Numbers in the General Form

We know how to expand numbers by placing them in the place value table. This is done by multiplying each digit by its place value and then adding them together.

For examples:

• Expanded form of 46 = 4 × 10 + 6 × 1

i.e., 46 = 4 × 10 + 6

• Expanded form of 53 = 5 × 10 + 3 × 1

i.e., 53 = 5 × 10 + 3

Is there any pattern in these expanded forms of digits? We expanded the two-digit numbers ‘ab’ and ‘ba,’ i.e.,

This form of numbers is called as generalised form of number. Generalised form of numbers is obtained by multiplying each digit by its place value and adding them together.

To three-digit numbers ‘abc’:

Similarly,

• The generalised form of the number ‘bca’ is 100b + 10c + a
• The generalised form of the number ‘cab’ is 100c + 10a + b

Let us take the generalised form of number 39, i.e., 39 = 10 × 3 + 9

The usual form of 10 × 3 + 9 is 39.

Usual form is obtained by simplifying the generalised form.

Reverse the Digits of Two- and Three- Digit Numbers

Arun and Rohan are playing with numbers by reversing the digits of the number. Arun asked Rohan to take any two-digit number. Rohan took 32.

Arun predicted that there would be no remainder and Rohan got the answer 5 with no remainder. How did he predict that there will be no remainder?

What is the trick behind this game?

• We know that the general form of two-digit number ab is 10a + b.

On reversing the digits, we get 10b + a. Now, add them.

i.e., 10a + b + 10b + a = 11a + 11b = 11 (a + b)

• We can see here, the sum of a two-digit number and its reverse is always the multiple of 11.

Arun knows this trick so he predicted that there will be no remainder.

• Also, observe that if we divide the sum by 11, the quotient is (a + b), which is exactly the sum of digits chosen from ab.

Now, Arun and Rohan are playing another game. Arun asked Rohan to take any two-digit number. Rohan took 67.

Rohan got the answer as 1 and there was no remainder. This time, Rohan explained the trick of this game.

What is the trick behind the game?

• You know that the general form of two-digit number ab is 10a + b.

On reversing the digits, we get 10b + a. Now, find the difference between two numbers

i.e., 10a + b – (10b + a) = 9a - 9b = 9 (a - b)

• We can see here, the difference of a two-digit number and its reverse is the multiple of 9.

Rohan knows this trick so he predicted that there will be no remainder.

• Observe that if we divide the difference by 9, the quotient is (a - b), which is exactly the difference between digits chosen from ab.

Also, note that we need to subtract the smaller number from bigger number.

Now, Arun asked Rohan to take a three-digit number. Rohan took 379. Then, Rohan asked Arun if there was any pattern in reversing a three-digit number.

Arun predicted that there would be no remainder and Rohan got the quotient as 6. How did he predict that there would be no remainder?

What is the trick behind the game?

• You know that the general form of three-digit number abc is 100a + 10b + c.

On reversing the digits, we get cba.

i.e., 100c + 10b + a

Now, find the difference of two numbers.

• If a > c, then the difference between the numbers is

(100a + 10b + c) – (100c + 10b + a)

= 100a + 10b + c – 100c – 10b – a

= 99a – 99c

= 99(a – c)

• If c > a, then the difference between the numbers is

(100c + 10b + a) – (100a + 10b + c)

= 99c – 99a

= 99(c – a)

And, of course, if a = c, the difference is 0.

• We can see here, the difference of a three-digit number and its reverse is a multiple of 99 in each case.

Arun knows this trick so he predicted that there would be no remainder.

Observe that if we divide the difference by 99, the quotient is (a - c) or (c - a).

Note that we need to subtract the smaller number from bigger number.

Remember these properties:

Property 1: the sum of a two-digit number and its reverse is always a multiple of 11.

Property 2: the difference of a two-digit number and its reverse is always a multiple of 9.

Property 3: the difference of a three-digit number and its reverse is always a multiple of 99.

Consider the three-digit number 429. Now, let us form another two numbers using the digits 4, 2 and 9. Take the first number by shifting one’s digit to the left end.

i.e., 942

Take the second number by shifting hundreds digit to the right end.

i.e., 294

Sum of 429, 942, and 294 is 429 + 942 + 294 = 1665

Now, let us take the general form of three-digit number abc, i.e., 100a + 10b + c.

• We can take other two numbers as cab and bca.

i.e., 100c + 10a + b and 100b + 10c + a

• Now, let us add the three numbers.

abc + cab + bca

= 100a + 10b + c + 100c + 10a + b + 100b + 10c + a

= 111a + 111b + 111c

= 111 (a + b + c)

= 37 × 3 (a + b + c)

From this we can see that their sum is always divisible by 37.

Property: the sum of three-digit numbers, abc, bca, and cab is always divisible by 37. If we divide the sum by 37, then the quotient obtained is 3(a + b + c).

Tanu was solving a problem to find the value of letters. The problem is given below.

From this, we can see that the addition of 6 and P gives a number whose ones digit is 2, i.e., 6 + P = 2. Therefore, the possible value of P is 6. If P = 6 then 6 + 6 = 12, 1 will be carried forward in the next step.

Now, RP = Q + 8 + 1

R6 = Q + 9

i.e., Q + 9 is a number whose ones digit is 6.

Hence, Q = 7 and R = 1.

