The Impossible Sum(Fixed)

Number Theory Level pending

I f If K n K_n (X)=mean how much can X be divided by n

For example

  • K 2 K_2 (8)=3

  • K 2 K_2 (5)=0

  • K 3 K_3 (9)=2

The Question is

  • Can ∑ i = 1 n m × X ( K n ( i ) ) − ∑ i = 1 X ( K n ( i ) ) X \dfrac{\displaystyle\sum_{i=1}^{n^{m}×X}(K_n(i))-\displaystyle\sum_{i=1}^{X}(K_n(i))} {X} = a positive integer?

m ≠ \neq {1,2}, n ≠ \neq {1}, X ≠ \neq {1}

All n, m, X are positive integers

some number is right no yes

Ahmed Pro by Ahmed Pro , 18, Egypt

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1 solution

Ahmed Pro
Aug 2, 2020
  • ∑ i = 1 n m × X ( K n ( i ) ) \displaystyle\sum_{i=1}^{n^{m}×X}(K_n(i)) = ∑ i = 1 X ( K n ( i ) ) \displaystyle\sum_{i=1}^{X}(K_n(i)) + ( ( n m − 1 ) × X n − 1 \frac{(n^m-1)×X}{n-1} ) My Equation and i proved it \fbox{My Equation and i proved it}

So ∑ i = 1 n m × X ( K n ( i ) ) − ∑ i = 1 X ( K n ( i ) ) \displaystyle\sum_{i=1}^{n^{m}×X}(K_n(i))-\displaystyle\sum_{i=1}^{X}(K_n(i)) = ( ( n m − 1 ) × X n − 1 \frac{(n^m-1)×X}{n-1} ) = A A

That means A ( n m − 1 ) × X n − 1 \dfrac{A}{\frac{(n^m-1)×X}{n-1}} = 1

And A X \frac{A} {X} = ( n m − 1 ) × X n − 1 X \frac{\frac{(n^m-1)×X}{n-1}}{X} = n m − 1 n − 1 \frac{n^m-1}{n-1}

And n m − 1 n − 1 \frac{n^m-1}{n-1} Always an integer, so A X \frac{A} {X} also an integer

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