How might you go about proving that N-M is divisible by 9 when N represents an integer (such as 6923) and M represents that integer in reverse order (such as 3296)?
so far, I have shown that 6923-3296=3627 is divisible by 9 by the sum of the digits, but how do you prove that for ALL N and M?
You use that when computing Modulo 9 you have:
10 = 1
So, 10^n = 1^n = 1
This means that Modulo 9 the numerical value of a digit is the same as the sum of its digits.
Then if you take a number N and subtract the number that has the digits in reverse order M, then Modulo 9 this is the same as adding up the digits in N and subtracting from that the sum of the digits in M, but M has the same digits as N, they are just in reverse order. So, Modulo 9, you will get zero.
To prove that N-M is divisible by 9 for any N and M, we can use a property of divisibility by 9 known as the "casting out nines" rule.
First, let's express N as a sum of its digits. For example, if N = 6923, we can write N = 6 + 9 + 2 + 3.
Next, let's express M (the integer in reverse order) as a sum of its digits. If M = 3296, we can write M = 6 + 9 + 2 + 3.
Now, let's consider N-M and express it as a difference of the sums of the digits.
N - M = (6 + 9 + 2 + 3) - (6 + 9 + 2 + 3)
= (6 - 6) + (9 - 9) + (2 - 2) + (3 - 3)
= 0 + 0 + 0 + 0
= 0
Since N - M equals 0, we can conclude that N - M is divisible by 9 because any number that equals 0 is divisible by any other number.
To summarize, to prove that N - M is divisible by 9 for all N and M, we expressed both N and M as sums of their digits, and then showed that the difference between these sums of digits resulted in 0, which is divisible by any other number.