculate it. (same method for all numbers)

  1. Raise 3, 6 and 9 each from 1 to 10.
  2. If the product is more than one digit then add up the digits until there is only one digit.
  3. Look at the result...

Number 3.

3^1 = 3

3^2 = 9

3^3 = 27 = 2+7 = 9

3^4 = 81 = 8+1 = 9

3^5 = 243 = 2+4+3 = 9

3^6 = 729 = 7+2+9 = 18 = 1+8 = 9

3^7 = 2187 = 2+1+8+7 = 18 = 1+8 = 9

3^8 = 6561 = 6+5+6+1 = 18 = 1+8 = 9

3^9 = 19683 = 1+9+6+8+3 = 27 = 2+7 = 9

3^10 = 59049 = 5+9+0+4+9 = 27 = 2+7 = 9*

Number 6. I'm not gonna write plus signs now, because I think you've got the Idea

6^1 = 6

6^2 = 36 = 9

6^3 = 216 = 9

6^4 = 1296 = 18 = 9

6^5 = 7776 = 27 = 9

6^6 = 46656 = 27 = 9

6^7 = 279936 = 36 = 9

6^8 = 1679616 = 36 = 9

6^9 = 10077696 = 36 = 9

6^10 = 60466176 = 36 = 9

Number 9.

9^1 = 9

9^2 = 81 = 9

9^3 = 729 = 18 = 9

9^4 = 6561 = 18 = 9

9^5 = 59049 = 27 = 9

9^6 = 531441 = 18 = 9

9^7 = 4782969 = 45 = 9

9^8 = 43046721 = 27 = 9

9^9 = 387420489 = 45 = 9

9^10 = 3486784401 = 45 = 9

So to my question, why is the sum 9 for number 3 and 6 and not only for the 9 itself? I know that 3, 6 and 9 is relative to each other but i'm still confused. (3+6+9 = 18 = 1+8 = 9)

And is there an equation for this? Thanks for reading :)

Luce newlen
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2 Answers2

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any number of the form :

$3^n$ for $n \ge 2$,

$6^n$ for $n \ge 2$,

$9^n$ for $n \ge 1$ is divisible by 9.

If a number is a multiple of 9, then its digit sum is 9. It is a easy result which you can prove.

GA316
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Note that it is not $9$ for $3^1$ and $6^1$ Any multiple of $9$ will have the sum of digits a multiple of $9$, so when you get to one digit it will be $9$. Then $3^2$ and $6^2$ (and all higher powers) have a factor of $9$, so will reduce to $9$. You can see Wikipedia on divisibility rules

Ross Millikan
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  • Of course most of us know the solution to Luce's observation. But aren't we sort of destroying Luce's fun to play with math when we provide Luce with a complete solution? Let Luce discover the insights by his/her own by only giving vague hints. Otherwise we would spoil the game. – Michael Hoppe Nov 08 '13 at 17:20
  • @MichaelHoppe: But Luce did ask the question specifically. – Ross Millikan Nov 08 '13 at 17:47
  • @MichaelHoppe I like your thinking, didn't even think about that :) But still, RossMillikan Thank you very much for the answer, really helped :) – Luce newlen Nov 08 '13 at 17:55