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# modular arithmetic

How would one use modular arithmetic to find the units digit of 7^7^7?

tens digit of 2^65 ?

Note by Alan Liang
4 years, 3 months ago

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FOR TENS DIGIT OF 2^65, FIND ITS MOD 100.

NOTE THAT 2^10=1024 = 24 (MOD 100) sO 2^20 = 576(MOD 100) = -24(MOD100) 2^40= (-24)^2 (MOD 100) = -24 (MOD100) AGAIN! 2^60 = 2^20 *2^40 (MOD 100) = 576(MOD100) = -24(MOD100)

ALSO, 2^5 = 32(MOD100) SO 2^65=2^60 * 2^5(MOD 100) = (-24)(32)(MOD100) = 32(MOD100)

SO THE TENS DIGIT IS 3 AND THE UNITS DIGIT IS 2. · 4 years, 3 months ago

The first problem :

$$7^{7^{7}} \equiv 7^{3} \equiv 3 \pmod {10}$$ and hence the unit digit is $$3$$. (The powers of $$7$$ form a cycle mod $$10$$. ) · 4 years, 3 months ago

how can you equate power 7 as power 3. I guess this is wrong. Please clarify. · 4 years, 3 months ago

$$7^{7} \equiv 7^{3}\cdot7^{4} \equiv 7^{3} \equiv 3 \pmod{10}$$. Since $$7^2 \equiv -1 \pmod {10} \Rightarrow 7^{4} \equiv 7^{2}\cdot7^{2} \equiv 1 \pmod{10}$$ · 4 years, 3 months ago

Just use euler's theorum for $$7^4$$, since phi(10) = 4 · 4 years, 3 months ago

Yes. · 4 years, 3 months ago

no..its perfectly alright.. · 4 years, 3 months ago

But unit digit would be 7. · 4 years, 3 months ago