How many digits does 3^100 have?

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Here's the solution to my viral math problem on my Instagram. How many digits does 3^100 have?
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#math #algebra #mathbasics

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11 сен 2024

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Комментарии : 301

@bprpmathbasics 26 дней назад

1 divided by 0 (a 3rd grade teacher & principal both got it wrong), Reddit r/NoStupidQuestions ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WI_qPBQhJSM.html

@ChavoMysterio 21 день назад

If I jot down 3¹⁰⁰ on some calculators, the answer will be in scientific notation.

@serae4060 15 дней назад

@@ChavoMysterioso now calculate the number of digits of 3^100 in binary, might make it more difficult

@yurenchu 26 дней назад

Without a calculator: 3^100 = = 3^(2*50) = (3^2)^50 = 9^50 < 10^50 {which has 51 digits} ==> 3^100 has _strictly less_ than 51 digits So answer options C and D are eliminated. 3^100 = = 3^(4*25) = (3^4)^25 = 81^25 > 80^25 = (8^25) * (10^25) = ((2^3)^25) * (10^25) = (2^75) * (10^25) = (2^5) * (2^70) * (10^25) = 32 * ((2^10)^7) * (10^25) = 32 * (1024^7) * (10^25) > 32 * (1000^7) * (10^25) = 32 * (10^21) * (10^25) = 32 * 10^(21 + 25) = 32 * 10^46 = 3.2 * 10^47 > 10^47 {which has 48 digits} ==> 3^100 has _at least_ 48 digits. So answer option A is eliminated. ==> Remaining answer option: B

@yurenchu 26 дней назад

Additionally, 3^100 = = ((3^5)^20) = (243^20) < (250^20) = (1000/4)^20 = (1000^20) / (4^20) = ((10^3)^20) / ((2^2)^20) = (10^60) / (2^40) = (10^60) / (1024^4) < (10^60) / (1000^4) = (10^60) / ((10^3)^4) = (10^60) / (10^12) = (10^60) * (10^(-12)) = 10^(60 - 12) = 10^48 {which is the smallest (positive) 49-digit integer} ==> 3^100 has _strictly less_ than 49 digits. (3.2 * 10^47) < (3^100) < (10^48) (3.2 * 10^47) < (3^100) < (10 * 10^47) ==> _Confirmed correct_ answer: 3^100 has 48 digits ==> option B is indeed correct.

@thefanboy3285 26 дней назад

clever boiii

@derwolf7810 25 дней назад

I suggest to drop some steps for improved readability: 3^100 = 81^25 > 80^25 = 2^75 10^25 = 1024^7.5 10^25 > 1000^7.5 10^25 = 10^(22.5+25) = 10^47.5

@jeffw1267 25 дней назад

But if you have log(3) memorized, it takes about three seconds to do the problem mentally.

@yurenchu 25 дней назад

@@jeffw1267 Yes, that's actually how I solved it initially; see also my first comment in this comment section. (Half a second to realize log(3^100) = 47.7... and then another 2.5 seconds to decide if we must round down or round up.)

@l9day 26 дней назад

oh no, he's doing a bait and switch. We see a cat so we stay, and then he teaches us math!

@Gremriel 26 дней назад

From his basement!

@jw8160 26 дней назад

😂

@ChandrasegaranNarasimhan 25 дней назад

Dude, you caught him. Just because math has mathe-ma-tics does not mean ma tics. Every ma should tick.

@richardhole8429 25 дней назад

It is a well known fact that cats know the value of log 3 to 5 decimals.

@dannysigurdson7108 24 дня назад

And not even a single cat recipe 😢

@andrerenault 26 дней назад

I wonder if logarithms would be better appreciated if we first taught them in this format, as “size of a number” rather than simply “inverse of exponential”

@yurenchu 25 дней назад

@@andrerenault The field of mathematics in general (and hence our understanding of the world) appears to have been developed around addition, multiplication, and (positive) exponential growth; and this seems to reflect (and also to have enforced) a reality of our (western) society that is based on acquiring, producing, and getting _more, more, more_ . I've honestly wondered if we would have lived in a better, less selfish, world if mathematics was rather built primarily around the inverses of addition, multiplication and exponentiation (i.e. around subtraction, division, and logarithms)...

