Algebra Play | IT

Question 13

Context: Mukta's trick: Choose a 2-digit number of different digits, reverse the digits to get another number, find their difference, and divide the result by 9.

Q. If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

Solution

Understand the Question

Any 2-digit number with tens digit $a$ and units digit $b$ can be written in generalized form as $10a + b$ .
Reversing its digits gives $10b + a$ .
The difference between the two numbers is always a multiple of $9$ , meaning dividing by $9$ will always give an integer quotient with no remainder.

Step 1 · Express the Numbers in General Form

Let the tens digit of the two-digit number be $a$ and the units digit be $b$ , where $a \neq b$ , $a \in \{1, 2, \dots, 9\}$ , and $b \in \{0, 1, \dots, 9\}$ .

Original number: $10a + b$

Reversed number: $10b + a$

Step 2 · Calculate the Difference

If $a > b$ :

\begin{aligned} \text{Difference} &= (10a + b) - (10b + a) \\ &= 10a + b - 10b - a \\ &= (10a - a) + (b - 10b) \\ &= 9a - 9b \\ &= 9(a - b) \end{aligned}

If $b > a$ :

\begin{aligned} \text{Difference} &= (10b + a) - (10a + b) \\ &= 10b + a - 10a - b \\ &= (10b - b) + (a - 10a) \\ &= 9b - 9a \\ &= 9(b - a) \end{aligned}

In both cases, the absolute difference is $9 \times |a - b|$ , which is always a multiple of $9$ .

Step 3 · Divide by 9

Dividing the difference by $9$ : $\dfrac{9 \times |a - b|}{9} = |a - b|$

Since $a$ and $b$ are whole-number digits, $|a - b|$ is always a non-negative integer. Hence, there is no remainder.

Answer

Yes, there will always be no remainder.

Common Mistakes

Relying Only on Examples: Testing specific cases (e.g., $72 - 27 = 45$ ) confirms the pattern, but an algebraic proof using $10a + b$ is required to show it holds for all 2-digit numbers.
Incorrect Place Value Form: Writing a 2-digit number as $ab$ (which denotes multiplication $a \times b$ ) instead of its expanded form $10a + b$ .

More questions in IT

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Q12

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Q13

Context: Mukta's trick: Choose a 2-digit number of different digits, reverse the digits to get another number, find their difference, and divide the result by 9.

Q. If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

Q14

Context: Suppose a two-digit number is $ab$ . When it is reversed, the new number is $ba$ . If $b > a$ , the difference is $ba - ab = 9(b - a)$ , which is divisible by 9.

Q. Can you work out what happens if $a > b$ ?