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

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2.3 Divisor is 33 or 99

Rule: Remainder when a number is divided by
33 or 99 is the remainder of the sum of digits of the number divided by 3 or 9.

Explanation

10=9+110 = 9 + 1

99 is divisible by 33 and 99.

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) = 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) = 1

If
n0n \ge 0, when 10n10^n is divided by 33 or 99, the remainder is always 11. The following example shows the extension of this concept for the sum of digits rule.

Example 4

What is the remainder when 3458534585 is divided by 33?

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)

Let the remainder be
xx.

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

=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)

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

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

Answer:
11


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
99s from the digits or sum of digits.

Example 5

What is the remainder when 4567834945678349 is divided by 99?

Solution

We can add 22 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)

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

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

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

Once again we apply digit sum.

1+0=11 + 0 = 1

Answer:
11

2.4 Divisor is 1111

Rule: Remainder when a number is divided by
1111 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 1111.

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) = 10

As noted above,
11 and 1010 are the remainders when even and odd powers of 1010 are divided by 1111 respectively.

As any number of divisors can be subtracted or added to a remainder, remainder for odd powers of
1010 can be written as 1011=110 - 11 = -1

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

Example 6

What is the remainder when 9426594265 is divided by 1111?

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})

Let the remainder be
xx.

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

=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)

As
Rem(10n11)=1\text{Rem} \left(\dfrac{10^n}{11} \right) = 1 if nn is even and -1 if nn 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)

=Rem(94+26+511)=Rem(611)=6= \text{Rem} \left(\dfrac{9 - 4 + 2 - 6 + 5}{11} \right)= \text{Rem} \left(\dfrac{6}{11} \right) = 6

Alternatively

The rule can directly be applied. Remainder when
9426594265 is divided by 1111,

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

Answer:
66
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