@theapexsurvivor9538 22 дня назад

@@yurenchu probably not, because losing a negative goat is still more desirable for survival than losing a goat. Acquisition based thought is derived from evolutionary sources, themselves derived from the physical laws of thermodynamics and will always be a flaw of a natural species in an entropic environment. Greed and Gluttony have always held sway in Every culture where it was not sufficiently monitored and penalized or where power was sufficiently centralised. The West just had a lot of cultures that had high centralisation and low accountability for the upper classes. This is also why communism tends toward collapse/dictatorship, as its proponents always decry Greed and Laziness as the products of capitalism rather than the product of us being an extension of the natural cycle that simply has broken the constraints that balanced us. If anything, the cultures where the most members of society prosper are the ones in which Greed was directly tied to the common good, where you couldn't prosper in the ways you wanted unless you helped the society as a whole prosper, and positive maths is equally supportive of that, as it's just basic investment strategies that were figured out repeatedly throughout history and lost again when power was centralised. It'd just mean a lot more deal negatives and be less intuitive. Now, if we were evolved in an negentropic universe where energy would always gather into larger clumps, then we'd probably benefit from a negative based system.

@cigmorfil4101 18 дней назад

What is log except to tell you the power of the base which equals the original number, ie the exponent of the base which gives you that number? Hence the link between exponention and logs. I was taught logs (back on the 1970s) as a calculation method to convert multiplication and dividision into addition and subtraction. When using log tables we first had to express the number in standard form and split it between the mantissa and exponent. The mantissa (between 1 and 10, 1

@piccolo64 26 дней назад

3^100, don't use calculator... a few moment later... log(3)=🤔🤔 hmm where is my calculator ? 🤣

@Ninja20704 25 дней назад

In some countries log(3)≈0.477 is a value that needs to be memorised for exams

@_crypt 25 дней назад

Log 1-10 is memorized bro

@rowlul 24 дня назад

@@_crypt nope

@QwertyUiop-ux2yh 24 дня назад

You are supposed to remember log values of first few primes.

@DakshHooda-x4d 24 дня назад

Bro you gotta memorise the values of log 2,3,5,7 etc at least

@msolec2000 26 дней назад

I immediately discarded C and D by doing 3^100 = (3^2)^50 = 9^50 < 10^50. But with A and B being so close, there's not much further progress to be made.

@michaelbujaki2462 20 дней назад

I immediately eliminated C and D by noting that 10^100 has 100 digits, so 3^100 has

@eragon78 19 дней назад

@@michaelbujaki2462 10^100 has 101 digits. 1 followed by 100 zeros, which is 101 digits. 10^1 = 10 which has 2 digits.

@eliteteamkiller319 13 дней назад

I did something similar and differentiated between 47 and 48 in two different ways: first, I figured, making it the smallest number is too easy. Then I tested it by entering 100 log (3) in my calculator ;)

@joeschmo622 25 дней назад

I didn't know we were allowed to use a calculator. I just "observed" that the progression is 1, 3, 9, 27, 81, 243, etc., so the number of places is roughly half the exponent. Hmm, so 47 *or* 48. I just worked out a "trend" and could extrapolate that it *should* be at least 47 but most likely 48, so picked out 48 as my final answer. Not quite scientific, but a good back-of-napkin estimate.

@cdkw8254 26 дней назад

I love cats!!!!!!

@kennethalbert5903 26 дней назад

so if i ask how many digits does x^y has, is it equal to floor(ylogx) + 1?

@RossPfeiffer 26 дней назад

ceil( ylogx )

@user-yv1qs7sy9d 26 дней назад

@@RossPfeiffer No, it is floor(ylogx)+1. ceil(ylogx) doesn't always work. For example, for x=10 and y=2: x^y=10^2=100 (3 digits). However, ceil(ylogx) = ceil(2log10) = ceil(2) = 2, whereas: floor(ylogx)+1 = floor(2log10)+1 = floor(2)+1 = 2+1 = 3.

@gdclemo 25 дней назад

only if x^y is a positive integer

@senpai12349 25 дней назад

@@gdclemoif y is a negative integer, you could still use it ig, it would just become 1/y logx

@BleuSquid 25 дней назад

Real world use case of this "size of a number" concept: I've seen this used multiple times in code for testing the primality of very large numbers, Mersenne numbers, which are of the format (2^n) - 1. How much memory to allocate, planning the algorithm, what size FFT should be used - all depend on the size of the number being tested.

