Numbers: Variables and Functions

Variables and Functions

To understand variables, we need to understand constants also.

So we can see two facts:

1. Constant      2. Variables

1. Constants: All the numbers are constants. 

For Example: 2 has a value 2. Its value can not be changed. So it is a constant. All real numbers (Positive-Negative, Rational-Irrational, Integers-Fractions, Prime-Composite-Co-Prime, etc) are constants.

2. Variable

Variables are the symbols represented by normally alphanumeric and sometimes special characters to express some numbers. The values of these assumed characters may change time to time and according to situation and conditions. 

For example: In a condition a = 5 

and in other condition it is changed to a = 22. 

Therefore these are known as variables.

The mathematics for variables is same as the mathematics for numbers. It means they follows same sets of rules for addition, subtraction, multiplication, division, brackets, BODMAS, LCM, HCF, etc. So it is not a tough job to understand the operations on variables.

Let there are two variables x and y then

Addition:            x + y

Subtraction:       x - y

Multiplication:     x × y = xy

Division:             x ÷ y = x/y

When we find any value of x and y, then we can find the answer of these operations

Like let x = 24 and y = 6

Addition = 30,     

Subtraction = 18,   

Multiplication = 144    and   

Division = 4

Similarly we can solve for other operations like power and other expressions.

Let a constant 4 is operated with variable y, then

Addition: y + 4,   

Subtraction: y - 4,   

Multiplication: y × 4 = 4y,       and

Division: y ÷ 4 = y/4

Let a constant 4 is operated with variable 4y - 3, then

Addition: 4y - 3 + 4 = 4y + 1,                    

Subtraction: 4y - 3 - 4 = 4y - 7,   

Multiplication: (4y - 3) × 4 = 4y×4 - 4×3 = 16y - 12,       

Division: (4y - 3) ÷ 4 = 4y/4 - 3/4 = y - 3/4

Rules for doing problems related to the variables in the equation and questions

Variables follow all the mathematical operations and formula. For solving a variable in a equation, we have certain rules

If we transfer a number or variable from one side of equal to another side then,

+ is changed to – 

– is changed to +

× is changed to ÷

÷ is changed to ×

For Example: if x + 3 = 5;                 or         x = 5 – 3 = 2

                      if x – 4 = 7;                  or         x = 7 + 4 = 11

                      if x × 7 = 9;                 or          x = 9 ÷ 7 = 9/7 =  1.286

                      if x ÷ 7 = x / 7 = 3;       or         x = 3 × 7 = 21

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Diode Resistances-Capacitances

Diode Resistances: 

It is defined as the resistance offered by a diode. There are two types of diode resistances: 

(1) Forward Resistance (2) reverse Resistance

Forward Resistance: It is defined as the resistance offered by the diode when connected in forward condition. It is represented by rf or Rf . Forward resistance of ideal diode is 0 𝛀 and that of practical diode is few 𝛀.

There are two types of forward resistances:

(a)  Static or DC Resistance (R_s or R_dc): It is a type of resistance which is defined as the ratio of voltage across diode to the corresponding current in the diode at any fixed voltage applied.

R_s = V/I     (at a constant voltage)

and I can be calculated by diode current equation

(b) Dynamic or AC Resistance (RD or  Rac ): It is a type of resistance which is defined as the ratio of change in the voltage across diode to the corresponding change in the diode at any fixed voltage applied.

Mathematically                  R_D = dV_D / dI          (at a constant voltage)

We can understand it by watching the figure.

Derivation:

(2) Reverse resistance: The resistance offered by the diode when connected in reverse bias condition, is called reverse resistance of diode. It is denoted by rr  or  Rr. It can be calculated as similar as the calculation of static resistance. It can also be expressed as static resistance of diode in reverse bias condition. 

In reverse bias condition reverse current is fixed and it does not need to be calculated.

So, reverse resistance = R_r = V_D / I₀           (At a constant Voltage)

Reverse resistance of ideal diode is ∾ 𝛀 and that of practical diode is few M𝛀 and onwards..

