Question 9 of 17intermediate🔍 AnalyzeLong Answer4 marks

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?

Correct Answer

Yes, the sum of any 2-digit number and its reverse is always divisible by 11.

Step 1: Let the 2-digit number be ab = 10a + b

Step 2: Its reverse is ba = 10b + a

Step 3: Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)

Step 4: Since the sum equals 11(a + b), it is always divisible by 11

∴ The sum is always divisible by 11.

The examples given (44, 110, 33) all confirm this pattern.

Exercise: Figure it Out (Section 3) | Q: 2 | (Page Page 11)
For More Understanding

Explanation

The chapter demonstrates algebraic justification for divisibility tricks. Using algebra, we can prove that the sum of a number and its reverse is always a multiple of 11, since the sum equals 11 times the sum of the digits.

Solution Steps

  1. Step 1: Let the 2-digit number be ab = 10a + b

  2. Step 2: Its reverse is ba = 10b + a

  3. Step 3: Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)

  4. Step 4: Since the sum equals 11(a + b), it is always divisible by 11

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