The how many digits in 2^1000 question (1 Viewer)

[Jase]

Im not sure if this has been posted before, and the search didnt show anything..

From the 2001 2U HSC, 5b)ii

well okay, Heres the question for those who dont know it

i) Find log10 (2^1000), correct to 3 decimal places

ii) We know that 2^10 = 1024, so that 2^10 can be represented by a four digit numeral. How many digits are there in 2^1000 when written as a numeral?

Before you get busy typing up the solution, i do know how to do it, but i would like to clear some strange discrepancies up.

right, so the answer to the first is 301 rounded down to integer..
now after some deducting, you get 2^1000 = 10^301

the deal is, from this, the answer should be 302 digits..

of course, this is 2U and we must consider those you wouldn't pick it up, so theres the other method which is arithmetic progression. From what i've heard, people also get the answer 301.
->
2^10 = 4 digits
2^20 = 7 digits
2^30 = 10 digits
2^40 = 13 digits

so if you'd say its an arithmetic progression; T_100 is 301
but 301 digits is not 302, and they technically arn't very close, so whats the real answer?? is arithmetic progression a viable method?

 

[withoutaface]

What you do to find it is go log[base2](10)=ln10/ln2, this finds what the index of 2 must be to get 10, then you divide 1000 by that and hey presto you have the number of digits after the 1 in the number, remembering that 10^1 has 2 digits, 10^2 has 3 digits and so on

this works because 2^1000=(2^log[2](10))^(1000/log[2](10))=10^(1000/log[2](10))

The answer is 302, because there are 301 digits AFTER the 1 in 10^(1000/log[2](10)), I don't know how arithmetic progression would work, but it seems that using the first answer and just adding 1 (for the reason I've already stated) would work just as well.

The problem with the arithmetic progression is that when you go log_10(2^10) and log_10(2^20), the decimal after the number given as the solution to these two logs changes. This suggests that with each addition of 10 to the index of 2 the solution is approaching gaining an extra digit which would throw the arithmetic progression out.

 

[Slidey]

2^1000

To find how many digits this number has, you need to perform the following operations:

Take base 10 log:

log_10(2^1000)=1000log_10(2)=301.03
.'.
2^1000=10^301.03
10^301<2^1000<10^302,
So 2^1000 is a 302-digit number.

EDIT: After actually READING the entire thread, it turns out I jsut showed you exactly what yuo didn't need to know. Sorry.

 

[mojako]

why exactly r u doing 2U paper?
btw how do u guys find 2003 HSC?

 

[speersy]

i found the 2003 HSC fairly hard. I got like 97/120
I have done the past six years and averaging like 104/120 so 2003 was definately harder than the past years.

 

[mojako]

2^1000

To find how many digits this number has, you need to perform the following operations:

Take base 10 log:

log_10(2^1000)=1000log_10(2)=301.03
.'.
2^1000=10^301.03
10^301<2^1000<10^302,
So 2^1000 is a 302-digit number.

EDIT: After actually READING the entire thread, it turns out I jsut showed you exactly what yuo didn't need to know. Sorry.

no
the reason it has 302 digit is because 10^301 has 302 digits.. 10^301.99999999999 has 302 digits.. and 10^302 has 303 digits, like what withoutaface mentioned
(although now.. he keeps changing faces.. hmm....)

 

[redslert]

the simplier explanation is that 10^301 had 301 zeros and then you have 1 infront of that, making 302 digits

 

[Slidey]

no
the reason it has 302 digit is because 10^301 has 302 digits.. 10^301.99999999999 has 302 digits.. and 10^302 has 303 digits, like what withoutaface mentioned
(although now.. he keeps changing faces.. hmm....)

That's exactly what this part meant:

2^1000=10^301.03
10^301<2^1000<10^302,
So 2^1000 is a 302-digit number.

10^1 has 2 digits. 10^2 has 3 digits. So obviously 10^301 has 302 digits. And since 2^1000 is between 10^301 and 10^302, it MUST have 302 digits.

OK?

 

[mojako]

Ok
(that tone sounds familiar... )

my apologies
EDIT: no, this is not a real apology.. coz I dont think I committed any sin against Slide Rule

 

[Slidey]

No, no, I didn't mean to sound annoyed, I meant to clarify, and if you disagreed, for you to explain why...

Sorry if my tone sounded caustic.