Question

The sum of the digits of a two-digit number is 9. 9 times of this number is equal to 2 times the number formed by reversing the digits. Find the number.

Hint:

For two-digit number problems, always assume digits as \(x\) and \(y\) and form equations using their sum and reversal relations.

Answer 1

Step 1: Let the digits of the number be. 
Let the tens digit be \( x \) and the units digit be \( y \). Then, the number is \( 10x + y \). |
Step 2: Write the given conditions. 
The sum of the digits is 9: \[ x + y = 9 \quad \text{(i)} \] 9 times the number is equal to twice the number formed by reversing the digits: \[ 9(10x + y) = 2(10y + x) \] Step 3: Simplify the equation. 
\[ 90x + 9y = 20y + 2x \] \[ 88x = 11y \Rightarrow 8x = y \quad \text{(ii)} \] Step 4: Substitute (ii) into (i). 
\[ x + 8x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1 \] \[ y = 8x = 8 \] Step 5: Find the number. 
\[ \text{Number} = 10x + y = 10(1) + 8 = 18 \] Step 6: Check the condition. 
\[ 9 \times 18 = 162, \quad 2 \times 81 = 162 \] Hence, the condition is satisfied. 
Step 7: Conclusion. 
The required number is 18.