Take a 2-digit number say, 27. Reverse its digits (72). Add them (99). Repeat for different 2-digit numbers.
Observation: When a 2-digit number is added to its reverse, the result is often a multiple of 11.
Examples: 27 + 72 = 99 36 + 63 = 99 45 + 54 = 99 12 + 21 = 33 89 + 98 = 187
Conclusion: The sum is always divisible by 11.
Proof: Let the 2-digit number be AB (where A is tens and B is ones). Number = 10A + B Reverse = 10B + A Sum: (10A + B) + (10B + A) = 11A + 11B = 11(A + B) ∴ The result is always a multiple of 11.
Explanation
The chapter introduces the 'Reverse and Add' activity. When you reverse a 2-digit number AB (where A is tens and B is ones), you get BA. Adding: (10A+B) + (10B+A) = 11A + 11B = 11(A+B). The result is always a multiple of 11.
Solution Steps
Let the 2-digit number be AB, where A is the tens digit and B is the ones digit.
Express the number as 10A + B.
Express the reversed number as 10B + A.
Add the two expressions: (10A + B) + (10B + A).
Simplify the sum to 11A + 11B.
Factor the expression to 11(A + B).
Conclude that the result is always a multiple of 11.