# Numbers Problems

#### Chapter 1

5 Steps - 3 Clicks

# Numbers Problems

### Introduction

Number System is all about finding the face value & place value of a digit, basic rules and different types of numbers, rules for divisibility, factors and multiples. Number System is one of the important topic in the Quantitative Aptitude section . Number system is a system to represent numbers using digits and symbols. Number system provides a unique representation of every number and represents arithmetic and algebraic structure of the figures.

### Methods

Face Value

Face Value of the digit in numeral is the value of digit itself, irrespective of its place in the numeral.

For example, In the numeral 36578,

• The face value of 8 is 8
• The face value of 7 is 7
• The face value of 5 is 5
• The face value of 6 is 6
• The face value of 3 is 3

Place value

Place value is the value of digit, depending on its place and we can relate place with ‘place value’.

For example, In the numeral 36578, Find the place values.

first write the given numerals in place-value chart, we get.

Crores Ten
Lakhs
Lakhs Ten
Thousands
Thousands Hundreds Tens Ones
3 6 5 7 8

• Place value of 8 = (8 x 1) = 8, (Here, to get the place value of 8, we multiply 8 by 1. Because 8 is at Ones place)
• Place value of 7 = (7 x 10) = 70, (Here, to get the place value of 7, we multiply 7 by 10. Because 7 is at Tens place)
• Place value of 5 = (5 x 100) = 500, (Here, to get the place value of 5, we multiply 5 by 100. Because 5 is at Hundreds place)
• Place value of 6 = (6 x 1000) = 6000, (Here, to get the place value of 6, we multiply 6 by 1000. Because 6 is at Thousands place)
• Place value of 3 = (3 x 10000) = 30000, (Here, to get the place value of 3, we multiply 3 by 10000. Because 3 is at Ten Thousands place)

Face Value and Place Value difference:

The image below explains the difference between face and place value.

For Clear understanding, let’s discuss about place value in detail. Look at the image given below, we can get the clear idea about place value.

For better understanding of the image, let’s consider an example.

Find out the place value of the digit “3” in numeral 36578

Explanation:

As per the Place, the location of “3” in 36578 is Ten Thousands place.

As per the Value, the value of “3” in 36578 is 30000.

Finally, the place value of “3” in 36578 is “30000

Shortcut to find out place value of a digit:

Find out the place value for 86791.

Step 1: First, Write the digit for the one you want to find the place value.

(Here in this case let’s find the place value for digit “7”)

Step 2: Now, Count the number of digits which come after the digit for the one you want to find the value.

(In 86791, after “7” we have two digits “91”)

Step 3: Since there are two digits (91) after 7, take two zeros after 7.

Then we get 700. This is the place value of digit 7 in 86791.

Example 1:

Find the face value and place value of the underlined digit in the number given below.

968352

Solution:

In the number above, 5 is underlined.

The face value of 5 is the same.

So, the face value of 5 is 5.

To get place value, count the number of digits after 5.

There is only one digit after 5.

So, the place value of 5 is 50.

Example 2:

Find the place value of “K” in the number given below.

K25937

Given that K = 2x and x = 4.

Solution:

From the given information, let us find the value of “K”.

K = 2x = 2(4) = 8 → K = 8

so, we have

K25937 → 825937

To get place value of “K”, we have to count the number of digits after “8”.

There are five digits after 8.

Hence, the place value of “K” is 800000.

Example 3:

Find the place value of “K” in the number given below.

58K32

Given that K is a number which is less than 10 and exactly divisible by both 2 and 3.

Solution:

From the given information, let us find the value of “K”.

K = 6

(Because 6 is the only number less than 10 and also exactly divisible by both 2 and 3)

So, we have

58K32 → 58632

To get place value of “K”, we have to count the number of digits after 6.

There are two digits after 6.

Hence, the place value of “K” is 600.

Types of numbers: There are various types of numbers. They are natural numbers, whole numbers, integers, even numbers, odd numbers, prime numbers, composite numbers, etc.

1. Natural numbers are defined as the numbers that occur commonly in nature. A natural number is a whole, non negative number and the set of natural numbers is denoted by letter ‘N’.
Set of natural numbers is N = {1,2,3…..}

2. Whole numbers are defined as all the numbers without fractions and no decimals.
Set of whole numbers is (W)={0,1,2,3…….}

3. Integers are defined as all the numbers i.e. zero, positive and negative numbers.
Set of integers = {……-3,-2,-1,0,1,2,3,…….}

4. Even numbers are defined as the numbers divisible by 2. i.e When an even number is divided by 2, the remainder is 0.
0,2,4,6,….. are even numbers.

5. Odd numbers are defined as the numbers which are not divisible by 2. i.e When an even number is divided by 2, the remainder is 1.
1,3,5,7,…… are odd numbers.

