Question

[PrudensOptimus]

How do you find the sum of the digits of number N?


Say N = 10

[BigRedDot]

If memory serves me correctly:


\sum_{k=1}^N k= \frac{N(N+1)}2


a fact that you would prove by induction.

[HallsofIvy]

Was that what was meant? The "sum of the digits" of 10 is 1+ 0= 1, of course.

[himanshu121]

Originally posted by BigRedDot
If memory serves me correctly:


\sum_{k=1}^N k= \frac{N(N+1)}2


a fact that you would prove by induction.

It will be true only if the digits form a Arithmetic Projection
for eg N=2468 etc {2+4+6+8}
but not in general say for N=12245 {1+2+2+4+5}

[BigRedDot]

The "sum of the digits"
You're right, I answered a completely different question. Well, I am sure someone asked my question somewhere, sometime. :)

[PrudensOptimus]

Originally posted by HallsofIvy
Was that what was meant? The "sum of the digits" of 10 is 1+ 0= 1, of course.


Yea what if N = 1231247839783924723840723084732084738274... + ... N?

[himanshu121]

Originally posted by PrudensOptimus
Yea what if N = 1231247839783924723840723084732084738274... + ... N? [g)]

Ofcourse it would be
as N=1+0
here it will be N=1+2+3+1+2+4+7+8+3+9+7+8+3+9+2+4+7+2+3+8+4+0+7+2+ 3+0+8+4+7+3+2+0+8+4+7+3+8+2+7+4+...+x+y+z+a+b+c+d [;)] [;)] [;)] [6)]