Example: A germanium diode is used in a rectifier circuit and is operating at a temperature of 25℃ with a reverse saturation current of 1000 𝜇A. calculate the value of forward resistance at a forward voltage 0.2 Volts.

Solution:

Given: VD = 0.2 V, I0 = 1000µA = 1000 x 10-6 A, η=1 (Ge diode)

T = 250 C = 273 + 25 K = 298 K

VT = 298 / 11600 = 0.026 volt

It is required to calculate forward resistance but static or dynamic, it is not clear, so we should calculate both the resistances.

Calculation of Static Resistance

putting all values in diode current equation

I = 1000 x 10-6 x (e(0.2/0.026) -1)

I = 2.19 A 

Rs = VD / I = 0.2 Volt / 2.19 A = 0.0913 𝛀                                 (Answer: Rs = 0.0913 𝛀)

Calculation of Dynamic Resistance

We know that dynamic resistance Rd = 𝜼VT / I          (Using third formula)

Rd = 1 × 0.026 / 2.19 = 0.012 𝛀                                                (Answer: Rd = 0.012 𝛀)

Diode Capacitance:

The capacitance offered by diode is known as diode capacitance. There are two types of diode capacitance. (1) Diffusion capacitance       (2) Transition Capacitance

Transition Capacitance: This capacitance is found when diode is reverse biased. The space charge region behaves as a dielectric material with ε dielectric constant. P side with positive charges and N side with negative charges behave like positive plate and negative plate respectively. This capacitance uses property of variation in depletion width (d) with applied reverse voltage. 

Transition capacitance can be calculated as                        C_T = εA / d

This is the general formula of capacitance of a material and we have studied it from high school even and in all onward classes so it is very easy to remember.

Sometimes depletion width is denoted by W.   Then           C_T = εA / W

In reverse bias condition, diode capacitance (Cj) can be calculated using the following formula:            C_T = εA / d           where d is the depletion width.

ε depends upon the doping level, material and space charge and charge carriers passing through the junction. These parameters cannot be changed once diode is designed. So ε cannot be varied. This is an independent parameter. A is the cross sectional area which also be not changed once diode is fabricated. ‘d’ is the only parameter which depends on external factor i.e. battery potential.

So we can say that           C_T     α    1/d

We know that d increase on increasing reverse voltage,

Therefore CT decreases on increasing reverse voltage.

VC curve or Voltage-Capacitance curve is shown in figure. C_T decreases on increasing reverse voltage. C0 is the voltage when diode is unbiased i.e. battery is not connected or battery voltage is zero. 

Diffusion capacitance: 

This capacitance is found when diode is forward bias. Since depletion width does not vary much more so capacitance due to depletion width is constant in this case.

As the forward voltage increases, majority charge carrier also increases passing through the junction which results in incrementing the value of dielectric constant which directly depends upon the number of charge carriers passing through the junction.

With this theory we can conclude that the diffusion capacitance directly depends on the existence of free charge carriers. So it depends on two factors

1. Forward current (I) and

2. Carrier life time (ꞇ): It is defined as the average time duration for which charge carries exists i.e. time between generation and recombination. 

Derivation of Diffusion Capacitance:

We know that the forward current exponentially increases with increasing voltage

Means        I        α      exp (VD)

But             CD     α      I

So,             CD     α      exp (VD)

VC curve or Voltage-Capacitance curve is shown in figure. CD  increases on increasing forward voltage.  C0  is the voltage when diode is unbiased i.e. battery is not connected or battery voltage is zero. 

Complete VC Curve: 

The capacitance offered by diode is known as diode capacitance. There are two types of diode capacitance. 

(1) Transition Capacitance          (2) Diffusion capacitance

        CT = εA / d                                   CD = ꞇI / ηVT

This is the complete voltage capacitance VC curve of a PN junction diode.

CT reciprocal decreases on increasing reverse voltage. 

CD exponentially increases on increasing forward voltage.

C₀ is the voltage when diode is unbiased i.e. battery is not connected or battery voltage is zero. 

Example: Example: A germanium diode is used in a rectifier circuit and is operating at a temperature of 25℃ with a reverse saturation current of 1000 𝜇A. Calculate the value of Diffusion Capacitance at a forward voltage 0.2 Volts. Mean carrier life time is given as 50ns.