6. Prime numbers are those numbers which have exactly two factors namely itself and 1.
Example: 2,3,5,7,11,13,17….etc.

7. Composite numbers are defined as the numbers which are not prime.
Example: 4,6,8,12,15,….. etc.

Rules for divisibility:

1. Divisibility by 2: A number is divisible by 2 if the last digit in a given numerical value is 0,2,4,6,8.
Example: 526492 is divisible by 2. (as last digit is 2)

2. Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
Example: 562185 is divisible by 3. (as sum of digits =27 is divisible)

3. Divisibility by 4: If the number formed by last two digits is divisible by 4 then the given numerical value is also divisible by 4.
Example: 1459872 is divisible by 4. (as last two digits 72 is divisible by 4)

4. Divisibility by 5: If the units digit is 0 or 5, then the given number is divisible by 5.
Example: 256480 is divisible by 5. (as units digit is 0)

5. Divisibility by 8: If the number formed by last three digits is divisible by 8 then the given numerical value is also divisible by 8.
Example: 45566568 divisible by 8. (as 568(the number formed by last 3 digits) is divisible by 8)

6. Divisibility by 9: If the sum of its digits is divisible by 9 then the given numerical value is also divisible by 9.
Example: 21421899 divisible by 9. (as sum of its digits = 36 is divisible by 9)

7. Divisibility by 10: If the units digit is 0, then the given number is divisible by 10.
Example: 65712590 is divisible by 10. (as units digit is 0)

8. Divisibility by 11: If the difference between sum of its digits at odd places and sum of its digits in even places is either zero or a number divisible by 11.
Example: 25784 is divisible by 11. (as (2+ 7 + 4) – (5+8) = 13 – 13 =0)

Explanation:

When dividing a number by another number, one will have the terms dividend, divisor, quotient and remainder.

• The number which we divide is called the dividend.

• The number by which we divide is called the divisor.

• The result obtained is called the quotient.

• The number left over is called the remainder.

Sample:

17 = 5 x 3 + 2

Where;

• 17 – Dividend
• 3 – Quotient
• 5 – Divisor
• 2 – Remainder

Example 1:

Divide 300 by 7, list out dividend, divisor, quotient, remainder and write division algorithm.

Solution:

If 300 divided by 7 using long division.

• Dividend – 300
• Divisor – 7
• Quotient – 42
• Remainder – 6

Division algorithm for the above division is.

⇒ 300 = 42 x 7 + 6

Example 2:

Divide 400 by 8, list out dividend, divisor, quotient, remainder and write division algorithm.

Solution:

If 400 divided by 8 using long division:

• Dividend = 400
• Divisor = 8
• Quotient = 50
• Remainder = 0

Division algorithm for the above division is

⇒ 400 = 50×8 + 0

Example 3:

Divide 750 by 16, list out dividend, divisor, quotient, remainder and write division algorithm.

Solution:

If 750 divided by 16 using long division:

• Dividend = 750
• Divisor = 16
• Quotient = 46
• Remainder = 14

Division algorithm for the above division is

⇒ 750 = 46×16 + 14

Explanation: This is the first formula in geometry and algebra of mathematics.

Introduction to Geometrical Approach:

• Line with a point = a+b
• Square Area = $$a^{2}$$
• Rectangle Area = ab

Prove $$(a+b)^2 = a^2 + b^2 +2ab$$ in geomentry:

• Draw a line with a point which divides a, b
• Total distance of this line = a+b
• Now find out the square of a+b ie. $$(a+b)^{2}$$
• The above diagram represents – a + b is a line and the square are is
$$a^{2}$$ + $$b^{2}$$ + 2ab

Hence proved $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab

Prove $$(a+b)^2 = a^2 + b^2 +2ab$$ in Algebra:

$$(a+b)^{2}$$ = (a+b)*(a+b)

= (a*a + a*b) + (b*a + b*b)

= ($$a^{2}$$ + ab) + (ba + $$b^{2}$$)

= $$a^{2}$$ + 2ab + $$b^{2}$$

Hence proved $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab

Example 1:

Expand the term $$(3x + 4y)^{2}$$ using the identity $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab

Solution:

$$(3x + 4y)^{2}$$ = $$(3x)^{2}$$ + $$(4y)^{2}$$ + 2(3x)(4y) = $$9x^{2}$$ + $$16y^{2}$$ + 24xy

∴ $$(3x + 4y)^{2}$$ = $$9x^{2}$$ + $$16y^{2}$$ + 24xy

Example 2:

Expand the term $$(\sqrt{2}x + 4y)^{2}$$ using the identity $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab

Solution:

$$(\sqrt{2}x + 4y)^{2}$$ = $$(\sqrt{2}x)^{2}$$ + $$(4y)^{2}$$ + 2$$(\sqrt{2}x)$$(4y) = $$2x^{2}$$ + $$16y^{2}$$ + 8$$\sqrt{2}$$xy

∴ $$(\sqrt{2}x + 4y)^{2}$$ = $$2x^{2}$$ + $$16y^{2}$$ + 8$$\sqrt{2}$$xy

Example 3:

