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 $2$ of $10^{4}, 5$ of $10^{3}, 6$ of $10^{2}, 7$ of $10^{1}$ and $8$ of $10^{0}$.

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

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
$5305680 = (5380 \times 10^{4}) + (5680 \times 10^{0})$ or $5305680 = (5 \times 10^{6}) + (305 \times 10^{3}) + (680 \times 10^{0})$

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

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

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 $2^{n}$ or $5^{n}$equals the remainder when the last n digits of the number is divided by $\bm{2^{n}}$ or $\bm{5^{n}}$.

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 $2^{n}$ or $5^{n}$ 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 $2$, $4$, $6$, $8$ or $0$.
2) $5$ only if it's last digit is $5$ or $0$.
3) $25$ only if it's last $2$ digits are $00$, $25$, $50$ or $75$.

$\bm{2.2}$ 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.