FINDING LAST TWO DIGITS

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LAST TWO DIGITS OF 3¹⁰⁰

3¹⁰⁰ = (3¹⁰)¹⁰ = ((3²)⁵)¹⁰

(3²)⁵ = 9x9x9x9x9 = 729x9x9// WE TAKE ONLY LAST 2 DIGITS OF 729
// AND OF OTHER 3 DIGIT NO.S

= 29x9x9 mod 100

= 261x9 mod100 = 61x9 mod 100 //ONLY 69 OF 269 TAKEN

= 49 mod 100

so 3¹⁰⁰ = 49¹⁰ mod 100

49¹⁰⁰ = ((49²)⁵)

49² = 2401 = 1 mod 100

3¹⁰⁰= (01⁵)¹ = 01 mod 100

last two digits of 3¹⁰⁰ are 01

#Mathematics #Fundamentals

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3 Answers

KR ·2009-03-22 11:17:14

HOPE THIS WAS USEFUL and i was correct

[32][32][33][32][32]

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Subash ·2009-03-23 01:12:26

thanks KR this is what i was looking for :)

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Hari Shankar ·2009-03-23 01:48:44

If you are using congruences its easier to check mod 4 and mod 25

We have 3 \equiv -1 \bmod 4 \Rightarrow 3^{100} \equiv 1 \bmod 4

By Euler Totient Theorem 3^{20} \equiv 1 \bmod 25 \Rightarrow 3^{100} \equiv 1 \bmod 25

Since gcd(4,25) = 1, we have 3^{100} \equiv 1 \bmod 100

In case you are not familiar using congruences use binomial theorem 3^{100} = (10-1)^{50} = 100k - \binom{50}{1} 10 + 1 = 100k-500+1 This immediately tells us that the remainder on division by 100 is 1.

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FINDING LAST TWO DIGITS

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