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Divisibility

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CAT 2025 Lesson : Divisibility - Divisibility - 2, 5, 10

We studied divisibility rules in secondary school. In this lesson, we will examine the basis behind these rules. This involves application of two vital concepts from Number Theory and Factors and Remainders lessons. These concepts are restated below. We will also look at concepts related to last digit and factorial.

1. Revisiting Basics

1.1 Decimal Number System

Numbers are used to measure and count. The number system used globally is the decimal number system, where the base is

10

. This just means that there are

10

digits in this number system. They are

0

1

2

3

4

5

6

7

8

9

.

In this number system, the placement of digits in a number creates the value for that number. In this lesson, we will be working with positive integers (or natural numbers) only.

From right to left, the places before the decimal points are units' place, tens' place, hundreds' place, thousands' place, etc. Each of these places have a certain value attached to it. They start with

10^{0}

and increase with increments of

1

in the power. The table below shows till the seventh digit from the right.

Position (Right to Left)	Value
Units' Place	$10^{0}$
Tens' Place	$10^{1}$
Hundreds' Place	$10^{2}$
Thousands' Place	$10^{3}$
Ten Thousands' Place	$10^{4}$
Hundred Thousands' Place	$10^{5}$
Millions' Place	$10^{6}$

∴ The number

25,678

means there are

2

10^{4}, 5

10^{3}, 6

10^{2}, 7

10^{1}

and

8

10^{0}

25678 = (2 \times 10^{4}) + (5 \times 10^{3}) + (6 \times 10^{2}) + (7 \times 10^{1}) + (8 \times 10^{0})

This forms the value of every number in the decimal system. The following is an example for a relatively larger number.

5305680 = (5 \times 10^{6}) + (3 \times 10^{5}) + (0 \times 10^{4}) + (5 \times 10^{3}) + (6 \times 10^{2}) + (8 \times 10^{1}) + (0 \times 10^{0})

We can also merge some of the places. For instance, the above number can also be expressed as

5305680 = (5380 \times 10^{4}) + (5680 \times 10^{0})

5305680 = (5 \times 10^{6}) + (305 \times 10^{3}) + (680 \times 10^{0})

1.2 Remainder rules

Sum Rule: Remainder of Sum of numbers = Sum of remainders when each of the numbers is divided by the divisor
When

n = a_{1} + a_{2} + a_{3} + ...,

\text{Rem} \left(\dfrac{n}{d} \right)

= \text{Rem} \left(\dfrac{a_{1} + a_{2} + a_{3} + ...}{d} \right)

= \text{Rem} \left(\dfrac{a_{1}}{d} \right)

+ \text{Rem} \left(\dfrac{a_{2}}{d} \right)

+ \text{Rem} \left(\dfrac{a_{3}}{d} \right) + ...

Product Rule: Remainder of product of numbers = Product of remainders when each of the numbers is divided by the divisor

When

n = a_{1} \times a_{2} \times a_{3} \times ...,

\text{Rem} \left(\dfrac{n}{d} \right)

= \text{Rem} \left(\dfrac{a_{1} \times a_{2} \times a_{3} \times ...}{d} \right)

= \text{Rem} \left(\dfrac{a_{1}}{d} \right)

\times \text{Rem} \left(\dfrac{a_{2}}{d} \right)

\times \text{Rem} \left(\dfrac{a_{3}}{d} \right) \times ...

Example 1

What is the remainder when

34585

is divided by

4

Solution

34585 = (3 \times 10^{4})

+ (4 \times 10^{3})

+ (5 \times 10^{2})

+ (8 \times 10^{1})

+ (5 \times 10^{0})

\text{Rem} \left(\dfrac{34585}{4} \right)

= \text{Rem} \left(\dfrac{(3 \times 10^{4}) + (4 \times 10^{3}) + (5 \times 10^{2}) + (8 \times 10^{1}) + (5 \times 10^{0})}{4} \right)

Applying sum rule, we write this as

= \text{Rem}\left(\dfrac{3 \times 10^{4}}{4} \right)

+ \text{Rem} \left(\dfrac{4 \times 10^{3}}{4} \right)

+ \text{Rem} \left(\dfrac{5 \times 10^{2}}{4} \right)

+ \text{Rem} \left(\dfrac{8 \times 10^{1}}{4} \right)

+ \text{Rem} \left(\dfrac{5 \times 10^{0}}{4} \right)

n \ge 2, 10^{n}

will perfectly divide

2^{2}

and leave a remainder of

0

. ∴ Applying Product rule,

= \text{Rem}\left(\dfrac{3 \times 0}{4} \right)

+ \text{Rem} \left(\dfrac{4 \times 0}{4} \right)

+ \text{Rem} \left(\dfrac{5 \times 0}{4} \right)

+ \text{Rem} \left(\dfrac{8 \times 10^{1}}{4} \right)

+ \text{Rem} \left(\dfrac{5 \times 10^{0}}{4} \right)

= \text{Rem} \left(\dfrac{80}{4}\right)

+ \text{Rem} \left(\dfrac{5}{4}\right)

= 0 + 1 = 1

Answer:

1

Note: This is the logic behind looking at the last two digits for divisibility by

4

2. Divisibility Rules / Remainder Rules

This section will provide the shortcuts to calculate the remainder for division by certain integers. A number is perfectly divisible by another, if the remainder equals

0

.

