Question

The sum of digits of a two-digit number is 7. If the digits are reversed and the resulting number is decreased by 2, twice of the original number is obtained. Find the original number.

Answer 1

Hint:
we will first let the number as \[xy\] and write the digit in the form of the sum by using their ones and tens places and on the other hand mark the sum of the digits as 7 which gives us one equation. Now, form an equation using, the number obtained by reversing the digits is \[yx\] that is also written as \[10y + x\]. Use further information to form another equation in \[x\] and \[y\]. Simplify the equations and find the values to find the original number.

Complete step by step solution:
First consider the given information that is the sum of digits of a two-digit number is 7 and if the digits are reversed and the resulting number is decreased by 2, twice the number is obtained.
Now, let the original number be \[xy\] which can be written by using ones and tens place as \[10x + y\].
We know that the sum of digits is 7,
Thus, we have,
\[ \Rightarrow y + x = 7\;\;\;\;\;\;\;\;\;\;\;\; \to \left( 1 \right)\]
Now, the number obtained by reversing the digits is \[yx\] which can also be written as \[10y + x\].
Thus, the using the second condition which says that if the digits are reversed and the resulting number is decreased by 2, twice the number is obtained forms another equation,\[ \Rightarrow \left( {10y + x} \right) - 2 = 2\left( {10x + y} \right)\]
Now, we will solve this equation and convert it in form of \[x + y = ab\],
Thus, we have,
\[
   \Rightarrow \left( {10y + x} \right) - 2 = 2\left( {10x + y} \right) \\
   \Rightarrow \left( {10y + x} \right) - 2\left( {10x + y} \right) = 2 \\
   \Rightarrow 10y + x - 20x - 2y = 2 \\
   \Rightarrow 8y - 19x = 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \to \left( 2 \right) \\
 \]
Now, we have 2 equations in terms of \[x\]and \[y\],
That are,
\[
  y + x = 7\;\;\;\;\;\;\; \to \left( 1 \right) \\
  8y - 19x = 2\;\;\;\;\;\;\; \to \left( 2 \right) \\
 \]
Simplifying the equations will find give us the values of \[x\]and \[y\].
Thus, we will multiply the equation (1) by 19, we have,
\[
  19y + 19x = 133\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \to \left( 1 \right) \\
  8y - 19x = 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \to \left( 2 \right) \\
 \]
Now, we will add both the equations and solve for \[y\].
\[
   \Rightarrow 19y + 8y + 19x - 19x = 133 + 2 \\
   \Rightarrow 27y = 135 \\
   \Rightarrow y = \dfrac{{135}}{{27}} \\
   \Rightarrow y = 5 \\
 \]
Further, we will put the value of \[y = 5\] in equation (1) to get the value of \[x\].
\[
   \Rightarrow 5 + x = 7 \\
   \Rightarrow x = 7 - 5 \\
   \Rightarrow x = 2 \\
 \]
Now, as we have both digits and our original number is \[10x + y\], we will determine the original number by putting the values of both \[x\]and \[y\],
We get,
\[ \Rightarrow 10(2) + 5 = 25\]

Hence, the original number is 25.

Note:
Sum of two-digit number can be written as \[10x + y\] where \[x\] is the first digit and is on the tens place and \[y\] is the second digit and is on the ones place, so do not substitute the value of \[x\] and \[y\] in the reverse number, otherwise we can get the wrong answer. We can use the substitution method also to solve the equations and get the values for \[x\] and \[y\] respectively. Digits are reversed implies that the position of ones and tens place get reversed.