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Kingzoolerius666

~~Time Number Theory~

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11:57PM EST 1 + 1 + 5 + 7= 14 and 1 + 4= 5 11 + 57= 68 and 6 + 8= 14 and 1 + 4= 5 ; This is true to every minute of every day. If X is a number of length n in base B, X can be written as X = B^0 * x0 + B^1 * x1 + ... + B^n * xn then S(X) is calculated as follows S(X) = S(x0+x1+...+xn) if X >= B (*) S(X) = X if X

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It''s a trivial result that for any base B the sum of the digits is congruent to the number modulus (B -1), so recursively summing digits until you can go no further, which you''ve called S(x), just gives you the number n, 0 < n < B which is congruent to your orginal number. Your theory is then just modulo arithmetic, in particlar the fact that.

A = B (mod n) and C = D (mod n) implies A + C = B + D (mod n)

Where ''='' represents congruence (the proper symbol is an equals sign but with three horizontals).

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quote:
Original post by johnb
It''s a trivial result that for any base B the sum of the digits is congruent to the number modulus (B -1), so recursively summing digits until you can go no further, which you''ve called S(x), just gives you the number n, 0 < n < B which is congruent to your orginal number. Your theory is then just modulo arithmetic, in particlar the fact that.

A = B (mod n) and C = D (mod n) implies A + C = B + D (mod n)

Where ''='' represents congruence (the proper symbol is an equals sign but with three horizontals).
Cool. I''m sure that King''s proof doesn''t actually show this though.

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quote:
Original post by NotAnAnonymousPoster
Cool. I''m sure that King''s proof doesn''t actually show this though.




King''s proof appears to overlook the case where addition creates a carry digit. For example, in base 10 (using | to separate digits)

7|5 + 5|7 = 1|3|2 not 0|12|12

It doesn''t take much work to show the modular result, but King''s proof doesn''t quite manage it.

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quote:
Original post by johnb
It''s a trivial result that for any base B the sum of the digits is congruent to the number modulus (B -1), so recursively summing digits until you can go no further, which you''ve called S(x), just gives you the number n, 0 < n < B which is congruent to your orginal number. Your theory is then just modulo arithmetic, in particlar the fact that.

A = B (mod n) and C = D (mod n) implies A + C = B + D (mod n)

Where ''='' represents congruence (the proper symbol is an equals sign but with three horizontals).

I believe it is pretty trivial, but it is still fun to show it:

Let x := x0 + x1b + ... + xnbn,
each xi and b are integers.

b = 1 (mod b - 1) implies
xibi = xi (mod b - 1) implies (summing over all i)
x = x0 + x1 + ... + xn (mod b - 1).

Note, the xi don''t even have to be less than b.

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