prove that the substraction of the two digit no and the no inter changing their's place is dividable by 9?

Answer 1

Let the first number be 10a + b
Thus the second number is 10b + a

Now (10a + b) - (10b + a) = 9a - 9b = 9(a - b) which is divisible by 9

Answer 2

2 digit number = 10x + y where x and y are integers
inversion of the digits = 10y + x

subtract the 2 = 10x+y - (10y +x) = 10x + y - 10y - x
= 9x-9y = 9(x-y)

therefore must be divisible by 9.

Answer 3

Let the digits be x and y

the number 10x+y
interchange the digits it becomes 10y+x
difference = (10x+y)-(10y+x)

= 9x - 9y = 9(x-y)
this is divisible by 9

Answer 4

All the listed proofs are good and true. Now, the basis of the question regarding 2 digit numbers is also applicable to ANY number that has more than 1 digit.

Answer 5

Let the no. be xy. The no. got by interchnaging the digits is yx.

Original value of no.= 10x+y
Second number = 10y+x

The difference between them= (10x+y)- (10y+x)= 9x-9y=9(x-y)
Since the differnce is a multiple of 9, it is divisible by 9

Answer 6

Let the first number be 10a + b where 1<=b<=a<=9
Thus the second number is 10b + a

Now (10a + b) - (10b + a) = 9a - 9b = 9(a - b) which is divisible by 9

Answer 7

Let the digits be x and y

the number 10x+y
interchange the digits it becomes 10y+x
difference = (10x+y)-(10y+x)

= 9x - 9y = 9(x-y)
this is divisible by 9
QED

Answer 8

so you let "a" be in the tens place and "b" be in the ones place for the first number.

so let's say if a>b
so (A*10)+B is the first integer
(B*10)+ A is the second integer.

[(A+10)+B] - [(B*10)+A] = 9A - 9B = 9(A-B) so it's divisible by 9. ******!

Answer 9

10a + b - 10b - a = 9a - 9b = 9(a-b)
9(a-b)/9 = a-b

Q.E.D

Answer 10

THERE ARE 98 LIVE EXAMPLES OF IT, WHAT ELSE TO PROVE? TAKE MY WORD & GO AHEAD