Introduction

Let's explore a fascinating problem involving two-digit numbers where, when added to 27, the result is the reversed form of the original number. This problem involves solving equations and analyzing the properties of digits, revealing the elegance of number theory.

Problem Statement

Consider a two-digit number ( xy ), where ( x ) is the tens digit and ( y ) is the units digit. The number can be expressed as ( 1 y ). If we add 27 to this number, we get the reversed form of the original number, which is ( 10y x ). Thus, the equation we need to solve is:

( 1 y 27 10y x )

Solution Method

Starting with the equation obtained from the problem statement:

( 1 y 27 10y x )

We can rearrange and simplify the equation:

( 1 y 27 - x - 10y 0 )

( 9x - 9y 27 0 )

( x - y 3 0 )

( x - y -3 )

Constraints on Digits

Since ( x ) and ( y ) are digits, their values are restricted as follows:

( x ) ranges from 1 to 9 (since ( x ) is the tens digit of a two-digit number). ( y ) ranges from 0 to 9.

Expressing ( x )

From the equation ( x - y -3 ), we can express ( x ) in terms of ( y ):

( x y - 3 )

Now, we need to find all pairs ((x, y)) that satisfy both the equation and the digit constraints:

If ( y 3 ), then ( x 0 ) (not valid). If ( y 4 ), then ( x 1 ) (valid: number is 14). If ( y 5 ), then ( x 2 ) (valid: number is 25). If ( y 6 ), then ( x 3 ) (valid: number is 36). If ( y 7 ), then ( x 4 ) (valid: number is 47). If ( y 8 ), then ( x 5 ) (valid: number is 58). If ( y 9 ), then ( x 6 ) (valid: number is 69).

Verification of Solutions

Let's verify each valid number to ensure the property holds:

( 14 27 41 ) ( 25 27 52 ) ( 36 27 63 ) ( 47 27 74 ) ( 58 27 85 ) ( 69 27 96 )

All these numbers, when added to 27, result in their reversed forms.

The valid two-digit numbers that satisfy the given condition are 14, 25, 36, 47, 58, and 69.

Alternate Method

We can also solve this problem using the ratio of the digits. Given that ( x : y 2 : 3 ), we can express ( x ) and ( y ) as:

( x 2k ) and ( y 3k ) for some constant ( k ).

Substituting these into the number form, we get:

( xy 10(2k) 3k 20k 3k 23k )

Adding 27 to this number:

( 23k 27 10(3k) 2k 30k 2k 32k )

Thus, we have:

( 23k 27 32k )

( 27 9k )

( k 3 )

Therefore, ( x 6 ) and ( y 9 ), confirming the number is 69.