Solution: 

Given: VD = 0.2 V, I0 = 1000µA = 1000 x 10-6 A, η=1 (Ge diode) and ꞇ=50 x 10-9 seconds

T = 250 C = 273 + 25 K = 298 K

VT = 298 / 11600 = 0.026 volt

Putting all values in diode current equation

I = 1000 x 10-6 x (e(0.2/0.026) -1)

I = 2.19 A 

Now 

CD = ꞇI / ηVT  

CD = 50 x 10-9 x 2.19 / (1 x 0.026)

CD = 4211.53 x 10-9 F = 4.212 x 10-6 F = 4.212 µF                        (Answer: CD = 4.212 µF)

Questions for Exercise

1. Forward resistance of a diode is 

(a) few 𝛀     (b) few k𝛀       (c) few M𝛀          (d) few T𝛀

2. A reverse biased diode has a resistance of 

(a) 10 𝛀 to 10 𝛀     (b) 1 k𝛀       (c)  0 𝛀           (d) few hundreds k𝛀  

3. Transition capacitance CT  ....................... with decrease in reverse voltage.

(a) decreases      (b) increases        (c) remains unchanged      (d) data not complete

4. At which of the following values CD  is maximum?

(a) VD  = -4 volt      (b) VD  = 0 volt    (c) VD  = 0.2 volt     (d) VD  = 0.31 volt 

5. The dynamic resistance is measured in

(a) Transient condition,    (b) DC condition,       (c) AC Condition     (d) can not say

Answer: 1 - a          2 - d           3 - b             4 - d         5 - c

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Working of PN Junction Diode and VI Characteristics

Working of PN Junction Diode under biasing:  

Once depletion region is formed, the diode is ready to operate voltage and current through it. When we analyze the working of PN Junction diode, we have three conditions

1. Diode is unbiased (VD = 0 Volt or anode and cathode terminals are shorted.)

2. Diode is biased with positive voltage at P side (VD > 0 Volt, Forward Bias)

3. Diode is biased with negative voltage at P side (VD < 0 Volt, Reverse Bias)

Case 1: Diode is unbiased 

When diode is unbiased, no energy is provided to the electrons and holes, therefore any electrons or holes will not be able to move across the junction.

So net current across junction will be 0 mA.

Case 2: Diode is forward bias  

When Vi is positive (Vi > 0 Volt), Diode is said as forward bias. We know that the junction behaves as a potential barrier with Vk voltage (knee voltage). It means diode will not allow any electron or hole movement if Vi < Vk .

When Vi ≥ Vk , electrons and holes have sufficient energy to cross the junction. So majority holes from P side and majority electrons from N side cross the junction due to repulsion force from the battery to the majority charge carriers as shown in the diagram. 

The movement of charge carriers due to repulsion force of the battery is known as drift process. So current is found due to drift process in biasing condition.

Junction in this case, becomes narrow because of the large concentration of charge carriers at the edge of the junction due to movement of majority charge carriers towards the junction. This also helps electrons and holes to cross the junction easily.

So current due to positive holes is in the same direction and current due to negative electrons is in opposite direction.

So current due to positive holes is in the same direction and current due to negative electrons is in opposite direction. Finally both the direction becomes same to each other.

So, total current becomes the sum of electron current and hole current I = Ih + Ie  

Where Ih = Current due to holes and Ie = Current due to electrons

Total current follows an exponential equation of input voltage across the diode using following equation

           (Diode Current Equation)

I = Net current

I0 = Reverse Saturation Current

VD = Voltage across diode

η  = 1 (for Ge)

    = 2 (for Si)                                                                   (η  is called material constant)

(VT is called thermal constant) and 

Here T is temperature in Kelvin)

At room temperature i.e. T = 27 C = 273 + 27 K = 300 K

VT = 300/11600 = 0.02586 Volt = 0.026 Volt = 26 mV

Case 3: Diode is Reverse Bias:  

When Vi < 0 volt, minority charge carriers are energized to move towards the junction. Since minority charge carriers are small in numbers therefore a small current is found in the circuit. In this case, negative electrons from P side and positive holes from N side, cross the junction due to repulsion force from the battery to the minority charge carriers. Junction is wide in this case. We can see in the diagram. 

