Asked by Laura
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?
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?
Answers
Answered by
Count Iblis
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.
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.
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