Revision Notes on Playing with Numbers

Numbers in General Form

The general form of a number is obtained by adding the product of the digits with there place values.

1. The General Form of a Two Digit Number

ab = a × 10 + b = 10a + b

Example

93 = 10 × 9 + 3

= 90 + 3

2. The General Form of a Three Digit Number

abc = 100 × a + 10 × b + c = 100a + 10b + c

Example

256 = 100 × 2 + 10 × 5 + 6

= 200 + 50 + 6

Remark: ab doesn’t mean a × b as we generally use it like this.

Games with Numbers

1. Reversing the Digits – Two Digit Number

a. The trick to divide by 11: If we have any two-digit number then we will always get the remainder zero if we add the original number with its reverse number and divide it by 11. The trick behind it:

Step 1: Let us choose any two digit number ab, which can be written as (10a + b).

Step 2: Its reverse number will be ba, which can be written as (10b + a).

Step 3: Sum of both the numbers is ab + ba, we can also write as (10a + b) + (10b + a) = 11a+ 11b = 11 (a + b)

Step 4: By dividing the number by 11 we get 11 (a + b)/11 = (a + b) .

This shows that the remainder will always be zero in case of any number. The quotient is a + b, which is the sum of the digits of the given number ab.

Example

Try for the number 78.

Solution:

The given number is 78.

The reverse number is 87.

Sum of both the numbers is 78 + 87 = 165, 165 ÷ 11 = 15 and the remainder are zero.

b. The trick to divide by 9: If we have any two digit number then we will always get the remainder zero if we subtract the original number with its reverse number and divide it by 9. The trick behind it:

Step 1: Let’s choose any two digit number ab, which can be written as (10a + b).

Step 2: Its reverse number will be ba, which can be written as (10b + a).

Step 3: By subtracting both the numbers i.e. ab – ba we get (10a + b) – (10b + a) = 9a – 9b = 9 (a – b)

Step 4: By dividing the number by 9 we get  9 (a – b)/9 = (a – b).

This shows that the remainder will always be zero in case of any number.

Example

Try for the number 78.

Solution:

The given number is 78.

The reverse number is 87.

Subtraction of both the numbers is 87 – 78 = 9, 9 ÷ 9 = 1 and the remainder is zero.

2. Reversing the Digits – Three Digit Number

The trick to divide by 99: If we have any three digit number then we will always get the remainder zero if we subtract the original number with its reverse number and divide it by 99. The trick behind it:

Step 1: Let’s choose any three digit number abc, which can be written as (100a + 10b + c).

Step 2: Its reverse number will be cba, which can be written as (100c + 10b + a).

Step 3: By subtracting both the numbers i.e. abc – cba we get (100a + 10b + c) – (100c + 10b + a) = 99a – 99c = 99 (a – c).

Step 4: By dividing the number by 99 we get 99 (a – c)/9 = (a – c).

This shows that the remainder will always be zero in case of any number. 

Example

Try with the number 456

Solution:

The given number is 456.

The reverse number is 654.

Subtraction of both numbers is 654-456=198

By dividing it by 99 we get 2 i.e. 198 ÷ 99 = 2 and the remainder is zero.

3. Forming Three-digit Numbers with given three-digits.

If we have a three digit number then we will always get the remainder zero if we rearrange the number in such a way that all the three numbers are different and if we add them all and then divide it by 37. The trick behind it:

Step 1: Let us take any three digit number abc, which can be written as (100a +10b + c).

Step 2: Rearrange the number in such a way that it forms two different numbers. One number can be bca which can be written as (100b + 10c + a) and other can be cab which can be written as (100c + 10a + b).

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

Step 4: By dividing the number by 37 we will always get the remainder zero .

Example

Try for the number 397.

Solution:

Given number is 397.

By rearranging we will have two other numbers i.e. 973 and 739.

Sum of all the three numbers is 397+ 973 + 739 = 2109.

By dividing 2109 by 37 we get 57 i.e. 2109 ÷ 37 = 57 and remainder will be zero.

Letters for Digits

In this type of puzzles, the letter represents the digits and we have to crack the code. Generally, it has puzzles related to addition and multiplication.

