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CAT 2025 Lesson : Factors & Remainders - Operations on Remainders

3.5 Sum of Remainders

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 + ...,

\mathrm{Rem} \left(\dfrac{n}{d}\right) = \mathrm{Rem} \left(\dfrac{a_{1} + a_{2} + a_{3} +...}{d}\right) = \mathrm{Rem} \left(\dfrac{a_{1}}{d}\right) + \mathrm{Rem} \left(\dfrac{a_{2}}{d}\right) + \mathrm{Rem} \left(\dfrac{a_{3}}{d}\right) + ...

If the sum of remainders is greater than

d

, this is once again divided by

d

to ascertain the final remainder.

Example 19

What is the remainder when

(3486 + 43975 + 245)

is divided by

4

Solution

\mathrm{Rem} \left(\dfrac{3486 + 43975 + 245}{4}\right) = \mathrm{Rem} \left(\dfrac{3486}{4}\right) + \mathrm{Rem} \left(\dfrac{43975}{4}\right) + \mathrm{Rem} \left(\dfrac{245}{4}\right)

= \mathrm{Rem} \left( \dfrac{2 + 3 + 1}{4} \right) = 2

Answer:

2

Note: To find the remainder when a number is divided by

4

, we divide the number's last

2

digits by

4

. This concept will be explained in greater detail in the Divisibility Lesson.

3.6 Product of Remainders

Rule: Remainder of the 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 ...

\mathrm{Rem} \left( \dfrac{n}{d} \right) = \mathrm{Rem} \left( \dfrac{a_1 \times a_2 \times a_3 \times...}{d} \right) = \mathrm{Rem} \left( \dfrac{a_1}{d} \right) \times \mathrm{Rem} \left( \dfrac{a_2}{d} \right) \times \mathrm{Rem} \left( \dfrac{a_3}{d} \right) \times ...

Explanation: Let's say there are three numbers

a_1

a_2

and

a_3

. The remainders obtained when each of

a_1

a_2

and

a_3

is divided by divisor

d

, are

r_1

r_2

and

r_3

respectively. And, the quotients are

q_1

q_2

and

q _3

respectively.

a_1 = dq_1 + r_1

;

a_2 = dq_2 + r_2

;

a_3 = dq_3 + r_3

\mathrm{Rem} \left( \dfrac{a_1 \times a_2 \times a_3}{d} \right) = \mathrm{Rem} \left( \dfrac{(dq_1 + r_1)(dq_2 + r_2)(dq_3 + r_3) }{d} \right)

In the expansion, all the terms in the numerator is a multiple of

d

, but for the last term, which is

r_1 \times r_2 \times r_3

\therefore

\mathrm{Rem} \left(\dfrac{a_1 \times a_2 \times a_3 }{d}\right) = \mathrm{Rem} \left(\dfrac{r_1 \times r_2 \times r_3}{d}\right)

Example 20

What is the remainder when

235 \times 237 \times 239

is divided by

9

Solution

The closest multiple of

9

for the three numbers is

234

\mathrm{Rem} \left(\dfrac{235 \times 237 \times 239}{9}\right) = \mathrm{Rem} \left(\dfrac{(234 + 1) \times (234 + 3) \times (234 + 5)}{9}\right) = \mathrm{Rem} \left(\dfrac{1 \times 3 \times 5}{9}\right) = 6

Answer:

6

3.7 Remainder of Exponents

(a + b)^n = ^{n}{c_0 a^n b^0} + ^{n}{c_1 a^{n - 1} b^1} + ... +

^{n}{c_{n-1} a^1 b^{n - 1}} +

^{n}{c_n a^0 b^n}

Let's say a number

a

when divided by

d

, results in a quotient of

q

and remainder of

r

a = dq + r

\mathrm{Rem} \left(\dfrac{a^n}{d}\right) = \mathrm{Rem} \left(\dfrac{(dq + r)^n}{d}\right)

= \mathrm{Rem} \left(\dfrac{^{n}c_0(dq)^n r^0 + ^{n}c_1(dq)^{n - 1}r^1 + .... + ^{n}c_{n - 1}(dq)^1 r^{n - 1} + ^{n}c_n (dq)^0 r^n}{d}\right)

All terms are divisible by

d

except for the last term.

\therefore

\mathrm{Rem} \left(\dfrac{(dq + r)^n}{d}\right) = \mathrm{Rem} \left(\dfrac{r^n}{d}\right)

Example 21

What is the remainder when

70^{795}

is divided by

23

Solution

69

is a multiple of

23

\therefore

\mathrm{Rem} \left(\dfrac{70^{795}}{23}\right) = \mathrm{Rem} \left(\dfrac{(69 + 1)^{795}}{23}\right) = \mathrm{Rem} \left(\dfrac{ 1^{795}}{23}\right) = 1

Answer:

1

The following example uses a combination of the rules.

Example 22

What is the remainder when

15^{125}

is divided by

8

Solution

16

is a multiple of

8

\therefore

\mathrm{Rem} \left(\dfrac{15^{125}}{8}\right) = \mathrm{Rem} \left(\dfrac{(16 - 1)^{125}}{8}\right)

\mathrm{Rem} \left(\dfrac{(- 1)^{125}}{8}\right) = \mathrm{Rem} \left(\dfrac{(- 1 + 8)}{8}\right) = 7

Answer:

7

Note: Adding or subtracting the divisor will not change the remainder. Therefore, while applying the rules if we get a negative value (like in this example), we can add a multiple of the divisor to arrive at the remainder.

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