How would you prove this works algebraically?

I just got this question today from my maths class, and frankly, I'm stumped;





a) Write down any 2 digit number with different digits.


b) Reverse the digits and write the larger number of the two on the top.


c) Subtract the smaller number from the larger number.


d) Divide the result by the difference between the first 2 digits.


The answer will be 9, prove how this works algebraically.





Any ideas? It will be _______ = 9, not sure what though :/


How would you prove this works algebraically?
Hi,





Let T = 10 digits of the larger number and U = units digit of the larger number. Then 10T + U is the larger number and 10U + T is the smaller number with reversed digits.





(10T + U) - (10U + T) = 9T - 9U = 9(T - U)





This shows that 9 is always a factor.





I hope that helps!! :-) How would you prove this works algebraically?
Let a and b be digits such that a %26gt; b and a is not equal to b.





10a + b is one number and 10b + a would be the number with its digits reversed. 10a + b would have to be the larger number since a %26gt; b.





[(10a + b) - (10b + a)]/(a - b) (largest - smallest)/the difference between the two digits


= [10a + b - 10b - a]/(a - b) distribute the negative


= [9a - 9b]/(a - b) combine like terms


= 9(a - b)/(a - b) factor out the nine on top


= 9 cancel common factor (a - b)
Can you please explain that more clearly.


It's a bit confusing.





Like in part b. it says ';write the larger number of the two on top'; but what am i suppose to do with the larger number that i just wrote on top?





In part d, it says divide the rest (';the rest'; of what?) by the difference (the difference from part c?)
The digits are x and y. x %26gt; y


The number is 10x + y (larger number )


The reverse is 10y + x





10x + y - (10y + x)


= 10x + y - 10y - x


= 9x - 9y


= 9(x - y)





9(x - y) / (x - y) = 9





(The reason x and y have to be different is to make sure you don't wind up with zero in the denominator)
o is the ones digit


t is the tens digit





So, your number will be N = 10t + o





Your second number, after reversing the digit order, will be n = 10o + t





Their difference is | N - n | = | (10t + o) - (10o + t) |


| N - n | = | 9t - 9o |


or


| N - n | = 9 | t - o |





Divide by the difference between digits | t - o |





| N - n | = 9
say your number is {ab} = 10a +b where a is not equal to b


{ab} - {ba} = (10a + b) - (10b + a) = 9a - 9b = 9(a - b) ----%26gt; This is a multiple of 9, so the sum of the 2 digits is always 9.
let the number be 10x + y and x %26gt; y


the other number is 10 y + x





difference = 10 x + y - (10y +x) = 9(x-y)


devide by diffence of dgits that is x-y





to get 9(x-y)/(x-y) = 9
9 is the alpha and the omega of numbers. if your teacher questions you, tell her she is a blasphemer and strike her down.
Let


t = tens digit of number


u = units digit of number


10t + u = number


10u + t = number with digit reveresed.


(10t + u) - (10 u + t) = 9t - 9u


(9t - 9u) / (t - u) = 9