Question

The difference of any 51 digit number and its reverse is always divisible by?

Answer 1

ANSWER

9

From a mathematical standpoint: For any N greater than 10 and less than 100, we can write it thusly:

N = 10x+y

...and its inverse, ~N as:

~N = 10y+x

You're asking for the calculation (N -~N). That can be simplified as below:

N -~N = (10x+y) - (10y+x)

N -~N = 10x + y - 10y - x

N -~N = 9x-9y

N -~N = 9(x-y)

So for any 10 < = N < 100 , N -~N will always be divisible by 9

If we use 41 * 14, we'd have x=4 and y=1:

N -~N = (10(4)+1(1)) - (10(1)+1(4))

N -~N = 10(4) + 1(1) - 10(1) - 1(4)

N -~N = 9(4)-9(1)

N-~N = 9(4-1) = 9(3) = 27

So for any 10<=N<=100, N-~N will always be divisible by 9

PLEASE MARK ME AS THE BRAINLIEST