Expand the term $$(x + \frac{1}{x})^{2}$$ using the identity $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab

Solution:

$$(x + \frac{1}{x})^{2}$$ = $$x^{2}$$ + $$\frac{1}{x^{2}}$$ + 2$$(x)(\frac{1}{x})$$ = $$x^{2}$$ + $$\frac{1}{x^{2}}$$ + 2

∴ $$(x + \frac{1}{x})^{2}$$ = $$x^{2}$$ + $$\frac{1}{x^{2}}$$ + 2

Explantion:

Step: 1

Prove for n = 1

1 = $$\frac{1}{2}$$n(n + 1)

1 = $$\frac{1}{2}$$.1.(1 + 1)

1 = $$\frac{1}{2}$$.1.2

1 = 1, proved to be true

Step: 2

Prove for n = k

1 + 2 + 3 + …. + K = $$\frac{1}{2}$$ k (k + 1), is true, so for n = k + 1 it is also true

1 + 2+ 3 + … + k + k + 1 = $$\frac{1}{2}$$ (k + 1) ((k+1) + 1)

(1 + 2 + 3 + … + k) + k + 1 = $$\frac{((k + 1) (k + 2))}{2}$$

( $$\frac{1}{2}$$ k (k + 1)) + K + 1 = $$\frac{((k + 1) (k + 2))}{2}$$

$$\frac{((k + 1) (k + 2))}{2}$$ = $$\frac{((k + 1) (k + 2))}{2}$$, proved to be true

Because Step 1 and Step 2 are true, so it can be concluded that the statement is true.

### Formulae

• (Divisor × Quotient) + Remainder = Dividend

• $$(a+b)^2 = a^2 + b^2 +2ab$$

• $$(a-b)^2 = a^2 + b^2 -2ab$$

• $$a^2 – b^2 = (a+b)(a-b)$$

• $$(1+2+3+….+n) = \frac{1}{2n(n+1)}$$

• $$1^2+2^2+3^2+….+n^2 = \frac{1}{6n(n+1)(2n+1)}$$

• $$1^3+2^3+3^2+….+n^3 = \frac{1}{4n^2(n+1)^2}$$

• For Arithmetic Progression
(i) nth term = $$a + (n – 1)d$$
(ii) sum of n terms = $$\frac{n}{2(2a+(n-1)d)}$$
(iii) sum of n terms = $$\frac{n}{2(a+l)}$$, where l is the last term

• For Geometric Progression
(i) nth term = $$ar^{(n-1)}$$
(ii) sum of n terms = $$\frac{a(1-r^n)}{(1-r)}$$ when r<1 ; $$\frac{a(r^n-1)}{(r-1)}$$ when r>1

### Samples

1. Find the unknown number 7429-?=4358-1587

Solution:

Take the unknown value as $$x$$

by substituting $$x$$,

7429 – $$x$$ = 4358 – 1587

7429 – $$x$$ = 2771

$$x$$ = 7429 – 2771

$$x$$ = 4658

Therefore, the unknown value $$x$$ = 4658

2. Face value of 7 and Place value of 9 from the given digit 38745962 is?

Solution:

As, Face value is the value of digit itself and should relate place with ‘place value’

Given digit is 38745962

Therefore, face value of 7 is 7

(100 is multiplied since 9 is in hundred’s place)

Place value of 9 is (9 × 100)=900

3. On dividing 132 by a certain number, 12 as a quotient and 0 as a remainder is obtained. Find the divisor?

Solution:

Given that Dividend = 132, Quotient = 12, Remainder = 0

[Divisor × Quotient] + Remainder=Dividend

by substituting the given values,

[Divisor × 12] + 0 = 132

Divisor = $$\frac{132}{12}$$

∴ Divisor = 11

4. What will be the units digit in the product (234 × 256 × 457 × 952)?

Solution:

Given product is (234 × 256 × 457 × 952)

Here units digits are 4,6,7,2

Therefore, product of Units digit in the given product = 4 × 6 × 7 × 2 = 336.

5. Test which of the following is not a prime number

a)19
b)11
c) 16
d) 13

Solution:

As Prime number is the number which has exactly two factors namely itself

Factors of 19 = 1, 19

Factors of 11 = 1, 11

Factors of 16 = 1, 2, 4, 16

Factors of 13 = 1, 13

Therefore, 16 is not a prime as it is divisible by more than twice numbers.

6. Does the number 23679715 divisible by 11?

Solution:

Given digit is 23679715

(Sum of digits at odd places) – (Sum of digits at even places)

Digits at odd places = 5, 7, 7, 3

Digits at even places = 1, 9, 6, 2

⇒ (5 + 7 + 7 + 3) – (1 + 9 + 6 + 2)

⇒ 22 – 18 = 4

as 4 is not divisible by 11, 23679715 also not divisible by 11