So current due to positive holes is in the same direction and current due to negative electrons is in opposite direction.

Net reverse saturation current in this case is Io = Ioe + Ioh   µA which is opposite to the forward diode current. This Io is used in diode current equation.

After understanding the working of PN diode, now we will try to understand its VI characteristics in detail.

Example: A germanium diode is used in a rectifier circuit and is operating at a temperature of 25℃ with a reverse saturation current of 1000 𝜇A. calculate the value of forward current at a forward voltage 0.2 Volts.

Solution:

Given: VD = 0.2 V, I0 = 1000µA = 1000 x 10-6 A, η=1 (Ge diode)

T = 250 C = 273 + 25 K = 298 K

VT = 298 / 11600 = 0.026 volt

putting all values in diode current equation

I = 1000 x 10-6 x (e(0.2/0.026) -1)

I = 2.19 A                                                                          (Answer: I = 2.19 A)

Example: A germanium diode has a reverse saturation current of 100  𝜇A. Calculate the voltage at which 10% of the rated current will flow through the diode, at room temperature if diode is rated for 1 A.

solution:

Given: I0 = 3 µA = 3 x 10-6 A, η=1 (Ge diode) and I = 10 % of rated 1 A = 0.1 A

T = 270 C = 273 + 27 K = 300 K

VT = 300 / 11600 = 0.026 volt,              To be calculated VD

putting all values in diode current equation

0.1 A = 100 x 10-6 x (e(VD/0.026) -1) A

e(VD/0.026) -1 = 0.1 x 10000  = 1000

e(VD/0.026)  = 1000 + 1 = 1001

VD/0.026 = ln(1001)         or              VD = 0.026 x ln(1001)

VD = 0.18 Volt                                                                        (Answer: VD = 0.18 Volt)

Symbol of Diode 

From the working of PN Junction Diode, we saw that a large current in diode is found from P side towards N side when diode is forward bias. However a small negligible current in diode is found from N side towards P side.

It means diode allows current in in forward direction however it restricts current in opposite direction. So the direction of arrow represents the allowed direction of current and a line like wall represents restriction of current from opposite side.

Terminal in P side is Anode and Terminal in N side is Cathode.

VI Characteristics of PN Junction Diode

Practical VI Characteristics

VI characteristics of PN junction diode is shown in the diagram. According to the working of diode

 In forward bias, For VD < Vk current is zero I = 0 mA

                           For VD > Vk current                                          

                   (Diode Current Equation)

In reverse bias, for VD < 0 current I = I0 (Opposite to I)

Here Vk = Knee Voltage: This is the minimum voltage required by the diode to make it ON or make it in forward bias.

VB = Breakdown Voltage: It is defined as the minimum reverse voltage at which junction gets broken down. After breakdown a large current is found in the circuit

Io = reverse saturation current in reverse bias diode. 

Ideal Characteristics

An ideal VI characteristic of a diode is shown in the diagram. This was the assumption about the correct usefulness of a diode. According to this concept

For VD > 0, diode becomes immediately ON and current tends towards infinite value.

For VD < 0, diode becomes immediately OFF and current becomes absolutely zero.

As diode voltage VD tends to 0+, current tends to its maximum value

So forward resistance Rf = VD / I = 0 / I = 0 Ω                 (Short Circuit)

As diode voltage becomes 0 or 0- , current also becomes 0.

So reverse resistance Rr = VD / I = VD / 0 = ꝏ Ω             (Open Circuit)

We always approach to fabricate a diode which has parameters very close to the Ideal diode. It is not possible to design an ideal diode.

Questions for Exercise

1. When PN Junction id forward biased:

(a) electrons from P side are injected into N side.

(b) holes from N side are injected into P side.

(c) Both of (a) and  (b)                                            (d) None of these.

