Counting up the Digits

Let N be $5^{500}$ .

$log N = log 5^{500} = 500*log 5 = 500 * (log \frac {10} {2})$

$= 500*(log 10 - log 2) = 500 - 500 * log 2 > 500 - 500 * 0.302 = 349$

Since $5^{10} = 9765625 < 10^7$ , so $log 5^{500} = log (5^{10})^{50} < log (10^7)^{50} = log 10^{350} = 350$ , so $349 < log 5^{500} < 350$ , thus $log 5^{500}$ has 350 digits.

The number of digits in the base b representation of a natural number N is $\lfloor log_b N+1 \rfloor$ so the number of decimal digits is $\lfloor log_{10} 5^{500} + 1 \rfloor = \lfloor 500 log_{10} 5 + 1 \rfloor = \lfloor 500(1 - log_{10} 2) + 1 \rfloor$ but using the given inequality $500(1-log_{10} 2) + 1 > 500(1-0.302) + 1 = 350$ so $\lfloor 500(1-log_{10} 2) + 1 \rfloor = 350$ which is the answer.

Let 5^500=y. Taking log base 10: 500 x \log 10 5 = \log 10 y So we get: \log 5 10 = 500 x \log y 10 i.e. 500 x \log y 10 = 1 + \log 5 2 ..................(i) Given \log 10 2 < 0.302 => 1/ [\log 10 2] > 1/0.302 => \log 2 10 > 1/0.302 => 1 + \log 2 5 > 1/0.302 => \log 2 5 > 1/0.302 - 1 = .698/.302 => 1/ [\log 2 5] < .302/.698 => \log 5 2 < .302/.698 => 1 + \log 5 2 < 1 + .302/.698 = 1/.698 Continuing (i): 500 x \log y 10 = 1 + \log 5 2 < 1/.698 => 349 x \log_y 10 < 1 => y > 10^349. Now 10^349 has 350 digits. So y = 5^500 has 350 digits.

$\log_{10} 5 = 1 - \log_{10} 2$

Now, $5^500 = 500 * 0.698 = 349$

Hence, the $5^500$ has $349+1 = 350$ digits

The number of digits of $N$ is equal to $\lfloor \log_{10} N \rfloor + 1$ , so we want to estimate $\log_{10} 5^{500}$ . Since $2^{10} = 1024 > 1000 = 10^3$ , we know that $10 \log_{10} 2 > 3 \Rightarrow \log_{10} 2 > 0.3$ . $\log_{10} 5 = \log_{10} 10 - \log_{10} 2$ , so $0.698 < \log_{10} 5 < 0.700$ , which gives $349 < \log_{10} 5^{500} < 350$ . Hence, there are $350$ digits in the decimal representation of $5^{500}$ .

My solution is a bit over complicated, but it's pretty cool, so I post it anyway. The amount of digits in 5^500 is log(5^500)+1
log(5^500)=x
10^x=5^500
(5 2)^x=5^500
5^x
2^x=5^500
It's given that 10^0.302=2
5^x 10^0.302x=5^500
5^x
5^0.302x 2^0.302x=5^500
5^(x+0.302x)
2^0.302x=5^500
It can go on forever so eventually its:
5^(x+0.302x+0.302²x+...+0.302^(inf)x) =5^500
Which means
x+0.302x+0.302²x+...+0.302^(inf)x=500
The left side is the geometrical progression where a1=x q=0.302 and n=inf
S=a1 (q^n-1)/(q-1)=
x (0.302^inf-1)/(0.302-1)=
x (0-1)/-0.698=x/0.698
So
X/0.698=500
X=500 0.698
X=349
Log(5^500)=349
Log(5^500)+1=350

Lemma: The number of digits in an integer x is ceiling(logx), where the base is 10. Proof: We can find an n such that 10^n<=x<=10^(n+1). 10^n has a 1 followed by n zeros which makes it have n+1 digits. 10^(n+1) has a 1 followed by n+1 zeros which makes it have n+1 digits. Take the log base 10 of the inequality to get that: n<=logx<=n+1. If the inequality is strict, then taking the log will give an integer in which case, just add one (as shown above). If the inequality is not strict, then the integer x will have the same number of digits as 10^n. Taking the log will give a number between n and n+1, in which case you must take the ceiling of the number to get the desired n+1 digits.

Going back to the problem, log5^500=500 log5=500 (1-log2)>500*(1-.302)=349. Since it is slightly bigger than 349, the number contains 350 digits.

all logarithm are in base 10 let A = 5^500

logA = 500log5 logA = 500*0.6989 logA = 349.45

A = 10^349.45 i.e. 350 digits

First, we need to realize that we can use logarithm to check the digits of very large number. As an example, I will use a small number, let's say 5^4 or 625. Log (5^4) will give me 2.795, which means that this number 5^4 is in between 100 (100 = 10^2) and 1000 (1000 = 10^3), it has 3 digits.

So to solve this problem, we just have to analyze the value of log (5^500)

log (5^500)

= 500 * log 5

= 500 * log (10/2)

= 500 * (1 - log 2)

= 500 * (1 - 0.3019) *here I just pick a value slightly smaller than 0.302

= 349.05

We can conclude that 5^500 is slightly more than 10^349 but lower than 10^350, hence it has 350 digits.

log5=0.69.. 5^500= (log[5^500])+1=(500log5)+1 500xlog5=349.5...=349+1=350

Let 5^500 = 10^x. Taking common logarithms of both sides gives 500log5=x. Therefore x=349.485... So 5^500 has 350 digits

Note that 5^500 = 10^500 / 2^500. There are 501 digits in 10^500 and (500*0.302+1)-1 = 151 digits in 2^500. Dividing yields 350 digits. A quick search on Google confirms the answer.

n=Log5^500+1=500Log5+1=349,48+1=350,48

Let, 5^500=x, 500=logx(base 5)=logx/log5(base 10); logx(base 10)=349.485; x=5^500=(10)^349; so 5^500 will have 350 digits.