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Factors & Remainders

Factors And Remainders

MODULES

HCF & LCM
Number of Factors
HCF with Remainders
LCM with Remainders
Operations on Remainders
Euler & Fermat
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ALL MODULES

CAT 2025 Lesson : Factors & Remainders - Operations on Remainders

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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=a1+a2+a3+...,n = a_1 + a_2 + a_3 + ...,n=a1​+a2​+a3​+...,

Rem(nd)=Rem(a1+a2+a3+...d)=Rem(a1d)+Rem(a2d)+Rem(a3d)+...\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) + ...Rem(dn​)=Rem(da1​+a2​+a3​+...​)=Rem(da1​​)+Rem(da2​​)+Rem(da3​​)+...

If the sum of remainders is greater than
ddd, this is once again divided by ddd to ascertain the final remainder.

Example 19

What is the remainder when (3486+43975+245)(3486 + 43975 + 245)(3486+43975+245) is divided by 444?

Solution

Rem(3486+43975+2454)=Rem(34864)+Rem(439754)+Rem(2454)\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)Rem(43486+43975+245​)=Rem(43486​)+Rem(443975​)+Rem(4245​)

=Rem(2+3+14)=2= \mathrm{Rem} \left( \dfrac{2 + 3 + 1}{4} \right) = 2=Rem(42+3+1​)=2

Answer:
222

Note: To find the remainder when a number is divided by
444, we divide the number's last 222 digits by 444. 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=a1×a2×a3×... n = a_1 \times a_2 \times a_3 \times ...n=a1​×a2​×a3​×...,
Rem(nd)=Rem(a1×a2×a3×...d)=Rem(a1d)×Rem(a2d)×Rem(a3d)×... \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 ...Rem(dn​)=Rem(da1​×a2​×a3​×...​)=Rem(da1​​)×Rem(da2​​)×Rem(da3​​)×...

Explanation: Let's say there are three numbers
a1a_1a1​, a2a_2a2​ and a3a_3a3​. The remainders obtained when each of a1a_1a1​, a2a_2a2​ and a3a_3a3​ is divided by divisor ddd, are r1r_1r1​, r2r_2r2​ and r3r_3r3​ respectively. And, the quotients are q1q_1q1​, q2q_2q2​ and q3q _3q3​ respectively.

a1=dq1+r1 a_1 = dq_1 + r_1 a1​=dq1​+r1​; a2=dq2+r2 a_2 = dq_2 + r_2a2​=dq2​+r2​; a3=dq3+r3a_3 = dq_3 + r_3a3​=dq3​+r3​

Rem(a1×a2×a3d)=Rem((dq1+r1)(dq2+r2)(dq3+r3)d)\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)Rem(da1​×a2​×a3​​)=Rem(d(dq1​+r1​)(dq2​+r2​)(dq3​+r3​)​)

In the expansion, all the terms in the numerator is a multiple of
ddd, but for the last term, which is r1×r2×r3r_1 \times r_2 \times r_3r1​×r2​×r3​.

∴\therefore∴ Rem(a1×a2×a3d)=Rem(r1×r2×r3d)\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)Rem(da1​×a2​×a3​​)=Rem(dr1​×r2​×r3​​)

Example 20

What is the remainder when 235×237×239235 \times 237 \times 239235×237×239 is divided by 999?

Solution

The closest multiple of 999 for the three numbers is 234234234.

Rem(235×237×2399)=Rem((234+1)×(234+3)×(234+5)9)=Rem(1×3×59)=6\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) = 6Rem(9235×237×239​)=Rem(9(234+1)×(234+3)×(234+5)​)=Rem(91×3×5​)=6

Answer:
666

3.7 Remainder of Exponents

(a+b)n=nc0anb0+nc1an−1b1+...+(a + b)^n = ^{n}{c_0 a^n b^0} + ^{n}{c_1 a^{n - 1} b^1} + ... +(a+b)n=nc0​anb0+nc1​an−1b1+...+ ncn−1a1bn−1+^{n}{c_{n-1} a^1 b^{n - 1}} +ncn−1​a1bn−1+ ncna0bn^{n}{c_n a^0 b^n}ncn​a0bn

Let's say a number
aaa when divided by ddd, results in a quotient of qqq and remainder of rrr.
a=dq+ra = dq + ra=dq+r

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

=Rem(nc0(dq)nr0+nc1(dq)n−1r1+....+ncn−1(dq)1rn−1+ncn(dq)0rnd)= \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)=Rem(dnc0​(dq)nr0+nc1​(dq)n−1r1+....+ncn−1​(dq)1rn−1+ncn​(dq)0rn​)

All terms are divisible by
ddd except for the last term.

∴\therefore∴ Rem((dq+r)nd)=Rem(rnd)\mathrm{Rem} \left(\dfrac{(dq + r)^n}{d}\right) = \mathrm{Rem} \left(\dfrac{r^n}{d}\right)Rem(d(dq+r)n​)=Rem(drn​)

Example 21

What is the remainder when 7079570^{795}70795 is divided by 232323?

Solution

696969 is a multiple of 232323.

∴\therefore∴ Rem(7079523)=Rem((69+1)79523)=Rem(179523)=1\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) = 1Rem(2370795​)=Rem(23(69+1)795​)=Rem(231795​)=1

Answer:
111

The following example uses a combination of the rules.

Example 22

What is the remainder when 1512515^{125}15125 is divided by 888?

Solution

161616 is a multiple of 888.

∴\therefore∴ Rem(151258)=Rem((16−1)1258)\mathrm{Rem} \left(\dfrac{15^{125}}{8}\right) = \mathrm{Rem} \left(\dfrac{(16 - 1)^{125}}{8}\right)Rem(815125​)=Rem(8(16−1)125​)

=
Rem((−1)1258)=Rem((−1+8)8)=7\mathrm{Rem} \left(\dfrac{(- 1)^{125}}{8}\right) = \mathrm{Rem} \left(\dfrac{(- 1 + 8)}{8}\right) = 7Rem(8(−1)125​)=Rem(8(−1+8)​)=7


Answer:
777

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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