∴ The same rules apply for finding the remainder and for checking divisibility.

2.1 Divisor is $2^{n}$ and $5^{n}$

Rule: Remainder when a number is divided by

2^{n}

5^{n}

equals the remainder when the last n digits of the number is divided by

\bm{2^{n}}

\bm{5^{n}}

.

Explanation

10 = 2 \times 5

. Likewise,

10^{n} = 2^{n} \times 5^{n}

Place value of

(n + 1)^{\text{th}}

digit from the right in a number is

10^{n}

, which would perfectly divide

2^{n}

and

5^{n}

. Digits to the left of it will have higher place values and perfectly divide

2^{n}

and

5^{n}

.

∴ Remainder of a number, when divided by

2^{n}

5^{n}

equals the remainder left by the number's last

n

digits.

So, if we divide a number by

125 \boldsymbol{(= 5^{3})}

, all place values that are

10^{3}

and more perfectly divide

125

\therefore

We look at the remainder for the last

\bm{3}

digits only, whose place values

10^2

10^1

and

10^0

do not perfectly divide

125

.

The following table provides the divisibility/remainder pattern for divisibility upto

2^{5}

and

5^{5}

Remainder when n is divided by x
x	Remainder Rule
$2$	Last digit of $n$ divided by 2
$4$	Last $2$ digits of $n$ divided by $4$
$8$	Last $3$ digits of $n$ divided by $8$
$16$	Last $4$ digits of $n$ divided by $16$
$32$	Last $5$ digits of $n$ divided by $32$
$5$	Last digit of $n$ divided by 5
$25$	Last $2$ digits of $n$ divided by $25$
125	Last $3$ digits of $n$ divided by 125
$625$	Last $4$ digits of $n$ divided by $25$

Additionally, a number divides
1)

2

only if it's last digit is

2

4

6

8

0

.
2)

5

only if it's last digit is

5

0

.
3)

25

only if it's last

2

digits are

00

25

50

75

Example 2

67485923576

is divisible by

(1)

4

but not

8

16

and

32

(2)

4

and

8

but not

16

and

32

(3)

4

8

and

16

but not

32

(4)

4

8

16

and

32

Solution

\text{Rem} \left(\dfrac{76}{4} = 0\right) ; \text{Rem} \left(\dfrac{576}{8} = 0 \right) ; \text{Rem} \left(\dfrac{3576}{16} = 8 \right)

As the number is not divisible by

16

, it will not be divisible by

32

.

∴ Number is divisible by

4

and

8

only.

Answer: (2)

4

and

8

but not

16

and

32

$\bm{2.2}$ Divisor is $\bm{10^{n}}$

Rule: Remainder when a number is divided by

\bm{10^{n}}

is the same as the last n digits of the number.

Likewise, if a number is divisible by

10^{n}

, then the last

n

digits will be zeroes.

Example 3

Radha, Geeta, Ram and Lakhan donated

41373

92324, 43288

and

13245

gold coins to a trust. The trust decided to distribute an equal number of these gold coins to each of their

100

beneficiaries. What is the minimum number of gold coins left after such distribution?

Solution

Minimum number of gold coins left will be the remainder when the total gold coins are divided by

100

\text{Rem} \left(\dfrac{41373 + 92324 + 43288 + 13245}{100} \right)

= \text{Rem}\left(\dfrac{73 + 24 + 88 + 45}{100} \right) = \text{Rem} \left(\dfrac{230}{100} \right) = 30

Answer:

30

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CAT 2025 Lesson : Divisibility - Divisibility - 2, 5, 10

1. Revisiting Basics

1.1 Decimal Number System

1.2 Remainder rules

Example 1

Solution

2. Divisibility Rules / Remainder Rules

2.1 Divisor is 2n2^{n}2n and 5n5^{n}5n

Example 2

Solution

2.2\bm{2.2}2.2 Divisor is 10n\bm{10^{n}}10n

Example 3

Solution

2.1 Divisor is $2^{n}$ and $5^{n}$

$\bm{2.2}$ Divisor is $\bm{10^{n}}$