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Question Number 206805 by BaliramKumar last updated on 26/Apr/24

Answered by A5T last updated on 26/Apr/24

$${\left(225\right)}^{40}={10}^x\Rightarrow x=40log225\approx 94.087<95$$$$\Rightarrow {10}^{94}<{\left(225\right)}^{40}<{10}^{95}\Rightarrow {225}^{40}has95digits$$$$$$$$Butif‘‘100{th}^{″}placemeansthethirdtothe$$$$lastdigit$$$${225}^{40}\equiv 0\left(mod125\right)\Rightarrow {225}^{40}=125k$$$${225}^{40}\equiv 1\left(mod8\right)\Rightarrow {225}^{40}=8q+1$$$$125k\stackrel{8}{\equiv }1\Rightarrow k\equiv 5\left(mod8\right)\Rightarrow {225}^{40}=125(8s+5)$$$$=1000s+625\Rightarrow {225}^{40}\equiv 625\left(mod1000\right)$$$$\Rightarrow Digitinthe‘‘100{th}^{″}placeis6.$$

Answered by A5T last updated on 26/Apr/24

$$17.2^b\equiv 2,4,-2,-4\left(mod10\right)$$$$3^c\equiv 3,-1,-3,1,\left(mod10\right)$$$$4^d\equiv 4,-4\left(mod10\right)$$$$\Rightarrow 1+2^b+3^c+4^d\equiv 0,6,4,8,2\left(mod10\right)$$$$\Rightarrow \left(c\right)$$

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