So, the values of P, Q, and R are 6, 7, and 1, respectively.

We know how to find the value of letters when the operation involved is addition. Now, Sanvi was solving problem to find the value of letters which involves the operation of multiplication. The problem is given below.

From this, we can see that the multiplication of B and 3 gives a number whose ones digit is B again. Hence, B must be 0 or 5.

Case 1: B = 5

First step of multiplication gives 3 × 5 = 15

1 will be carried forward in the next step.

In the next step, we get 3 × A + 1 = CA

This is not possible for any values of A.

Hence, B must be 0 only.

Case 2: B = 0

If B = 0 then 0 × 3 = 0 there will be no carrying for the next step.

In the next step, 3 × A = CA

i.e., one’s digit of 3 × A should be A. So, A should be 0 or 5.

We cannot take A as 0, since AB is a two-digit number.

Hence, A = 5.

So, we obtain the multiplication as given below.

Hence, A = 5, B = 0, and C = 1.

Divisibility Rules

A number is divisible by 10 if its ones digit is 0. In other words, if the ones digit of a number is ‘0’ then the number is multiple of 10 and if the ones digit of a number is not ‘0’ then the number is not multiple of 10.

Now, let us see the reason for the divisibility of numbers by 10 when numbers are written in general form. Suppose the number is cba.

We can write this number in general form as 100c + 10b + a

Clearly, ‘a’ is at ones place, ‘b’ is at tens place and ‘c’ is at hundreds place. We know that 10, 100 are divisible by 10. So 10b, 100c are also divisible by 10. To say the number is divisible by 10, ‘a’ should be divisible by 10. What is the possible value of ‘a’ so that it should be divisible by 10?

he only possible value of ‘a’ is 0.

This is the reason behind the divisibility test for 10.

A number is divisible by 5 if its ones digit is 0 or 5. In other words,

• If the ones digit of a number is ‘0’ or 5 then the number is a multiple of 5.
• If the ones digit of a number is other than ‘0’ or 5 then the number is not a multiple of 5.

Now, let us see the reason for the divisibility of numbers by 5 when numbers are written in general form. Suppose the number is cba. We can write this number in general form as 100c + 10b + a. Clearly, ‘a’ is at ones place, ‘b’ is at tens place and ‘c’ is at hundreds place.

We know that 10, 100 are divisible by 5 since they are multiple of 5. So 10b, 100c are also divisible by 5.

To say the number is divisible by 5, ‘a’ should be divisible by 5. What is the possible value of ‘a’ so that it should be divisible by 5?

The possible value of ‘a’ is either ‘0’ or 5.

This is the reason behind the divisibility test for 5.

A number is divisible by 2 if it has any of the digits 2, 4, 6, 8 or 0 in its ones place. In other words, if the ones digit of a number is 2, 4, 6, 8 or 0 then they are even numbers. All even numbers are multiples of 2.

Now, let us see the reason for the divisibility of numbers by 2 when numbers are written in general form. Suppose the number is cba. We can write this number in general form as 100c + 10b + a. Clearly, ‘a’ is at ones place, ‘b’ is at tens place and ‘c’ is at hundreds place.

We know that 10, 100 are divisible by 2 since they are multiples of 2. So 10b, 100c are also divisible by 2. To say the number is divisible by 2, the ones digit ‘a’ should be divisible by 2.

What is the possible value of ‘a’ so that it should be divisible by 2? The possible value of ‘a’ is 2, 4, 6, 8 or 0.

This is the reason behind the divisibility test for 2.

Divisibility by 3:

• A number is divisible by 3 if the sum of the digits is divisible by 3

Divisibility by 9:

• A number is divisible by 9 if the sum of the digits is divisible by 9.

We saw that the divisibility rules for 2, 5 and 10 by writing the numbers in general form. Now, let us see the reason for the divisibility of numbers by 3 and 9 when numbers are written in general form. Suppose the number is cba. We can write this number in general form as 100c + 10b + a

i.e., 100c + 10b + a

= 99c + c + 9b + b + a

= 9 (b + 11c) + (a + b + c)

To say the number is divisible by 9 (or 3), the sum of the digits should be divisible by 9 (or 3). Clearly, the first term 9(b + 11c) is divisible by 9. Also, 9(b + 11c) is divisible by 3. Then what will you say about the rule for divisibility of 3 and 9?

We can say the number 100c + 10b + a is divisible by 9 (or 3) if a + b + c is divisible by 9(or 3).

There are some questions in which we know that the given number is divisible by 3 or 9, but we need to find the unknown digit in the number. Consider the question.

If 1y16 is a multiple of 9, where y is a digit then what is the value of y?

Given that, 1y16 is a multiple of 9 and y is a digit. We know that if a number is a multiple of 9, then the sum of the digits will be divisible by 9.

Sum of digits of 1y16 = 1 + y + 1 + 6 = 8 + y

So, 8 + y should be multiple of 9.

This is possible when 8 + y is any of the numbers 0, 9, 18, 27 and so on.

i.e., 8 + y = 0, 9, 18,,,,,

• If 8 + y = 0, then y = (-8)
• If 8 + y = 9, then y = 1
• If 8 + y = 18 then y = 10

Since y is a single-digit number, 8 + y can only be 9.

Therefore, y should be 1 and the number is 1116.