CAT 2025 Lesson : Divisibility - Concepts & Cheatsheet

1) List of divisibility rules

Number	Divisibility Rule
$2$	Last digit divisible by $2$
$3$	Sum of digits divisible by $3$
$4$	Last two digits divisible by $4$
$5$	Last digit is $0$ or $5$
$6$	Divisible by $2$ and $3$
$7$	Difference between sum of alternate sets of digits is divisible by $7$
$8$	Last three digits divisible by $8$
$9$	Sum of digits divisible by $9$
$10$	Last digit is $0$
$11$	Difference between sum of alternate sets of digits is divisible by $11$
$12$	Divisible by $3$ and $4$
$13$	Difference between sum of alternate sets of digits is divisible by $13$
$14$	Divisible by $2$ and $7$
$15$	Divisible by $3$ and $5$
$16$	Last $4$ digits divisible by $16$
$18$	Divisible by $2$ and $9$
$20$	Divisible by $4$ and $5$
$24$	Divisible by $3$ and $8$
$25$	Last $2$ digits divisble by $25$
$32$	Last 5 digits divisible by $32$
$100$	Last $2$ digits are $0$
$125$	Last $3$ digits divisible by $125$

2) For all composite numbers,
Divisibility Rule: Check for divisibility by each of the prime factors raised to their respective powers.
Remainder Rule: Apply Chinese Remainder theorem.

3) Where $p$ is a prime number, greatest power of p that divides n! is the sum of quotients when $n$ is successively divided by $p$.

4) Where $a$, $b$ and $c$ are prime numbers and $x = a^{p}b^{q}c^{r}$, to find the greatest power of x that divides n!,
(a) find the largest powers for each of $a$, $b$ and $c$ that can divide $n!$ ;
(b) divide these respective powers by $p$, $q$ and $r$ and write down the quotients.
(c) The least quotient is the highest power of $x$ that can perfectly divide $n!$.

5) To find the last $n$ digits in the product of certain numbers, we can simply multiply the last $n$ digits of each of these numbers.

6) Cyclicity of units digit is
(a) $\bm{1}$ for $0$, $1$, $5$, $6$
(b) $\bm{2}$ for $4$, $9$
(c) $\bm{4}$ for $2$, $3$, $7$, $8$

$7$) Last $2$ digits of $x^{n}$ when

$x$ is a	Last $2$ digits
number that ends in $5$	If power is even, then $25$. If power is odd, then $25$ if tens digit of base is even and $75$ if tens digit of base is odd.
number that ends in $0$	Last $2$ digits are $00$
multiple of $2$ but not $4$	Last $2$ digits of $x^{40k + 1} = x + 50$
multiple of $4$	Last $2$ digits of $x^{40k + 1} = x$
number that ends in $1$	Last digit is $1$, tens digit is $U(\text{tens digit of} x \times \text{units digit of} n)$
number that ends in $3$, $7$ or $9$	Raise it by a power so that the number ends in $1$. Then, apply the above rule for 'number that ends in $1$'.