find the last 2 digits of 7^(7^1000)

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kabi (1665)

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7¹⁰⁰⁰ = (50-`1)⁵⁰⁰=⁵⁰⁰C₀ 50⁵⁰⁰-.......................................-500*50+1

=1000(k)+1

ab dekh 7 ki powers ke unit digit periodic hain 4 ke . e.g . 7 , 49 , 243 ,2401,then again 16807

So u can say without any doubt that unit digit of 7^{7^(1000)} would be 7 .

7 ^1000k+1=7 ( 7^1000k) = 7 ( 50 -1)^500k =7.(^500kC₀50^500k-................-500*50k+1)

all the first 500k numbers would be having atleast 3 zero on the last three places . 07 would be last two digit

Suggestion ::

Yaar have a knowlegde of number theory . It helps a lot .

Thank u

The time u guys take to find the derivative of a function or for finding the equilibrium constant of a reaction or for finding the angle of dispersion of prism or for standing from ur seat to congratulate our team after their win almost in that time one kid die because of poverty.

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