Some important rules related to such types of puzzles:

Every letter can represent only one digit and the vice-versa.

The first digit cannot be zero Like 25 cannot be written as 025 and 0025.

Every puzzle must have only one solution.

Example

Find the value of P & Q.

Solution:

Here, at the ones place the addition of P & Q is given 9. This shows that it is the sum of two single digits as the sum of two single digit numbers cannot be 19. Hence there is not any carry in this step. For the next step, 2 + P = 0 and as there was not any carry from the last step it is possible only if P = 8.
Now, 2 + 8 = 10 and 1 will be carry for the next step.
So, 1 + 1 + 6 = P. Hence, clearly, P is 8 and Q can only be 1 to satisfy the given addition.

Hence, the value of P is 8 and Q is 1.

2: – A number is divisible by 2 if the last digit (unit’s digit) is even (or 0, 2, 4, 6, 8). 

Ex- 32, 459992

5: – A number is divisible by 5, if the last digit of the number is 0 or 5. 

Ex- 35, 200 and 1005

10: – A number is divisible by 10, if the last digit is 0, then 10 divides that number. 

Ex– 2520, 4320 and2190

3: – A number is divisible by 3, if the sum of digits of a number is divisible by 3.

Ex–    252 = 2 + 5 + 2 = 9 ÷ 3 = 3 ∴ 252 is divisible by 3
9: – A number is divisible by 9, if sum of digits of the number is divisible by 9.

Ex– 252 = 2 + 5 + 2 = 9 ÷ 9 = 1 ∴ 252 is divisible by 9
4: – A number is divisible by 4, if the last two digits of the number is divisible by 4. 

Ex- 81924 = since last two digits 24 is divisible by 4, hence the number is divisible by 4.

8: –    A four digits or higher number is divisible by 8, if its last 3 digits are divisible by 8.

Note-       If 3^rd digit is even then check only 2-digits.

Ex– 34064 since last three digits forms 064 which is divisible by 8.

6: –    A number is divisible by 6, if the number is divisible by both 2 and 3.

Ex- 252 is divisible by 2 and 3 because its unit digit is 2 and sum of digit = 2+5+2=9 is divisible by 3 hence the number 253 is divisible by 6.

7: – A number is divisible by 7, to check for this follow these steps:

Take the last digit (right most) and double it.

Subtract the doubled number from the remaining number.

Check the answer if it divisible by 7 and Repeat the above two steps.

Ex-   

47292 = 4729 – (2 × 2) = 4729 – 4 = 4725
= 472 – (2 × 5) = 472 – 10 = 462

= 46 – (2 × 2 ) = 46 – 4 = 42

Since 42 is divisible by 7. ∴ Number 47292 is divisible by 7.

11: –  A number is divisible by 11, to check this take the sum of all the digits at odd places and the sum of digits at even places. If the difference of these two sums is zero or multiple of 11, then the number is divisible by 11.

Ex-    8679 = Sum of odd place digits- Sum of even place digits

= (8 + 7) – (6+9) = 15 – 15 = 0.

246906 = (2+6+0) – (4+ 9 + 6) = |8 – 19| = 11.

13: –  To check if a number is divisible by 13, take four times of the last digit and add it to the remaining number. Repeat the step if necessary. If the result number is multiple of 13, the number is divisible by 13.

Ex-3172 = 317 + (4×2) = 317 + 8 = 325 = 32 + (4 × 5) = 32 + 20 = 52 which multiple of 13.
19: –  To check if a number is divisible by 19, take the double of last digit and add it to remaining    number. Repeat the steps. If the result is divisible by 19, hence the original number.

Ex-    1292 = 129 + (2 × 2) = 133 = 13 + (2 × 3) = 19. ∴ Number 1292 is divisible by 19

12: –  A number is divisible by 12, if the number is divisible by 4 and 3 both.

24: –  A number is divisible by 24, if it is divisible by 3 and 8 both.
25: –  A number is divisible by 25, if the last two digits are divisible by 25.

125: – A number is divisible by 125, if the last three digits are divisible by 125.