@archangecamilien1879 15 дней назад

Yeah, it's going to be one of 47 or 48, lol, it comes down to whether 3^100 is larger than 10^48 or not...if it isn't, it's 47, if it is, then it's 48...I mean, lol, by the theorem of multiple choice...

@blankspace178 26 дней назад

FYI: log(3^100) and 100*log(3) are computationally identical in terms of resources. It doesn't matter how large or small the number is, what matters is the factorization, and since relative larger and smaller numbers have equal factorization, the computational resources required are identical.

@461weavile 26 дней назад

I'm pretty sure he was saying you won't get an accurate answer because most calculators won't show that many digits. I could've misinterpreted him, though.

@saaah707 25 дней назад

depends on the computer's architecture

@JubilantJerry 25 дней назад

If the calculator can find 3^100 to N decimal places, it can find the log of that to a similar level of precision.

@blankspace178 24 дня назад

@461weavile Kind of....at each step of factorization, a calculator will round off based on its limitations to varying degrees. So the true result gets further and further away from the calculators final result. This limitation extends to all forms of programming with integer, float, etc. It's actually impossible for a computer to return an accurate result, even a supercomputer, to some very basic calculations due to funny quirks in arithmetic.

@JubilantJerry 24 дня назад

@@blankspace178 I think it's likely that the calculator already implements 3^100 as exp(100 * ln(3)), and yeah it's true it will accumulate some error in doing the exp but the relative error is certainly very small. I think the calculator also likely computes log_10(3) as ln(3) / ln(10) so it also does extra steps that way too. Either method will definitely be precise enough to compute the number of digits. Though on the other hand, sure you can't count the number of digits in 3^1000 directly because it will overflow, you have to compute it with 1000 * log_10(3)

@Nikioko 26 дней назад

That's why you use lg in science. The mantisse (the part after the decimal comma) determines the digits, while the number before the decimal comma determines the magnitude in powers of 10. E.g., if 0,3010 = lg(2), then 1,3010 = lg(20). That's how you calculated before calculators were invented. If you wanted to multiply two numbers, you added their logs. And since log tables were just from 1 to 10, you had to multiply with powers of 10 by adding an integer.

@i_am_aladeen 25 дней назад

I did it in my head before the video started: "3 times 3, is still 1 digit. Times 3, adds one digit. Times 3, stays the same number of digits. Times 3, adds one digit. And so on..." So the result must be either 47 or 48. Your video made it even easier. I didn't know that trick.

@user-ui7hk6zm1r 25 дней назад

Fun fact, this method of using logarithms is actually used in deriving the general formula for finding the numbet of digits of any positive integer n. N(n) = ceil(log10(n+1)) N : Z+ -> Z+

@imonkalyanbarua 24 дня назад

Another quick rough estimation one can make is that in every power of 3 which is a multiple of 2, the number of digits increase by 1. For eg. 3, 9, 27, 81, 243, 729 etc. In that way for the number 3^100, the number of digits should be close to 50.

@charlesabju907 21 день назад

For the first time I had this correct, for the right reason, and it only took me about 20 seconds, including looking up log 3

@m.h.6470 24 дня назад

Quick conversion in the head gives you 9^50, which is close to 10^50, which is a 1 with 50 zeros. That alone eliminates C and D. Now, is 10^47 - the smallest number with 48 digits - smaller or larger than this? 10^47 ? 9^50 |*10³ 10^50 ? 9^50 * 10³ |: 9^50 (10/9)^50 ? 10³ 1.12 > 10/9 1.12^5 ≅ 1.76 1.76^5 ≅ 16.89 16.89² ≅ 285.27 so 1.12^50 < 300 therefore (10/9)^50 < 10³ which means, that 3^100 has at least 48 digits

@utsab_us 25 дней назад

number = 3 ** 100 number_in_string = str(number) print(len(number_in_string)) # output : 48 The answer is option : B

@bprpmathbasics 25 дней назад

Amazing! 😆

@NaplesFC 24 дня назад

I thought the same 😂😂

@JOHN-WICK-3-6-9 24 дня назад

Python use 🗿

@jasejj 22 дня назад

Of course that very much depends on whether your machine can handle numbers that large, which some can't.

