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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 - Euler & Fermat

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4. Euler's Totient

Euler's Totient of a number is a function that provides a count of the natural numbers less than the number that would form a co-prime pair with it.

ϕ(n)\phi (n)ϕ(n) is the function that represents the Euler's totient for the number nnn.

If the prime factors of
nnn are x1,x2,x3,...x_1, x_2, x_3, ...x1​,x2​,x3​,...

ϕ(n)\bm{ \phi (n)}ϕ(n) = n×(1−1x1)×(1−1x2)×(1−1x3)×... n \times \left(1 - \dfrac{1}{x_1}\right) \times \left(1 - \dfrac{1}{x_2}\right) \times \left(1 - \dfrac{1}{x_3}\right) \times ... n×(1−x1​1​)×(1−x2​1​)×(1−x3​1​)×...

Example 23

How many natural numbers less than 242424 are co-prime with 242424?

Solution

24×23×3 24 \times 2^3 \times 324×23×3

∴\therefore∴ The prime factors of 242424 are 222 and 333.

ϕ(24)= \phi (24) =ϕ(24)= 24×(1−12)×(1−13) 24 \times \left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right)24×(1−21​)×(1−31​)

=24×12×23=8= 24 \times \dfrac{1}{2} \times \dfrac{2}{3} = 8=24×21​×32​=8

Number of co-primes
=8= 8=8

Answer:
888

Note: The co-primes of
242424 that are less than it are 111, 555, 777, 111111, 131313, 171717, 191919 and 232323.

4.1 Fermat's Little Theorem of Remainders

For any given number
nnn and divisor ddd, if
(
111) nnn and ddd are co-prime, and
(
222) ϕ(d)\phi(d)ϕ(d) is the Euler's totient of ddd

Rem(nϕ(d)d)=1 \mathrm{Rem} \left(\dfrac{n^{\phi(d)}}{d}\right) = 1Rem(dnϕ(d)​)=1

Example 24

What is the remainder when 79198379^{1983}791983 is divided by 121212?

Solution

We first reduce the base of 797979, applying the exponent rule.

Rem(79198312)=Rem((72+7)198312)=Rem(7198312)\mathrm{Rem} \left(\dfrac{79^{1983}}{12}\right) = \mathrm{Rem} \left(\dfrac{(72 + 7)^{1983}}{12}\right) = \mathrm{Rem} \left(\dfrac{7^{1983}}{12}\right) Rem(12791983​)=Rem(12(72+7)1983​)=Rem(1271983​)

777 and 121212 are co-prime. ∴\therefore∴ We can now reduce the power by applying Fermat's little theorem.

ϕ(12)\phi (12)ϕ(12) = 12×(1−12)×(1−13)=4 12 \times \left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right) = 412×(1−21​)×(1−31​)=4

Fermat's little theorem states
Rem(7412)=1\mathrm{Rem} \left(\dfrac{7^4}{12}\right) = 1Rem(1274​)=1

Rem(7198312)=\mathrm{Rem} \left(\dfrac{7^{1983}}{12}\right) =Rem(1271983​)= Rem((74)495×7312)=\mathrm{Rem} \left(\dfrac{(7^4)^{495} \times 7^3}{12}\right) =Rem(12(74)495×73​)= Rem(1495×34312)=7\mathrm{Rem} \left(\dfrac{1^{495} \times 343}{12}\right) = 7Rem(121495×343​)=7

Alternatively

When the base and the divisor are co-prime, you can directly apply the following to improve your speed:

1) Exponent Rule: In the base, write the remainder when base is divided by the divisor
2) Fermat's Theorem: In the power, write the remainder when the power is divided by the
ϕ\phiϕ(divisor).

Rem(7912)=7\mathrm{Rem} \left(\dfrac{79}{12}\right) = 7Rem(1279​)=7 ; Rem(19834)=3\mathrm{Rem} \left(\dfrac{1983}{4}\right) = 3Rem(41983​)=3

∴\therefore∴ Rem((79)198312)=Rem(7312)=7\mathrm{Rem} \left(\dfrac{(79)^{1983}}{12}\right) = \mathrm{Rem} \left(\dfrac{7^3}{12}\right) = 7Rem(12(79)1983​)=Rem(1273​)=7

Answer:
777

Example 25

What is the remainder when 434743^{47}4347 is divided by 232323?

