+91 9600 121 800

Plans

Dashboard

Daily & Speed

Quant

Verbal

DILR

Compete

Free Stuff

calendarBack
Quant

/

Numbers

/

Divisibility

Divisibility

MODULES

bookmarked
Divisibility - 2, 5, 10
bookmarked
Divisibility - 3, 9, 11
bookmarked
7, 13 and Composite
bookmarked
Divisibility of Factorial
bookmarked
Last Digit
bookmarked
Last 2 Digits
bookmarked
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Divisibility 1
-/10
Divisibility 2
-/10

PRACTICE

Divisibility : Level 1
Divisibility : Level 2
Divisibility : Level 3
ALL MODULES

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

bookmarked

2.3 Divisor is 333 or 999

Rule: Remainder when a number is divided by 333 or 999 is the remainder of the sum of digits of the number divided by 3 or 9.

Explanation

10=9+110 = 9 + 110=9+1

999 is divisible by 333 and 999.

Rem(10n3)=Rem((9+1)n3)=Rem(1n3)=1\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) = 1Rem(310n​)=Rem(3(9+1)n​)=Rem(31n​)=1

Rem(10n9)=Rem((9+1)9)=Rem(1n9)=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) = 1Rem(910n​)=Rem(9(9+1)​)=Rem(91n​)=1

If n≥0n \ge 0n≥0, when 10n10^n10n is divided by 333 or 999, the remainder is always 111. The following example shows the extension of this concept for the sum of digits rule.

Example 4

What is the remainder when 345853458534585 is divided by 333?

Solution

34585=(3×104)+(4×103)+(5×102)+(8×101)+(5×100)34585 = (3 \times 10^4) + (4 \times 10^3) + (5 \times 10^2) + (8 \times 10^1) + (5 \times 10^0)34585=(3×104)+(4×103)+(5×102)+(8×101)+(5×100)

Let the remainder be xxx.

x=Rem(345853)x = \text{Rem} \left(\dfrac{34585}{3} \right)x=Rem(334585​)

=Rem((3×104)+(4×103)+(5×102)+(8×101)+(5×100)3) = \text{Rem} \left(\dfrac{(3 \times 10^4) + (4 \times 10^3) + (5 \times 10^2)+ (8 \times 10^1) + (5 \times 10^0)}{3} \right)=Rem(3(3×104)+(4×103)+(5×102)+(8×101)+(5×100)​)

As Rem(10n3)=1\text{Rem} \left(\dfrac{10^n}{3} \right) = 1Rem(310n​)=1,

x=Rem((3×1)+(4×1)(5×1)(8×1)(5×1)3)x = \text{Rem} \left(\dfrac{(3 \times 1) + (4 \times 1) (5 \times 1) (8 \times 1) (5 \times 1)}{3}\right)x=Rem(3(3×1)+(4×1)(5×1)(8×1)(5×1)​)

=Rem(3+4+5+8+53)=Rem(253)=Rem(2+53)= \text{Rem} \left(\dfrac{3 + 4 + 5 + 8 + 5}{3} \right) = \text{Rem} \left(\dfrac{25}{3} \right)= \text{Rem} \left(\dfrac{2 + 5}{3} \right)=Rem(33+4+5+8+5​)=Rem(325​)=Rem(32+5​)

=Rem(73)=1= \text{Rem}\left(\dfrac{7}{3} \right)= 1 =Rem(37​)=1

Answer: 111


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 999s from the digits or sum of digits.

Example 5

What is the remainder when 456783494567834945678349 is divided by 999?

Solution

We can add 222 digits at a time and then add the digits of the results. This is a relatively faster way to find the remainder.

(4+5)+(6+7)+(8+3)+4+(9)(4 + 5) + (6 + 7) + (8 + 3) + 4 + (9)(4+5)+(6+7)+(8+3)+4+(9)

=9+13+11+4+9= 9 + 13 + 11 + 4 + 9=9+13+11+4+9

'999's can be removed and rest of the digits can be added again.

1+3+1+1+4=101 + 3 + 1 + 1 + 4 = 101+3+1+1+4=10

Once again we apply digit sum.

1+0=11 + 0 = 11+0=1

Answer: 111

2.4 Divisor is 111111

Rule: Remainder when a number is divided by 111111 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 111111.

Explanation

Rem(10011)=1;Rem(10111)=10;Rem(10211)=1;Rem(10311)=10\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) = 10Rem(11100​)=1;Rem(11101​)=10;Rem(11102​)=1;Rem(11103​)=10

As noted above, 111 and 101010 are the remainders when even and odd powers of 101010 are divided by 111111 respectively.

As any number of divisors can be subtracted or added to a remainder, remainder for odd powers of 101010 can be written as 10−11=−110 - 11 = -110−11=−1

∴ Rem(10n11)=1\text{Rem} \left(\dfrac{10^n}{11} \right) = 1Rem(1110n​)=1 if nnn is even and -1 if nnn is odd.

Example 6

What is the remainder when 942659426594265 is divided by 111111?

Solution

94265=(9×104)+(4×103)+(2×102)+(6×101)+(5×100)94265 = (9 \times 10^{4}) + (4 \times 10^{3}) + (2 \times 10^{2}) + (6 \times 10^{1}) + (5 \times 10^{0})94265=(9×104)+(4×103)+(2×102)+(6×101)+(5×100)

Let the remainder be xxx.

x=Rem(9426511)x = \text{Rem} \left(\dfrac{94265}{11} \right) x=Rem(1194265​)

=Rem((9×104)+(4×103)+(2×102)+(6×101)+(5×100)11)= \text{Rem} \left(\dfrac{(9 \times 10^4) + (4 \times 10^3) + (2 \times 10^2) + (6 \times 10^1) + (5 \times 10^0)}{11} \right)=Rem(11(9×104)+(4×103)+(2×102)+(6×101)+(5×100)​)

As Rem(10n11)=1\text{Rem} \left(\dfrac{10^n}{11} \right) = 1Rem(1110n​)=1 if nnn is even and -1 if nnn is odd. ,

x=Rem((9×104)+(4×−1)+(2×1)+(6×−1)+(5×1)3)x = \text{Rem} \left(\dfrac{(9 \times 10^4) + (4 \times -1) + (2 \times 1) + (6 \times -1) + (5 \times 1)}{3} \right)x=Rem(3(9×104)+(4×−1)+(2×1)+(6×−1)+(5×1)​)

=Rem(9−4+2−6+511)=Rem(611)=6= \text{Rem} \left(\dfrac{9 - 4 + 2 - 6 + 5}{11} \right)= \text{Rem} \left(\dfrac{6}{11} \right) = 6=Rem(119−4+2−6+5​)=Rem(116​)=6

Alternatively

The rule can directly be applied. Remainder when 942659426594265 is divided by 111111,

=(5+2+9)−(6+4)=6= (5 + 2 + 9) - (6 + 4) = 6=(5+2+9)−(6+4)=6

Answer: 666
Loading...Loading Video....