@yurenchu 26 дней назад

3^100 = = 10^[log10(3) * 100] = 10^[(0.477...) * 100] = 10^[47.7...] ==> 48 digits.

@xavierburval4128 26 дней назад

I came up with something similar, but instead of finding log10(3) I found log3(10), and then rewrote the function as 10^[100/log3(10)]. 100/log3(10) = 47.7… so 48 digits.

@Osirion16 25 дней назад

@@xavierburval4128 log10(3)=1/log3(10) so it is trivial

@xavierburval4128 25 дней назад

@@Osirion16 huh, good to know

@gdclemo 25 дней назад

3^100 can also be four digits, depending on how you interpret the question.

@yurenchu 25 дней назад

@@gdclemo Or three digits, because 3^100 = 9^50 . 😋

@Wildcard71 26 дней назад

To border it in: 3²=9, so the number of digits has to be fewer than 50.

@notsure6218 26 дней назад

Pretty much what I did, i was like, it takes a few powers to jump up a digit

@emiliapeabea 26 дней назад

wdym the number of digitshas to. be fewer than 50? 3^2=9 seems to work according to this method so imconfused

@Wildcard71 26 дней назад

@@emiliapeabea 9^50 is less than 10^50, and there are several steps a digit is dropped.

@yurenchu 26 дней назад

You mean "fewer than 51". 10^50 has 51 digits.

@zachansen8293 26 дней назад

@@emiliapeabea 3^100 is the same as 9^50. Since 9^50 is less than 10^50 and 10^50 has 51 digits... it has to have less than 51 digits. Not sure where they got 50.

@JayTemple 12 дней назад

0.47 < common-log 3 < 0.48. That means 10^0.47 < 3 < 10^0.48. Taking their 100th powers, we get 10^47 < 3^100 < 10^48. Those bounds are 1 followed by 47 zeroes and 1 followed by 48 zeroes, which have 48 and 49 digits respectively. Anything less than 10^48 will have at most 48 digits, so the answer is 48.

@SirRebrl 21 день назад

Roughing it out in my head I got 3^9 is roughly 2e4. (2e4)^11 is (2^11)e44 is roughly 2e47. (because 2^10 is approximately 1e3) 2e47 * 3 is 6e47. 48 digits.

@Mateusz-yp3wq 12 дней назад

Assuming this can be obtained without calculator: 3^7=2187 , 2^11=2048 , 10^3=1000 , 2^10=1024, 10^2=100, 2^7=128, 2^3=8, 3^2=9 We have: 10^47 = (10^3)^15 * 10^2 < (2^10)^15 * 2^7 = 2^157 => 10^47 < 2^157 2^157 = (2^11)^14 * 2^3 < (3^7)^14 * 3^2 = 3^100 Therefore: 10^47 < 2^157 < 3^100 So 3^100 has at least 48 digits

@theresaotoole9141 16 дней назад

It's 3 multiplied by 3 a 100 times plus ''an extra digit is added every 2 and a small bit times'' - meaning the answer is approx less than 50 - so 47 or 48 and when I checked it was ~48. Took 15 sec in my head.

@Haha_Yes0 25 дней назад

log(3¹⁰⁰) =100log(3) =100×0.4771 =[47.71] =47+1 Thus 48 digits are present in 3¹⁰⁰ (by using characteristic and mantissa)

@barik111_ 24 дня назад

I didn't understand bro , why is 1 added in the last step for ? I don't know matissa .

@ald6980 25 дней назад

3^7> 2*10^3; 2^10>10^3; 3^70>(2^10)(10^30)>10^33; 3^100>10^(33*100/70)>10^47 --> at least 48 digits. 3^100 less than 51 digit. So options C and D excluded and answer is 48. If option C were 50 than you should work a little bit more: 3^100=9^50=(10^50)*(0.9^50). If 0.9^50 answer is 48. 0.9^50 = ((1-1/10)^10)^5 < e^(-5)

@eliteteamkiller319 13 дней назад

Here’s how I did it: 3^1 has 1 digit. 3^2 has 1 digit 3^3 has 2 digits 3^4 has 2 digits 3^5 and 3^6 has 3 digits. But I also noticed that the number was going down relative to the nearest larger 10s place. So I figured for every power of then there are just less than half as many digits as there are of the power in question, which means it has to be either 47 or 48, because these are near half of 100. Then I figured the smallest number would be too easy and guessed B. Afterward I cheated and did 100*log(3) to test :D.

