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How do you prove algebraically that if you write down a two-digit number then reverse the order of the digits and take it away from the first number that you end up with a multiple of 9?
Wiki User
∙ 14y ago
Suppose the two digits of the first number are x and y where 0<=x,y<10.
[Strictly speaking, if it is a two digit number then x>0, but the proof still works.]
Then the first number is 10x + y
and the second number, therefore, is 10y + x
Taking the second number away from the first gives: (10x + y) - (10y + x)
= 10x + y - 10y - x = 9x - 9y = 9(x-y)
Now, since x and y are integers, x-y must be an integer.
That is, the result of the subtraction is 9*an integer ie it is a multiple of 9.
Wiki User
∙ 14y ago
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The answer is below:
