Back

Factors & Remainders

ALL MODULES

CAT 2025 Lesson : Factors & Remainders - Special Types

6. Special Types

Following are some of the common types of questions asked in the entrance tests.

6.1 Difference of squares of positive integers

1) Where

\bm{n}

is an odd natural number and

x

is the number of factors of

n

, then the number of ways in which

n

can be expressed as the difference of squares of positive integers is

\dfrac{x}{2}

x

is even and

\dfrac{x - 1}{2}

x

is odd.

2

) Where

\bm{n}

is divisible by 4 and

x

is the number of factors of

\dfrac{n}{4}

, then the number of ways in which

n

can be expressed as the difference of squares of positive integers is

\dfrac{x}{2}

x

is even and

\dfrac{x - 1}{2}

x

is odd.

3) Where

\bm{n}

is an even number not divisible by 4, it can never be written as the difference of squares of positive integers.

Explanation

ab = \left(\dfrac{a + b}{2}\right)^2 - \left(\dfrac{a - b}{2}\right)^2

(

a + b

) and (

a - b

) have to be even numbers so that they are divisible by

2

. This requires both

a

and

b

to be even or both to be odd.

If

\bm{n}

is odd, we find the number of ways

\bm{n}

can be written as a product of two numbers.

If

\bm{n}

is even, we have to ensure that it is written as product of

2

even numbers.

As

a

and

b

have to be multiples of

2

, we find the number of ways

\bm{\dfrac{n}{4}}

can be written as a product of two numbers.

Note that a number such as

18 = 2^1 \times 3^2

, can never be expressed in this form. (As the power of

2

1

, one of

a

and

b

will be even, while the other will be odd)

\therefore

Integers of the form

\bm{4k} + 2

can never be expressed as difference of squares.

All other numbers, including prime numbers can be expressed in the form of difference of squares.

Example 27

In how many different ways can

48

be expressed as the difference of squares of two positive integers?

Solution

48 = 2^4 \times 3

We assign one

2

each to

a

and

b

to make both of them even.

Therefore, we are left with

2^2 \times 3

to be split across the two numbers.

Number of ways

2^2 \times 3

can be written as a product of

2

numbers

= \dfrac{\mathrm{Number \ of \ factors}}{2}

= \dfrac{(2 + 1)(1 + 1)}{2} = 3

Answer:

3

6.2 Sum of 2 or more consecutive natural numbers

Rule: Number of ways in which

x

can be expressed as the sum of

2

or more consecutive natural numbers is the number of odd factors of $\bm{x}$ other than 1.

Explanation
Sum to

n

terms of Arithmetic Progression

= \dfrac{n}{2}[2a + (n - 1)d]

n

is the number of terms,

a

is the first term and

d

is the common difference between two consecutive terms.

Let the number

x

be expressed as the sum of consecutive natural numbers. In the case of consecutive natural numbers,

d = 1

. We do not know

n

a

\dfrac{n}{2}[2a + (n - 1)1] = x

⇒

n(2a + n - 1) = x

n

is an even number, then (

2a + n - 1

) is an odd number and vice–versa.

\therefore

For every odd factor of a number (other than 1), there is 1 way in which the number can be expressed as the sum of consecutive natural numbers.

The numbers which cannot be expressed in this form are those without odd factors, which are of the sequence

2, 2^2, 2^3, 2^4

, ...

Example 28

In how many ways can you write

120

as the sum of

2

or more consecutive natural numbers?

Solution

120 = 2^3 \times 3 \times 5

No. of odd factors of

120 =

No. of factors for

3 \times 5 = (1 + 1) \times (1 + 1) = 4

As odd numbers other than

1

should be considered, there are

4 - 1 = 3

ways in which

120

can be written as the sum of

2

or more consecutive natural numbers.

Answer:

3

6.3 Remainder Theorem for Algebraic Expressions

Where

a

is a constant, the remainder when a polynomial

f(x)

is divided by (

x - a

), is

\bm{f(a)}

Example 29

What is the remainder when

(x)^4 - 3(x)^3 + 2(x)^2 - 8

is divided by

x + 3

Solution

Let

f(x) = (x)^4 - 3(x)^3 + 2(x)^2 - 8

x + 3 = x - k

, then

k = - 3

f(- 3) = (- 3)^4 - 3(- 3)^3 + 2(- 3)^2 - 8

= 81 + 81 + 18 - 8

= 172

Answer:

172

Example 30

(x)^3 + 9(x)^2 - 8x - 2

when divided by

(x - 2)

, leaves a remainder of

0

, then a possible value of

x

is

(1)

12

(2)

13

(3)

15

(4)

16

Solution

Substituting the values, even with a calculator, will be time consuming.

Let

f(x)

(x)^3 + 9(x)^2 - 8x - 2

f(x)

when divided by

(x - 2)

, leaves a remainder of

f(2) = 8 + 9 \times 4 - 8 \times 2 - 2 = 26

\therefore

Where Q is the quotient,

f(x) = Q \times (x - 2) + 26

If (

x - 2

) is a factor of

26

, then (

x - 2

) will perfectly divide

f(x)

.

If

x - 2 = 13

⇒

x = \bm{15}

Answer: (3)

15

6.4 Wilson's Theorem

Where

\bm{n}

is a prime number,

\mathrm{Rem} \left(\dfrac{(n - 2)!}{n}\right) = 1

and

\mathrm{Rem} \left(\dfrac{(n - 1)!}{n}\right) = n - 1

.

(Note: The second equation above can be derived from the first by applying remainder theorem)

For instance, where

\bm{n}

\bm{29}

\mathrm{Rem} \left(\dfrac{27!}{29}\right) = 1

and

\mathrm{Rem} \left(\dfrac{28!}{29}\right) = 28

Additionally, note that where

n

is a composite number and

n \neq 4

\mathrm{Rem} \left(\dfrac{(n - 2)!}{n}\right) = \mathrm{Rem} \left(\dfrac{(n - 1)!}{n}\right) = 0

.

For instance, where

\bm{n = 28}

\mathrm{Rem} \left(\dfrac{27!}{28}\right) = \mathrm{Rem} \left(\dfrac{26!}{28}\right) = 0

Where

\bm{n = 4}

\mathrm{Rem} \left(\dfrac{2!}{4}\right) = \mathrm{Rem} \left(\dfrac{3!}{4}\right) = 2

The following example is of very high difficulty and is unlikely to appear in entrance tests. However, the concepts applied here will enhance your understanding of remainders which would help ace a test.

Example 31

What is the remainder when

25!

is divided by

29

Solution

Let

\mathrm{Rem} \left(\dfrac{25!}{29}\right) = x

Applying Wilson's theorem we get

\mathrm{Rem} \left(\dfrac{27!}{29}\right) = 1

⇒

\mathrm{Rem} \left(\dfrac{(27 \times 26 \times 25!)}{29}\right) = 1

⇒

\mathrm{Rem} \left(\dfrac{(29 - 2) \times (29 - 3) \times x}{29}\right) = 1

⇒

\mathrm{Rem} \left(\dfrac{(- 2) \times (- 3) \times x}{29}\right) = 1

⇒

\mathrm{Rem} \left(\dfrac{6x}{29}\right) = 1

Where

k

is a positive integer,

6x

can be expressed as

6x = 29 k + 1

⇒

x = \dfrac{(29k + 1)}{6}

The smallest integral solution is when

k = 1

and

x = 5

\therefore

The remainder is 5.

Answer:

5

6.5 Remainder when terms are in GP

This is a specific type of question where we need to express the divisor as sum of terms in GP in order to find the remainder.

Example 32

What is the remainder when

2 + 2^{2} + 2^{3} + ... + 2^{50}

is divided by

7

Solution

7 = 1 + 2 + 4 = 2^{0} + 2^{1} + 2^{2}

There are a total of

50

terms in

2 + 2^{2} + 2^{3} + ... + 2^{50}

Sum of any three consecutive terms will be divisible by

( 2^{0} + 2^{1} + 2^{2} )

and will therefore leave a remainder of

0

.

For instance,

\mathrm{Rem} \left(\dfrac{2^{35} + 2^{36} +2^{37}}{2^{0} + 2^{1} + 2^{2}}\right) =

\mathrm{Rem} \left(\dfrac{2^{35}(2^{0} + 2^{1} +2^{2}}{2^{0} + 2^{1} + 2^{2}}\right) = 0

As it is easier to work with the smaller terms, the sum of the last

48

terms (

16

sets of

3

terms each) will leave a remainder of

0

. And, we will be left with just the first

2

terms.

\mathrm{Rem} \left(\dfrac{2^{1} + 2^{2} + 2^{3} + ... + 2^{50}}{2^{0} + 2^{1} + 2^{2}}\right)

= \mathrm{Rem} \left(\dfrac{2^{1} + 2^{2} + 2^{3}(2^{0} + 2^{1} + 2^{2}) + 2^{6}(2^{0} + 2^{1} + 2^{2}) + ... + 2^{48}(2^{0} + 2^{1} + 2^{2})}{2^{0} + 2^{1} + 2^{2}}\right)

= \mathrm{Rem} \left(\dfrac{2^{1} + 2^{2}}{2^{0} + 2^{1} + 2^{2}}\right) = \mathrm{Rem} \left(\dfrac{6}{7}\right) = 6

Answer:

6

Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now