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26 Dec 2009 06:55:35 IST

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what will bethe remainder when 4raised to power 101 is divided by 101.......?(power of 4 is 101)

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what will bethe remainder when 4raised to power 101 is divided by 101.......?(power of 4 is 101)

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krishma

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Joined: 15 Dec 2009

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29 Dec 2009 06:37:22 IST

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hey anyone reply to my query...tell me if the ques. isnt clear............

Millind Gupta

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Joined: 7 Oct 2009

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31 Dec 2009 16:11:55 IST

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I am posting the 2 ways to solve this question "

Ist way :

4¹⁰¹ = (1 + 3)¹⁰¹ = ¹⁰¹C₀ + ¹⁰¹C₁3 + ¹⁰¹C₂3² + ¹⁰¹C₃3³ + ............+¹⁰¹C₁₀₁3¹⁰¹

4¹⁰¹ = 1 + ¹⁰¹C₁3 + ¹⁰¹C₂3² + ¹⁰¹C₃3³ + .....................+3¹⁰¹

we know ⁿC_r is always an integer and each term conatins a multiple of 101 apart from 1st and last one.

so 4¹⁰¹ when divided by 101 gives the same remainder what 1 + 3¹⁰¹divided by 101 will give.

so,

1 + 3¹⁰¹ = 1 + (1 + 2)¹⁰¹ = 1 + ¹⁰¹C₀ + ¹⁰¹C₁2 + ¹⁰¹C₂2² + ¹⁰¹C₃2³ + ............+¹⁰¹C₁₀₁2¹⁰¹

1 + 3¹⁰¹ = 1 + 1 + ¹⁰¹C₁2 + ¹⁰¹C₂2² + ¹⁰¹C₃2³ + ............+2¹⁰¹

same logic goes here as well,

as every term conatins a multiple of 101 apart from 1st and last one.

so 1 + 3¹⁰¹when divided by 101 will give the same remainder what 1 + 1 + 2¹⁰¹ will give.

2 + 2¹⁰¹ = 2 + (1 + 1)¹⁰¹ = 2 + ¹⁰¹C₀ + ¹⁰¹C₁ + ¹⁰¹C₂ + ¹⁰¹C₃ + ............+¹⁰¹C₁₀₁

= 2 + 1 + ¹⁰¹C₁ + ¹⁰¹C₂ + ¹⁰¹C₃ + ............ + 1

= 4 + ¹⁰¹C₁ + ¹⁰¹C₂ + ¹⁰¹C₃ + ............

this when divided by 101 will give 4 as the remainder

as every other term contains 101.

Ans = 4

2nd way :

From Fermat's Little theorem, we know the following result

If a is an integer and p is a prim number, then a^p - a is divisible by p

Here a = 4 (an integer)

p = 101 (a prime number)

hence according to theorem , 4¹⁰¹ - 4 is divisible by 101

i.e. when 4¹⁰¹ is divided by 101, it gives 4 as the remainder.

4¹⁰¹ - 4 = 101k where k is a positive integer

4¹⁰¹ = 101k + 4 i.e. 4 is the remainder. Ans

akki ~~ unlucky forever ~~

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31 Dec 2009 17:12:53 IST

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$2^9\equiv7\ mod(101)\\ \\ \\ 2^{18}\equiv49\ mod(101)\\ \\ \\ 2^{19} \equiv 98\ mod(101),98 \equiv -3\ mod(101)\\ \\ \\ so,2^{19} \equiv -3\ mod(101)\\ \\ \\ 2^{190} \equiv 3^{10}\ mod(101)\\ \\ \\ 2^{12}.2^{190} \equiv 4.2^{10}.243^{2}\ mod(101)\\ \\ \\ 2^{10} \equiv 14\ mod(101),243^2 \equiv (41)^2 \ mod(101),41^2 \equiv 65\ mod(101)\\ \\ \\ => 2^{202} \equiv 4.2^{10}.3^{10}\ mod(101),4.2^{10}.3^{10} \equiv 4.14.65\ mod(101)\\ \\ \\ 4^{101} \equiv 3640\ mod(101) \equiv 4\ mod(101)\\ \\ \\ 4^{101} \equiv 4\ mod(101)\\ \\ \\ hence\ the\ remainder\ left\ is\ 4$

But, try to remember Fermat's Little Theorem, it's of good help.

krishma

New kid on the Block

Joined: 15 Dec 2009

Posts: 25

1 Jan 2010 03:26:50 IST

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thanks for all the three solutions................we haven't done any of thse still in tenth

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