Solution

434343 and 232323 are co-prime. ϕ(23)=\phi (23) =ϕ(23)= 23×(1−123)=2223 \times \left(1 - \dfrac{1}{23}\right) = 2223×(1−231​)=22

Applying Fermat's little theorem,

we get
Rem((43)2223)=1\mathrm{Rem} \left(\dfrac{(43)^{22}}{23}\right) = 1Rem(23(43)22​)=1

Rem((43)4723)=Rem(((43)47)2×(43)323)=Rem((43)323)\mathrm{Rem} \left(\dfrac{(43)^{47}}{23}\right) = \mathrm{Rem} \left(\dfrac{((43)^{47})^2 \times (43)^3}{23}\right) = \mathrm{Rem} \left(\dfrac{(43)^3}{23}\right) Rem(23(43)47​)=Rem(23((43)47)2×(43)3​)=Rem(23(43)3​)

=
Rem((43−3)323)=Rem((−3)323)=Rem(−2723)\mathrm{Rem} \left(\dfrac{(43 - 3)^3}{23}\right) = \mathrm{Rem} \left(\dfrac{(- 3)^3}{23}\right) = \mathrm{Rem} \left(\dfrac{- 27}{23}\right)Rem(23(43−3)3​)=Rem(23(−3)3​)=Rem(23−27​)

The remainder is unchanged when a multiple of the divisor is added to the dividend.

∴\therefore∴ Rem((−27+46)23)=19\mathrm{Rem} \left(\dfrac{(- 27 + 46)}{23}\right) = 19Rem(23(−27+46)​)=19

Answer:
191919

5. Chinese Remainder

If the numerator and denominator are not co-prime, then Fermat's Little Theorem cannot be directly applied.

The following steps are then followed,
1) The divisor is prime factorised
222) The remainders when the number is divided by each of the prime factors (with their respective powers) is found.
333) The number is expressed in dq+rdq + rdq+r for the different divisors to find the common remainder.

Example 26

What is the remainder when 250340250^{340}250340 is divided by 444444?

Solution

44=22×11 44 = 2^2 \times 1144=22×11

222^222 perfectly divides 250340250^{340}250340 ⇒ Rem((250)34022)=0\mathrm{Rem} \left(\dfrac{(250)^{340}}{2^2}\right) = 0Rem(22(250)340​)=0

∴\therefore∴ Where aaa is an integer, 250340=4a+0\bm{250^{340} = 4a + 0}250340=4a+0 ⟶\longrightarrow⟶(111)

Rem((250)34011)=Rem((242+8)34011)=Rem((8)34011)\mathrm{Rem} \left(\dfrac{(250)^{340}}{11}\right) = \mathrm{Rem} \left(\dfrac{(242 + 8)^{340}}{11}\right) = \mathrm{Rem} \left(\dfrac{(8)^{340}}{11}\right)Rem(11(250)340​)=Rem(11(242+8)340​)=Rem(11(8)340​)

ϕ(11)=11×(1−111)=10\phi(11) = 11 \times \left(1 - \dfrac{1}{11}\right) = 10ϕ(11)=11×(1−111​)=10

∴\therefore∴ Rem(834011)=Rem((810)3411)=1\mathrm{Rem} \left(\dfrac{8^{340}}{11}\right) = \mathrm{Rem} \left(\dfrac{(8^{10})^{34}}{11}\right) = 1Rem(118340​)=Rem(11(810)34​)=1

∴\therefore∴ Where bbb is an integer, 250340=11b+1\bm{250^{340} = 11b + 1}250340=11b+1 ⟶\longrightarrow⟶(222)

Equating (
111) and (222),

4a+0=11b+14a + 0 = 11b + 14a+0=11b+1

⇒
a=11b+14a = \dfrac{11b + 1}{4}a=411b+1​

bbb = 111 satisfies. Substituting in (222),

Remainder =
11×1+1=1211 \times 1 + 1 = 1211×1+1=12

Answer:
121212

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