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What is the tens digit in the expansion of 234th powe of 81?
4 months ago

we want to know the tens digit of x= 81^234.
so we can simply find the remainder when x is divided by 100.
using binomial theorem, remainder when 81^234 is divided by 100= remainder when 19^234 is divided by 100.
= remainder when (361)^117 is divided by 100= remainder when 61^117 is divided by 100
= remainder when 61*61^116 is divided by 100
= remainder when 61*(61^2)^58 is divided by 100
= remainder when 61*121^58 is divided by 100= remainder when 61*21^58 is divided by 100
= remainder when 61*441^29 is divided by 100
= remainder when 61*41^29 is divided by 100
= remainder when 61*41*(41^2)^14 is divided by 100
= remainder when 61*41*81^14 is divided by 100
= remainder when 61*41*19^14 is divided by 100
= remainder when 61*41*361^7 is divided by 100
= remainder when 61*41*61^7 is divided by 100
= remainder when 41*61^8 is divided by 100
= remainder when 41*(61^2)^4 is divided by 100
= remainder when 41*(121)^4 is divided by 100
= remainder when 41*(21)^4 is divided by 100
= remainder when 41*(441)^2 is divided by 100
= remainder when 41*(41)^2 is divided by 100
= remainder when 41^3 is divided by 100
= remainder when 40^3+1+3*40*1(40+1) is divided by 100
= remainer 1+3*40= 1+120
= remainder of 121
= 21
hence, the tens digit would be 2.
kindly approve :)
4 months ago
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