The sum of the digits of N*9 is 9,Where 0<N<10

Example:

2*9 = 18, the sum of the digits is 9

3*9= 27, the sum of the digits is 9, etc.

Proof:

Define sumdig(I2) as the sum of the digits of a 2 or 1 digit integer number I2.

Note that a 2 digit number can be expressed as

I2= 10*N2 + 1*N1 (Definition 1).

where N1 and N2 are an integer 0<N1,N2<10.

Therefore:

sumdig(10*N2+1*N1) = N2+N1. (Equation 1).

We must show that: sumdig(N*9)= 9 where 0<N<10.

sumdig(N*9)= sumdig(N*(10-9))= sumdig(10N-N)= sumdig((10N-10) + (10 -9N))

= sumdig( 10*(N-1) + (10-N)) (Equation 2).

Since 0<N<10, then 0<= N-1<9, Equation 2 fits the form of Definition 1, where (N-1) is N2 and (10-N) is N1.

sumdig( 10*(N-1) + (10 -N)) = N-1+ 10-N = 9

QED.

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