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

We can use algebra to represent any 2-digit number and its reversed form.

Step 1 — Setting up the numbers

Let us choose a 2-digit number. Let its tens digit be $a$ . Let its units digit be $b$ . The value of this number is $10a + b$ . For example, if the number is 72, then $a=7$ and $b=2$ . The problem states that the digits must be different, so $a \neq b$ . The tens digit $a$ cannot be zero. So, $a$ can be any digit from 1 to 9. The units digit $b$ can be any digit from 0 to 9. When we reverse the digits, the new number becomes $10b + a$ . For example, if the original number is 72, the reversed number is 27.

Step 2 — Finding the difference

Now, let us find the difference between these two numbers. We subtract the smaller number from the larger number. Let us assume $10a + b$ is larger than $10b + a$ . This means that $a$ is greater than $b$ .

$\text{Difference} = (10a + b) - (10b + a)$

$= 10a + b - 10b - a$

$= (10a - a) + (b - 10b)$

$= 9a - 9b$

$= 9(a - b)$

If $10b + a$ is larger than $10a + b$ , then $b$ is greater than $a$ .

$\text{Difference} = (10b + a) - (10a + b)$

$= 10b + a - 10a - b$

$= (10b - b) + (a - 10a)$

$= 9b - 9a$

$= 9(b - a)$

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

Step 3 — Dividing by 9

The last step of Mukta's trick is to divide this difference by 9. Let us divide the absolute difference by 9.

$\frac{9 \times |a - b|}{9}$

$\boxed{|a - b|}$

The result is $|a - b|$ . Since $a$ and $b$ are digits, they are whole numbers. The difference between two whole numbers is always a whole number. For example, if $a=7$ and $b=2$ , then $|a-b| = |7-2| = 5$ . If $a=2$ and $b=7$ , then $|a-b| = |2-7| = |-5| = 5$ . Since the result of the division is always a whole number, it means there is no remainder.

Answer

Yes, there will always be no remainder.

More questions in IT

Context: Think of a number trick:

Think of a number.
Double it.
Add four.
Divide by two.
Subtract the original number you thought of.

Q. I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why?

Context: Consider the following number trick:

Think of a number.
Double it.
Add four.
Divide by two.
Subtract the original number you thought of. (This trick always results in 2).

Q. How would you change this game to make the final answer 3? What about 5?

Context: Consider the following number trick:

Think of a number.
Double it.
Add four.
Divide by two.
Subtract the original number you thought of. (This trick always results in 2).

Q. Can you come up with more complicated steps that always lead to the same final value?

Context: In the date trick, the final answer is given by $100M + 165 + D$ , where $M$ is the month and $D$ is the day.

Q. Find the dates if the final answers are the following:

(i) 1269 (ii) 394 (iii) 296

Context: In the date trick, the final answer is given by $100M + 165 + D$ , where $M$ is the month and $D$ is the day.

Q. Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number.

Try to devise your own 'Think of a Number' trick.

Use the same rule to fill these pyramids:

Fill the following pyramids:

What is the relationship between the numbers in the bottom row and the number at the top?

Let us start with the simplest pyramid.

Q10

What about a pyramid with three rows?

Using letter numbers for the bottom row, we can write an expression for the top row.

Q11

In the following grids, find the values of the shapes and fill in the empty squares:

Q12

Context: 6.5 The Largest Product

Q. Fill the digits 2, 3, and 5 in $\square\square \times \square$ , using each digit once. What is the largest product possible?

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 $a > b$ ?