@go_beyond_the_sky 25 дней назад

That's great! , this was my college admission test question, the question was 2^100

@morselclash6414 18 дней назад

w/o calculator (ik someone already posted this but here are my thoughts): log(3^100)=100log(3) sqrt(10) is slightly larger than 3 (3^2=9log(3) 1/2>log(3) 50>100log(3) so now we have our upper bound. We can say with further accuracy that since log(3)

@DaHaiZhu 25 дней назад

18. The answer is 18. He's holding a cat. A cat has 18 toes (digits). Therefore the answer is 18. Can I run for POTUS now?

@user-ox4ii2bw6x 26 дней назад

my function is basically identical, floor(log(3^100))+1 -> floor(100*log(3))+1 -> floor(47.7...)+1 -> 47+1 -> 48

@landfillbaby 25 дней назад

oh oops i just wrote out the same thing in another comment. i'll delete mine

@Mrtheunnameable 25 дней назад

Floor + 1 is a cieling

@Serghey_83 24 дня назад

@@landfillbaby Не надо. Зачем?

@KingGisInDaHouse 26 дней назад

whatever floor(100*log 3)+1 is.

@major__kong 26 дней назад

Isn't floor(x) + 1 just ceiling(x)?

@zachansen8293 26 дней назад

@@major__kong it's different for integers

@ericgoldman7533 26 дней назад

@@major__kong Like @zachansen8293 says, they are the same only for numbers with a non-zero value to the right of the decimal. For example, floor(3.21)+1 = ceiling(3.21) = 4, but floor(3)+1 = 4 while ceiling(3) = 3

@RyanLewis-Johnson-wq6xs 25 дней назад

Input 3^100 = 515377520732011331036461129765621272702107522001 Result True Logarithmic form 100 log(3, 3) = log(3, 515377520732011331036461129765621272702107522001) B) 48 digits final answer

@kelvinle8662 23 дня назад

3^x goes up an order of magnitude roughly every 2nd interation of ×3 and sometimes every third. Which means 46 or 47 digit is the only reasonable possible answerr.

@NaplesFC 24 дня назад

a=3**100 b=str(a) print(len(b)) Done 😂

@kfitch42 26 дней назад

Just a bit of multiple choice/guess analysis can work pretty well on this one. Clearly we can see that A and B are related. C and D both look like they are trying to catch someone who thinks this is trivial and trying to match numbers in the question to numbers in the answer. If this were something like how many digits are in 10^x, then C and D would be interesting... but applying a tiny modicum of math knowledge you can say those are VERY unlikely. Looking at A vs B we can say it is either trying to catch someone inappropriately rounding down or rounding up. Again applying a tiny modicum of math knowledge we can say that counting digits usually involves rounding up (as the little example in the video showed), so we probably want the value that is 1 larger. As a result we can make a very fast moderate confidence guess that B is the answer. Of course if you want high confidence then you need to apply a bit more effort (as shown in the video). But if you are running out of time on a test ... this type of meta-analysis of the choices can help.

@ronald3836 24 дня назад

The question would have been a lot better without 47 (e.g. replace with 31) and with the other two answers not being round numbers.

@stevenwilson5556 10 дней назад

Great, great, great video!! Thank you

@rogerkearns8094 25 дней назад

Over sixty years ago, I learned the logs base ten of the numbers 2 to 9 and at last this has come in useful.

@jerry2357 23 дня назад

Using logarithms to the base 10 (common logarithms): log(3^100)=100*log(3). Looking up log(3)=0.477, so 100*log(3)=47.7. Round up to get the number of digits. Thus 3^100 must have 48 digits. I'm old, and wasn't allowed to use a calculator in my O level exams, so I had to use log tables instead. So using logs is automatic for me, for this sort of question.

@jasejj 22 дня назад

Depends on the base of the number system you present the answer in. In base-3 arithmetic it will have 101 digits.

