CAT 2025 Lesson : Divisibility - Divisibility - 3, 9, 11

2.3 Divisor is $3$ or $9$

Rule: Remainder when a number is divided by $3$ or $9$ is the remainder of the sum of digits of the number divided by 3 or 9.

Explanation

$10 = 9 + 1$

$9$ is divisible by $3$ and $9$.

$\text{Rem} \left(\dfrac{10^{n}}{3} \right) = \text{Rem} \left(\dfrac{(9 + 1)^{n}}{3} \right) = \text{Rem} \left(\dfrac{1^n}{3} \right) = 1$

$\text{Rem} \left(\dfrac{10^{n}}{9} \right)= \text{Rem} \left(\dfrac{(9 + 1)}{9} \right) = \text{Rem} \left(\dfrac{1^n}{9} \right) = 1$

If $n \ge 0$, when $10^n$ is divided by $3$ or $9$, the remainder is always $1$. The following example shows the extension of this concept for the sum of digits rule.



The above example was to explain the rationale for the sum of digits rule. Going forward, successive digit sums. should be used to obtain the remainder. The following will help to improve your speed.

1) Keep adding the digits till you're left with a single digit number.
2) Remove all $9$s from the digits or sum of digits.

2.4 Divisor is $11$

Rule: Remainder when a number is divided by $11$ equals the remainder when the sum of alternating digits starting from the units' place is subtracted from the sum of the remaining digits and then divided by $11$.

Explanation

$\text{Rem} \left(\dfrac{10^0}{11} \right) = 1 ; \text{Rem} \left(\dfrac{10^1}{11} \right) = 10 ; \text{Rem} \left(\dfrac{10^2}{11} \right) = 1 ; \text{Rem} \left(\dfrac{10^3}{11} \right) = 10$

As noted above, $1$ and $10$ are the remainders when even and odd powers of $10$ are divided by $11$ respectively.

As any number of divisors can be subtracted or added to a remainder, remainder for odd powers of $10$ can be written as $10 - 11 = -1$

∴ $\text{Rem} \left(\dfrac{10^n}{11} \right) = 1$ if $n$ is even and -1 if $n$ is odd.