# Playing with Numbers - Class 8 : Notes

**Types of Numbers:
**

**1. Natural numbers**Â : The group of the positive numbers which are countable are known as natural numbers. The examples of natural numbers will be 1, 2, 3, 4, 5,....... etc.

**2. Whole numbers**Â : The group of natural numbers with inclusion of zero in it, are known as whole numbers. The examples of whole numbers will be 0, 1, 2, 3, 4, 5,........... etc.

**3. Integer numbers**Â : The group of positive and negative numbers along with zero are known as integer numbers. The examples of integer numbers will be ........-3, -2, -1, 0, 1, 2, 3, 4, 5,...... etc.

**4. Rational numbers**Â : The numbers which can be expressed as ratio of integers are known as rational numbers. The examples of rational numbers will be 1/4, 2/7, - 3/10, 34/7, etc.

**General Form of Numbers:
**

**1. For two digit number:**

Any two digit number ab can be written as ab = 10 x a + b = 10a + b

**91 can be written as 10 x 9 + 1 = 90 + 1**

*Example 1*:**34 can be written as 10 x 3 + 4 = 30 + 4**

*Example 2*:**2. For three digit number
**Any three digit number abc can be written as abc = 100 x a +10 x b + c = 100a +10b + c

**123 can be written as 100 x 1 + 10 x 2 + 3 = 100 + 20 + 3**

*Example 1*:**277 can be written as 100 x 2 + 10 x 7 + 7 = 200 + 70 + 7**

*Example 2*:

Number Games:

**1. Reversing the digits â€“ two digit number:
**

**(a) Trick for dividing by 11:**

For any given two digit number, reverse the number and add it to the original number. Then on dividing it by 11 will always result in a zero remainder.

Look at steps below to understand the trick

Step 1: Select any two digit number ab, which can be expressed as (10a + b).

Step 2: Reverse its digits i.e. number ab will become ba, which can be expressed as (10b + a).

Step 3: Add both the numbers i.e. ab + ba, so we get (10a + b) + (10b + a) = 11a+ 11b = 11 (a + b)

Step 4: Divide the number by 11 i.e. 11 (a + b)/11 = (a + b). Hence, remainder will be zero under any case.Â Â The quotient is a + b, which is exactly the sum of the digits of chosen number ab.

**Example****:** Try the reversing two digit trick and dividing by 11 for 34.

*Solution*:The given number is 34. On reversing it we have 43.

Now, let us add both the numbers, so we have 34 + 43 = 77.

Dividing 77 by 11 will result into a zero remainder.

**
(b) Trick for dividing by 9:
**For a given two digit number, reverse the number and subtract it from original number. Then on dividing it by 9, will always result in a zero remainder.

Look at steps below to understand the trick:

Step 1: Select any two digit number ab, which can be expressed as (10a + b).

Step 2: Reverse its digits i.e. number ab will become ba, which can be expressed as (10b + a).

Step 3: Subtract both the numbers i.e. ab - ba, so we get (10a + b) - (10b + a) = 9a- 9b = 9 (a - b)

Step 4: Divide the number by 9 i.e. 9 (a - b)/9 = (a - b). Hence, remainder will be zero under any case.Â

** Example:** Try the reversing two digit trick and dividing by 9 for 43.

*Solution*:The given number is 43. On reversing it we have 34.

Now, let us subtract both the numbers, so we have 43 - 34 = 9.

Dividing 9 by 9 will result into a zero remainder.

2. Reversing the digits â€“ three digit number

**(a) Trick for dividing by 99
**For a given three digit number, reverse the number and subtract it from original number. Then on dividing it by 99 will always result in a zero remainder.

Look at steps below to understand the trick

Step 1: Select any three digit number abc, which can be expressed as (100a +10b + c).

Step 2: Reverse its digits i.e. number abc will become cba, which can be expressed as (100c + 10b + a).

Step 3: Subtract both the numbers i.e. abc - cba, so we get (100a + 10b + c) - (100c + 10b + a) = 99a- 99c = 99 (a - c).

Step 4: Divide the number by 99 i.e. 99 (a - c)/9 = (a - c). Hence, remainder will be zero under any case andÂ quotient is a â€“ c or c â€“ a.

** Example:** Try the reversing three digit trick and dividing by 99 for 987.

*Solution*:The given number is 987. On reversing it we have 789.

Now, let us subtract both the numbers, so we have 987 - 789 = 198.

Dividing 198 by 99 will result into a zero remainder.

**
3. Forming three-digit numbers with given three-digits:
**For a given three digit number, rearrange the number such that all the three numbers are distinct and if we add them all. Then dividing it by 37 will always result in a zero remainder.

Look at steps below to understand the trick

Step 1: Select any three digit number abc, which can be expressed as (100a +10b + c).

Step 2: Rearrange the same number to form two other distinct numbers i.e. numbers can be bca which can be expressed as (100b + 10c + a) and cab which can be expressed as (100c + 10a + b).

Step 3: Add all the three numbers i.e. abc +bca + cab, so we get (100a + 10b + c) + (100b+ 10c + a) + (100c + 10a + b) = 111 (a + b + c)

Step 4: Divide the number by 37. The resultant remainder will be zero under any case.Â

** Example:** Try the reversing three digit trick and dividing by 37 for 123.

*Solution*:Â The given number is 123. On rearranging it we will have two numbers 231 and 312.

Now, let us add all the numbers, so we have 123 + 231 + 312 = 666.

Dividing 666 by 37 will result into a zero remainder.

