Gabriel's digits

Let $k=6^{20}$

Taking $log$ to the base 10 on both sides, we get-

$log_{10}k=log_{10}6^{20}=20\times log_{10}6 = 20\times (log_{10}2+log_{10}3)$

Now $0.30<log_{10}2<0.31$ and $0.47<log_{10}3<0.48$

$\Rightarrow 0.77<(log_{10}2+log_{10}3)<0.79$

$\Rightarrow 15.4<20\times (log_{10}2+log_{10}3)<15.8$

$\Rightarrow 15.4<log_{10}k<15.8$

So, we get - $10^{15.4}<k<10^{15.8}$

Since, $10^{15}$ has 15 zeroes in its end, we have $16$ digits in $10^{15.x}$ that is 16 digits in $k$

So, in the expansion of $6^{20}$ will have $\boxed{16}$ digits

Lemma 1

Note that the decimal expansion of any integer $N$ is given by: $a_k10^{k}+a_{k-1}10^{k-1}+\cdots+a_010^0=N$

Where $0< a_i \le9$ for integers $a_i$ .

With this, we have, then, that $k+1$ is the number of digits. To show that $\lfloor\log_{10}N+1\rfloor=k+1$ we first take the log of both sides: $\log_{10}(a_k10^{k}+a_{k-1}10^{k-1}+\cdots+a_010^0)=\log_{10}N$

We then have that: $\log_{10}(a_k10^{k}+a_{k-1}10^{k-1}+\cdots+a_010^0)=\log_{10}(10^{k})+\log_{10}(a_k)+C$

Where $0\le C < 1-\log_{10}9$

Then, we see that $0\le C+\log_{10}a_k<1$ . With this in mind, we floor both sides: $\lfloor\log_{10}(10^{k})+\log_{10}(a_k)+C\rfloor=\lfloor\log_{10}(10^{k})\rfloor=\lfloor\log_{10}N\rfloor$

So: $\lfloor k\rfloor=k=\lfloor\log_{10}N\rfloor$

Adding one to both sides gives us our final case: $k+1=\lfloor\log_{10}N\rfloor +1$

Q.E.D

Now, we are given bounds on the logarithms, so, by our previous lemma, the number of digits should be: $\lfloor\log_{10}6^{20}\rfloor+1=\lfloor20\log_{10}6\rfloor+1$

But we know: $\log_{10} 6=\log_{10}(3)+\log_{10}(2)$

And, given that $.3<\log_{10}2<.31$ and $.47<\log_{10}3<.48$ , we add both together to get a new inequality: $.77<\log_{10}(2)+\log_{10}(3)=\log_{10}6<.79$

Multiplying the inequality by 20, we have: $15.4<20\log_{10}6<15.8$

Hence: $\lfloor20\log_{10}6\rfloor=15$

We add unity to both sides to receive: $\lfloor20\log_{10}6\rfloor+1=16$

Which is our answer.

The answers to $log_{10}2$ and $log_{10}3$ are provided. This means you know that you have to use logs in some way.

Let $6^{20}=y$ .

Then take the log of both sides getting $log_{10}6^{20} = log_{10}y$

Expand the equation into $20(log_{10}3+log_{10}2) = log_{10}y$

Now the equation can be put into the form $10^{20(log_{10}3+log_{10}2)} = y$

We know that the number of digits in $10^{n}$ is $n + 1$ , and that $20(log_{10}3+log_{10}2)$ is approximately 15, so the number of digits in this number is $15 + 1 = \boxed{16}$

If we can find an $x$ such that $10^x=6^{20}$ then we're done. For instance, we know that $10^3$ has $3+1=4$ digits in its decimal representation, and sure enough $10^3=1000$ which has 4 digits.

Taking log base 10 of both sides of $10^x=6^{20}$ we obtain $x=\log_{10} (6^{20})=20\log_{10} 6=20[\log_{10} 2+\log_{10} 3]$ where the last 2 equalities follows by rules of logs. Now, using the bounds in the "details and assumptions" we get that $0.77<\log_{10} 2+\log_{10} 3<0.79$ , and consequently $15.4<x=20[\log_{10} 2+\log_{10} 3]<15.8$ . So clearly $6^{20}$ must have $15+1=\boxed{16}$ digits.

The number of digits in a number $x$ in base 10 is given by $\lfloor{ \log_{10} {x} }\rfloor+1$ . From the given inequalities, we can also conclude that $0.77<\log_{10}(6)<0.79$ . From this we conclude that $\log_{10}(6) \approx 0.78$ . Then the number of digits in $6^{20}$ is:

$\lfloor{ \log_{10} {6^{20}} }\rfloor+1$ $=\lfloor{ 20\log_{10} {6} }\rfloor+1=$ $\lfloor{ 20 \times 0.78 }\rfloor+1=$ $\lfloor{ 15.6}\rfloor+1=$ $15+1= \boxed{16}$

In decimal representation total digits of a number is: floor( log10(N)) + 1 In this case floor( 20xlog10(6))+1=16.

$log_{10}$ 6 = 778, 20 x 778= 15.56.... rounding up (ceiling) = 16

The useful information at the bottom suggests that perhaps we should consider using logarithms to the base 10 to solve this problem. Thus we will take $\log_{10}6^{20}$ Using the power rule gives: $=20\log_{10}6$ Using the product rule gives: $=20(\log_{10}2+log_{10}3)$ We know from the details the values of these logarithms. So, this equation is greater than: $=20(0.30+0.47)=20*0.77=15.4$ And less than: $=20(0.31+0.48)=20*0.79=15.8$ This means that $6^{20}=10^{x}, 15.4<x<15.8$ $10^{15} and 10^{16}$ Have 16 digits (1 followed by 15 0's) and 17 digits (1 followed by 16 0's) respectively. Since x falls in between these exponents, this means that the decimal expansion will fall between these values. Since 10^16 is the first 17 digit number, then the expansion will have 16 digits, since 16 is the lowest integer greater than these values. $\boxed{16}$

Since e15 means 10^15 you move the decimal place down to the right 15 times after I solved 6^20