Let's analyze the problem step-by-step.
Problem:
The sum of a two-digit number and the number obtained by reversing its digits is what?
Step 1: Represent the two-digit number
Let the two-digit number be 10x+y10x+y10x+y, where:
- xxx is the tens digit (1 through 9),
- yyy is the units digit (0 through 9).
Step 2: Represent the reversed number
The number obtained by reversing the digits is 10y+x10y+x10y+x.
Step 3: Find the sum
Sum = (10x+y)+(10y+x)=10x+y+10y+x=11x+11y=11(x+y)(10x+y)+(10y+x)=10x+y+10y+x=11x+11y=11(x+y)(10x+y)+(10y+x)=10x+y+10y+x=11x+11y=11(x+y).
Conclusion:
The sum of a two-digit number and the number obtained by reversing its digits is always 11 times the sum of its digits.
Example:
If the number is 23:
- Sum of digits = 2 + 3 = 5
- Sum of the number and its reverse = 23 + 32 = 55
- 11×5=5511\times 5=5511×5=55, which matches perfectly.
If you want, I can help with more examples or related problems!