**
Letters for Digits:
**In this trick for given addition or multiplication problem, we need to find the equivalent digits for the letters.

Rules for letter representation

(i) Each digit must stand for just one letter i.e. a single letter represents a single digit not multiple digits.

(ii) The digit represented by a letter cannot have its first digit as zero i.e. the first digit must be a non-zero number.

Follow simple arithmetic and multiplication rules to find the answers.

**Find the value of A for the following addition.**

*Example 1*:Â Â Â Â Â Â Â AÂ Â Â Â 2

Â Â Â Â Â

__+ AÂ Â Â Â 9Â Â Â__

Â Â Â Â Â Â Â Â Â Â Â Â 3Â Â Â Â A

*Solution*:Â We know that, here there is only one possible value of A.

Hence, for the given addition to be true, A = 1 is the only possible value.

Â Â Â Â Â Â Â Â Â Â Â Â 1Â Â Â Â 2

Â Â Â Â Â Â

__+1Â Â Â Â 9Â Â Â__

Â Â Â Â Â Â Â Â Â Â Â Â 3Â Â Â Â 1

Thus, A = 1.

**Example 2:** Find the value of A & B.

Â Â Â Â Â Â Â Â Â Â Â 1 2 A

__+Â Â Â Â Â Â Â Â Â 6 A B
__Â Â Â Â Â Â Â Â Â Â Â A 0 9

Solution:Â Here, addition of A & B is giving 9. The sum can be 9 only as the sum of two single digit numbers cannot be 19. Therefore, there will not be any carry in this step.

For next step, 2 + A = 0 and this possible only if A = 8.

Now, 2 + 8 = 10 and 1 will be carry for the next step.

So, 1 + 1 + 6 = A.

Hence, clearly A is 8 and B can only be 1 to satisfy the given addition.

Â Â Â Â Â Â Â Â Â Â Â 1 2 8

__+Â Â Â Â Â Â Â Â Â 6 8 1__

Â Â Â Â Â Â Â Â Â Â Â 8 0 9

Hence, value of A is 8 and B is 1.

Divisibility Tests:

**1. Divisibility by 10:
**Any number whose unitâ€™s digit is â€˜0â€™ will always be divisible by 10.

**Check divisibility by 10 test for (i) 340 Â Â Â (ii) 231 Â Â Â (iii) 12**

*Example*:*Solution*:Â On checking the unitâ€™s digit for each given number, we can see that

(i) 340 is divisible by 10 (ii) 231 is not divisible by 10 (iii) 12 is not divisible by 10

**2. Divisibility by 5:
**Any number whose unitâ€™s digit is either â€˜0â€™ or â€˜5â€™ will always be divisible by 5.

**Check divisibility by 5 test for (i) 210 Â Â Â (ii) 1215 Â Â Â (iii) 17**

*Example*:*Solution*:Â On checking the unitâ€™s digit for each given number, we can see that

(i) 210 is divisible by 5 (ii) 1215 is divisible by 5 (iii) 17 is not divisible by 5

**3. Divisibility by 2:
**Any number whose unitâ€™s digit is even (0, 2, 4, 6 and 8) will always be divisible by 2.

**Check divisibility by 2 test for (i) 230 Â Â Â (ii) 215 Â Â Â (iii) 18**

*Example*:*Solution*:Â On checking the unitâ€™s digit for each given number, we can see that

(i) 230 is divisible by 2 (ii) 215 is not divisible by 2 (iii) 18 is divisible by 2

**4. Divisibility by 3:
**For any given number, if the sum of its digits is divisible by 3, then that number will always be divisible by 3.

**Check divisibility by 3 test for (i) 630 Â Â Â (ii) 1211 Â Â Â (iii) 19**

*Example*:*Solution*:Â On checking the sum of digits for each given number, we can see that

(i) Sum of digits of 630 is 6+3+0 = 9. Now, we know that 9 is divisible by 3.Â Thus, 630 is divisible by 3.

(ii) Sum of digits of 1211 is 1+2+1+1 = 5. Now, we know that 5 is not divisible by 3.Â Thus, 1211 is not divisible by 3.

(iii) Â Sum of digits of 19 is 1+9 = 10. Now, we know that 10 is not divisible by 3.Â Thus, 19 is not divisible by 3.

**5. Divisibility by 9:
**For any given number, if the sum of its digits is divisible by 9, then that number will always be divisible by 9.

**Check divisibility by 9 test for (i) 740 Â Â Â (ii) 81 Â Â Â (iii) 1818**

*Example 1*:*Solution*:Â On checking the sum of digits for each given number, we can see that

(i) Sum of digits of 740 is 7+4+0 = 11. Now, we know that 11 is not divisible by 9.Â Thus, 740 is not divisible by 9.

(ii) Sum of digits of 81 is 8+1 = 9. Now, we know that 9 is divisible by 9.Â Thus, 81 is divisible by 9.

(iii) Sum of digits of 1818 is 1+8+1+8 = 18. Now, we know that 18 is divisible by 9.Â Thus, 1818 is divisible by 9.

** Example 2:** If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

*Solution*:Â For any number to be the multiple of 3, the sum of its digit must also be multiple of 3.

Thus, 3 + 1 + z + 5 = 9 + z must be multiple of 3.

Here, z is a single digit number, now since z is added with 9, so any multiple of 3 can take place i.e. z can be 3 or 6 or 9. Moreover, z can also be 0 as 9 itself is multiple of 3.

Hence, value of z can be 0, 3, 6 or 9.

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