# If a two digit number increases by 54 when the digits are reversed, what is the number?

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Let the digits of the number be x and y.

So we have

10*y + x - (10*x + y) = 54

=> 10y + x – 10x – y = 54

=> 9y – 9x = 54

=> y – x = 54 / 9 = 6

=> y = 6 + x

Now, x can be 1, 2 or 3, it cannot take on higher values as that would make y a 2-digit number. The corresponding values of y for x equal to 1, 2 and 3 are 7, 8 and 9.

**So the required number can be 17 , 28 and 39.**

If a two digit number increases by 54 when the digits are reversed, what is the number.

Let the 2 digits in order be x and y.

Then the value of the number = 10x+y.

When the digits are reversed , the digits in order are y and x.

So the valuse of the number is 10y+x.

Since 10y+x -(10x+y) = 54,

9y -9x = 54.

Or y-x= 6.

Therefore y = x+6.

So the digits in the solution are when x= 1, y = 1+6 = 7.

When x=2, y = 2+6 = 8.

When x= 3, y = 3+6 = 6 = 9.

Therefore there are 3 solutions : 17 ; 28 and 39 which, when reversed, become 71 ; 82 and and 93 each increase by 54 after reversing the digits.