2. When a reverse bias is applied to a junction diode, it

(a) rises in potential barrier,                                        (b) lower the potential barrier

(c) generally increases the majority carrier current    (d) None of these

3. The minimum voltage which is required to switch diode in ON state is known as

(a) Threshold voltage     (b) Knee voltage       (c) Cut in voltage      (d) All of these

4. Breakdown occurs when 

(a) Battery is not connected to the diode

(b) positive terminal of battery is connected to P side of diode and vice versa

(c) negative terminal of battery is connected to P side of diode and vice versa

(d) all of the above.

Answer: 1 – d,   2 - a ,     3 - d,      4 - c 

Numerical 1- A silicon diode is used in a rectifier circuit and is operating at a temperature of 36℃ with a reverse saturation current of 10 𝜇A. calculate the value of forward current at a forward voltage 0.0.18 Volts.

Numerical 2- A germanium diode has a reverse saturation current of 15  𝜇A. Calculate the voltage at which 1% of the rated current will flow through the diode, at room temperature if diode is rated for 100 mA.

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Divisibility Rule By Prime Numbers (2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37)

In many competitive questions, it becomes very important to find whether given number is divisible by another given number or not. It is required when we solve HCF (Highest Common Factor), LCM (Least Common Multiple), Composite numbers, Co-Prime numbers, multiples, factors, etc.

So this topic is extensively important in understanding many parts of number mathematics and algebra as well because all operations in number mathematics are always true for algebra and other mathematics also.

Divisibility Rule (Divisibility Test - 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37) By Prime Numbers

Divisibility test means testing of a number whether it is divisible completely by another given number. It is always means that the first number (Dividend) and the second number (Divisor) are integers. Result of a division is called Quotient and if something is left, it is called a remainder.

For Example:

456382 ÷ 12 = (Q=38031 and R=10)

Important:

"If in a division, quotient is an integer, and remainder is zero, it means that dividend is completely divisible by divisor."

"If in a division, either quotient is a fraction with remainder zero or quotient is integer with remainder non-zero, it means that dividend is not completely divisible by divisor."

1. Divisible By 2: An integer having 0, 2, 4, 6, 8 at its unity place or Last digit, is divisible by 2.

Steps of Test: There is only one step of test i.e. check the last digit is 0,2,4,6,8 or number is even number

For Example: 

2340, 12, 34, 56, 35798, etc

Note: The integers which are completely divisible by 2 are called even numbers.

2. Divisible By 3: If sum of all digits in an integer is divisible by 3, then the number is divisible by 3.

Step for Test: 1. Sum all the digits 2. If sum found is more than one digit number, then again sum all the digits of the previous sum. 3. Repeat the process until we get sum of digits as a one digit number. 4. If final sum is divisible by 3, then the given number is divisible by 3.     If final sum is not divisible by 3, then the given number is not divisible by 3.

For Example:

Example 1: Let us check whether 162738951 is divisible by 3.

Sum of All digits = 1 + 6 + 2 + 7 + 3 + 8 + 9 + 5 + 1 

                           = 42

Again Sum all digits = 4 + 2 = 6      (Divisible by 3)

We can see that the sum of all digits in the given number is divisible by 3, so the given number 162738951 is divisible by 3.

Verification: Now divide the number by 3 using conventional division method, we get that the given number 162738951 is divisible by 3 and quotient is 54246317‬ with remainder 0.

Example 2: Now let us take another example: Check whether 7635428743 is divisible by 3?

Sum of all digits = 7 + 6 + 3 + 5 + 4 + 2 + 8 + 7 + 4 + 3

                          = 49

Again sum all digits = 4 + 9 = 13           (Not divisible by 3)

We can see that sum of all digits in the given number is not divisible by 3, so the given number 7635428743 is not divisible by 3.

Verification: Now divide the number by 4 using conventional division method, we get that the given number 7635428743 is not divisible by 3 and quotient is 2545142914 with remainder 1 (≠ 0).

3. Divisible By 5: If Unit place (Last place) of number has either 0 or 5 then number is divisible by 5.

Steps of Test: There is only one step of test i.e. check the last digit is 0, 5.

For Example: 7520, 23456785, 975315, etc.

4. Divisible By 7: A number is divisible by 7 if result obtained from the subtraction of double of last digit (one's place) from the remaining part of the number is divisible by 7.

Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Double it (Means multiply it by 2) 3. Subtract this number from the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 70. 5. If final result is divisible by 7, then the given number is divisible by 7.

For Example:

Example 1: Let us check whether 98994 is divisible by 7.

The unit place of this number is 4

Doubling 4, we get 4 × 2 = 8

Remaining Portion of number = 9899

Subtract double number            9899  -  8  =  9891

Repeat the process: The unit place of new result is 1

Doubling 1, we get 2

Subtract it from new remaining portion      989  -  2  =  987

Repeat the process: The unit place of new result is 7

Doubling 7, we get 14

Subtract it from new remaining portion      98  -  14  =  84

Repeat the process: The unit place of new result is 4

Doubling 4, we get 8

Subtract it from new remaining portion      8  -  4  =  0        (Divisible By 7)

So the given number 98994 is  divisible by 7.

Verification: Now divide the number by 7 using conventional division method, we get that the given number 98994 is divisible by 7 and quotient is 14142 ‬with remainder 0.

Example 2: Let us check whether 2268 is divisible by 7.

The unit place of this number is 8

Doubling 8, we get 8 × 2 = 16

Remaining Portion of number = 226

Subtract double number            226  -  16  =  210

Repeat the process: The unit place of new result is 0

Doubling 0, we get 0

Subtract it from new remaining portion      21  -  0  =  21        (Divisible By 7)

So the given number 2268 is  divisible by 7.

Verification: Now divide the number by 7 using conventional division method, we get that the given number 2268 is divisible by 7 and quotient is 324 ‬with remainder 0.

5. Divisible By 11: If difference of sums of alternative digits of a given number is divisible by 11, then the given number is divisible by 11.

If difference of the sum of odd digits and the sum of even digits from a given number is divisible by 11, then the given number is divisible by 11.

First we will understand alternate places or even and odd places in a given number

(Figure: Alternative Digits in a Number)

Step for Test: 1. Sum all odd alternate digits from the given number. Say it Odd Alternative Sum  2. sum all even alternate digits from the given number. Say it Even Alternative Sum 3. Subtract Even Alternative Sum from Odd Alternative Sum. Say it subtracted result. 4. If Subtracted Result is divisible by 11, then the given number is divisible by 11.

For Example:

Let us check whether 9182701 is divisible by 11.

First sum alternative odd digits = 9 + 8 + 7 + 1 = 25

Then Sum alternative even digits = 1 + 2 + 0 = 3

difference of two results = 25 -3 = 22

We can see that 22 is divisible by 11.

So the given number 9182701 is divisible by 11.

Verification: Now divide the number by 11 using conventional division method, we get that the given number 9182701 is divisible by 11, and quotient is 834791 along ‬with remainder 0.

6. Divisible By 13: A number is divisible by 13 if result obtained from the addition of four times the last digit (one's place) from the remaining part of the number is divisible by 13.

Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Multiply it by 4. 3. Add this number to the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 130. 5. If final result is divisible by 13, then the given number is divisible by 13.

For Example:

Let us check whether 262743 is divisible by 13.

The unit place of this number is 3

Multiply it by 4 = 3 × 4 = 12

Remaining Portion of number =         26274

add the number multiplied by 4 =       26274  +   12  =  26286

Repeat the process: The unit place of new result is 6

Multiply it by 4 = 6 × 4 = 24

add this number in remaining portion =   2628  +   24  =  2652

Repeat the process: The unit place of new result is 2

Multiply it by 4 = 2 × 4 = 8

add this number in remaining portion =   265  +   8  =  273

Repeat the process: The unit place of new result is 3

Multiply it by 4 = 3 × 4 = 12

add this number in remaining portion =   27  +   12  =  39              (Divisible By 13)

So the given number 262743 is  divisible by 13.

Verification: Now divide the number by 13 using conventional division method, we get that the given number 262743 is  divisible by 13 and quotient is 20211 ‬with remainder 0.

7. Divisible By 17: A number is divisible by 17 if result obtained from the subtraction of five times the last digit (one's place) from the remaining part of the number is divisible by 17.

Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Multiply it by 5. 3. Subtract this number from the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 170. 5. If final result is divisible by 17, then the given number is divisible by 17.

For Example:

Let us check whether 27897 is divisible by 17.

The unit place of this number is 7

Five times the 7, we get 7 × 5 = 35

Remaining Portion of number = 2789

Subtract double number            2789  -  35  =  2754

Repeat the process: The unit place of new result is 4

Five times the 4, we get 4 × 5 = 20

Subtract it from new remaining portion     275  -  20  =  255

Repeat the process: The unit place of new result is 5

Five times the 5, we get 5 × 5 = 25

Subtract it from new remaining portion     25  -  25  =  0         (Divisible By 17)

So the given number 27897 is  divisible by 17.

Verification: Now divide the number by 17 using conventional division method, we get that the given number 27897 is divisible by 17 and quotient is 1641 ‬along with remainder 0.

8. Divisible By 19: A number is divisible by 19 if result obtained from the addition of double of last digit (one's place) with the remaining part of the number is divisible by 19.

Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Double it (Means multiply it by 2) 3. Add this number from the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 190. 5. If final result is divisible by 19, then the given number is divisible by 19.

For Example:

Let us check whether 27968 is divisible by 19.

The unit place of this number is 8

Doubling 4, we get 8 × 2 = 16

Remaining Portion of number = 2796

adding double number               2796  +  16  =  2812

Repeat the process: The unit place of new result is 2

Doubling 2, we get 4

Adding it from new remaining portion      281  +   4  =  285

Repeat the process: The unit place of new result is 5

Doubling 5, we get 10

adding it from new remaining portion      28  +  10  =  38        (Divisible By 19)

So the given number 27968 is  divisible by 19.

Verification: Now divide the number by 19 using conventional division method, we get that the given number 27968 is divisible by 19 and quotient is 1472 ‬with remainder 0.

9. Divisible By 23: There are two rules for divisibility by 23. Whichever you get easy, you should apply in exams

Rule A: A number is divisible by 23 if result obtained from the addition of seven times the last digit (one's place) to the remaining part of the number is divisible by 23.

Rule B: A number is divisible by 23 if result obtained from the addition of three times the last two digits (combination of ten's place and one's place) to the remaining part of the number is divisible by 23.

Rule A: Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Multiply it by 7 3. Add this number to the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number in two digits. 5. If final result is divisible by 23, then the given number is divisible by 23.  Rule B: Steps for Test: 1. Write the last two digits of number i.e. combination of ten's and one's place of the number. 2. Multiply it by 3 3. Add this number to the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 230. 5. If final result is divisible by 23, then the given number is divisible by 23.

For Example:

Let us check whether 93173 is divisible by 23.

Rule A: First Write last digit of the number i.e. 3

              Multiply it by 7 = 3 × 7 = 21

              Add this number in remaining portion = 9317 + 21 = 9338

              Repeat above steps

              Last digit of the new result = 8

              Multiply it by 7 = 8 × 7 = 56

              Add this number in remaining portion = 933 + 56 = 989

              Repeat above steps

              Last digit of the new result = 9

              Multiply it by 7 = 9 × 7 = 63

              Add this number in remaining portion = 98 + 63 = 161

              Repeat above steps

              Last digit of the new result = 1

              Multiply it by 7 = 1 × 7 = 7

              Add this number in remaining portion = 16 +7 = 23                   (Divisible By 23)

              Hence the given number 93173 is divisible by 23.

Rule B: First Write last two digits of the number i.e. 73

              Multiply it by 3 = 73 × 3 = 219

              Add this number in remaining portion = 931 + 219 = 1150

              Repeat above steps

              Write last two digits of the new number i.e. 50

              Multiply it by 3 = 50 × 3 = 150

              Add this number in remaining portion = 11 + 150 = 161            (Divisible By 23)

              Hence the given number 93173 is divisible by 23.

Verification: Now divide the number by 23 using conventional division method, we get that the given number 93173 is divisible by 23, and quotient is 4051 along ‬with remainder 0.

10. Divisible By 29: A number is divisible by 29 if result obtained from the addition of three times the last digit (one's place) from the remaining part of the number is divisible by 29.

Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Multiply it by 3. 3. Add this number to the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 290. 5. If final result is divisible by 29, then the given number is divisible by 29.

For Example:

Let us check whether 61857 is divisible by 29.

The unit place of this number is 7

Multiply it by 3 = 7 × 3 = 21

Remaining Portion of number =         6185

add the number multiplied by 4 =       6185 + 21 = 6206

Repeat the process: The unit place of new result is 6

Multiply it by 3 = 6 × 3 = 18

add this number in remaining portion =   620 + 18 = 638

Repeat the process: The unit place of new result is 8

Multiply it by 3 = 8 × 3 = 24

add this number in remaining portion =   63  +   24  =  87              (Divisible By 29)

So the given number 61857 is  divisible by 29.

Verification: Now divide the number by 29 using conventional division method, we get that the given number 61857 is  divisible by 29 and quotient is 2133 ‬with remainder 0.

11. Divisible By 31: A number is divisible by 31 if result obtained from the subtraction of three times the last digit (one's place) from the remaining part of the number is divisible by 31.

Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Multiply it by 3. 3. Subtraction this number to the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number less than 310. 5. If final result is divisible by 29, then the given number is divisible by 31.

For Example:

Let us check whether 55149 is divisible by 31.

The unit place of this number is 9

Multiply it by 3 = 9 × 3 = 27

Remaining Portion of number =         5514

add the number multiplied by 4 =       5514 - 27 = 5487

Repeat the process: The unit place of new result is 7

Multiply it by 3 = 7 × 3 = 21

add this number in remaining portion =   548 - 21 = 527

Repeat the process: The unit place of new result is 7

Multiply it by 3 = 7 × 3 = 21

add this number in remaining portion =   52  -   21  =  31              (Divisible By 31)

So the given number 55149 is  divisible by 31.

Verification: Now divide the number by 31 using conventional division method, we get that the given number 55149 is  divisible by 31 and quotient is 1779 ‬with remainder 0.

12. Divisible By 37: There are two rules for divisibility by 23. Whichever you get easy, you should apply in exams

Rule A: A number is divisible by 37 if result obtained from the subtraction of eleven times the last digit (one's place) to the remaining part of the number is divisible by 37.

Rule B: A number is divisible by 37 if result obtained from the addition of numbers by taking sets of three digits from right to left without reforming three digit numbers is divisible by 37.

Rule A: Steps for Test: 1. Write the last digit of number i.e. one's place of the number. 2. Multiply it by 11 3. Subtract this number to the remaining portion of the number. We will obtain a result. 4. Repeat above three steps for the result obtained many times until we get a number in two digits. 5. If final result is divisible by 37, then the given number is divisible by 37.  Rule B: Steps for Test: 1. Write all the numbers of three digits taking from right to left from the given number 2. Add all the numbers 3. Repeat above two steps for the result obtained many times until we get a number less than 1000. 4. If final result is divisible by 37, then the given number is divisible by 37.

For Example:

Let us check whether 310837 is divisible by 37.

Rule A: First Write last digit of the number i.e. 7

              Multiply it by 11 = 7 × 11 = 77

              Add this number in remaining portion = 31083 - 77  = 31006

              Repeat above steps

              Last digit of the new result = 6

              Multiply it by 11 = 6 × 11 = 66

              Add this number in remaining portion = 3100 - 66 = 3034

              Repeat above steps

              Last digit of the new result = 4

              Multiply it by 11 = 4 × 11 = 44

              Add this number in remaining portion = 303 - 44 = 259            (Divisible By 37)

              Hence the given number 310837 is divisible by 37.

Rule B: Write 310837 as sets of three digits numbers 310,837

              Adding all sets 310 + 837 = 1147

              Repeat above steps: Write 1147 as sets of three digits numbers 1,147

              Adding all sets 1 + 147 = 148                                                  (Divisible By 37)

              Hence the given number 310837 is divisible by 37.

Verification: Now divide the number by 37 using conventional division method, we get that the given number 310837 is divisible by 37, and quotient is 8401 along ‬with remainder 0.