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**vjdm** · January 4th 2013, 11:47 AM

what would be the no of digits in 4⁶¹+4⁶²+4⁶³+4⁶⁴+4⁶⁵

**HallsofIvy** · January 4th 2013, 12:29 PM

All of those numbers is less than or equal to $4^{65}$ so their sum is less than $5(4^{65})$ . $65 log(4)= 39.1339$ so $4^{65}$ has 39 digits and 5 times that has 40 digits.

**Soroban** · January 4th 2013, 07:09 PM

Hello, vjdm!

$\text{What is the number of digits in: }\:N \;=\;4^{61}+4^{62}+4^{63}+4^{64}+4^{65}$

We have: . $N \;=\;4^{61}(1 + 4 + 4^2+4^3+4^4) \;=\;4^{61}\cdot341$

Take logs: . $\log N \;=\;\log\left(4^{61}\cdot341\right) \;=\;\log\left(4^{61}\right) + \log(341)$

m . . . . . . . $\log N \;=\;61\log(4) + \log(341) \;=\;39.25841385$

. . . . . . . . . . . $N \;=\;1.813066986 \times 10^{39}$

Therefore, $N$ has 40 digits.

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