@TrossachsPhoto 24 дня назад

3^100 =3 ^ 2(50) =9^50 If this were 10^50, it would be 51 digits long. 9 is less than 10, so number of digits < 51 which means it's not 100 digits long. 9 is close to 10, so the right number feels like 48. For speed calculation multiple choice, I'd be happy with my gut feeling

@jsfbr 17 дней назад

(1) I thought the number should be a multiple of 3, so 47 and 100 were out. (2) Then I thought 300 would be extremely high, so I concluded it must be 48. (3) Then I realized that reasoning # 1 has nothing to do with the number of digits of a multiple of 3! (4) Although my reasoning was wrong, I got the correct answer. 😊

@abhishekam5525 21 день назад

can also do this by binomial ( probably not exactly) 3^100 = 9 ^ 50 (10-1)^50 then proceed with expansion

@RyanLewis-Johnson-wq6xs 25 дней назад

Log(541)=Log(5.41)+2

@kznsq77 24 дня назад

I solved it the same way. The log10 works here because it exactly shows us how many tens we have in this number. 10 here is because we have a decimal system. If the system is hexadecimal, for example, we will use log16

@michaelbyrd1674 26 дней назад

I think a better way to think about it is floor(log_10 (x)) +1. Suppose log_10 (x)=n, a natural number, then the ceiling of n would still be n, but x would have n+1 digits.

@nicholasscott3287 25 дней назад

Here's how i did it: 3x3 = 9, 9 is about 10. 3^2^50 is therefore about 10^50, so it has about 50 digits. 48 is the closest to 50 from the options given, so that's the one I picked.

@Viper36P 25 дней назад

It could have any number of digits you like, depending on the base of the number system.

@sashagornostay2188 25 дней назад

Blind: 3^100 = x log(3^100) base 10 = log(x) base 10 100×log(3) base 10 = log(x) base 10 100×0.4771212... = log(x) base 10 47.71212...=log(x) base 10 10^47.71212...=x 47 digits, closer to 48 than to 46

@RyanLewis-Johnson-wq6xs 25 дней назад

Log(10)=1 Log(100)=2 Log(1000)=3

@bagaj6431 17 дней назад

I don't wanna be picky, but I think there is a (small) mistake in the reasoning. 10 doesn't have 1 digit, it has 2: 1 AND 0. The number of digits is not ceil(log_10(x)), but this number +1. It means the final answer you gave (48) is not the number of digits of 3^100, but this number-1. Proof: Formalization of number of digits: #digits(10)=2=log_10(10)+1 for a general number x: #digits(x)=ceil(log_10(x)) +1 Then #digits(3^100) = ceil(47.7)+1 = 48+1 = 49, NOT 48.

@flothus 17 дней назад

Try to think about how the digit count changes at every step of the geometric progression (checking left and right of it) and how you can express that in your generalisation. E.g. check log(10), log(10-1), and log(10+1)

@yurenchu 14 дней назад

10^47 is a 48-digit number: it is a 1 followed by 47 zeroes. 10^47 is the smallest 48-digt number. 10^48 is the smallest 49-digit number. Therefore, 10^(47.7) is a 48-digit number. By the way, the correct formula is #digits(N) = floor(log_10(N)) + 1 for any (positive) integer N .

@vampire_catgirl 18 дней назад

Before watching the video, intuitively it has to be less than 100 because multiplying any number by 10 or less can never add more than one digit, and the first multiplication, 3×3 doesn't even add a digit so it can't reach 100

@JobBouwman 24 дня назад

3^100 = 9^50 = 81^25 ~ 80^24*100 = 2^72*10^26 = 4*(2^10)^7*10^26 ~ 4*10^21*20^26 = 4*10^47 so 48 digits.

@bhuwaneshbisht8195 25 дней назад

Let S = {4,6,9} and T = {9,10,11,...,1000}. If A={a1+a2+⋯+ak:k∈ N,a1,a2,a3,⋯,ak∈S}, then the sum of all the elements in the set T−A is equal to . A guys please upvote this q so that he will solve it or u can also solve but dont give same texted sol.

@snowjix 25 дней назад

I guessed right, but i think my method was somewhat inefficient. Thought about 100, but only 10^100 would get you there so 3^100 must be less. I went with 48 because i liked the 8, as 18 has three as a factor. Im sure this mess wouldnt work generally, but i find it funny it worked now.

@black_890real3 26 дней назад

I thinked by another form: 3^1=3, 3^2=9, 3^3=27, 3^4=81... if we continue, we always will see that for each pair of power, we have one more digit, then with 50 pairs we would have 50 digits for 3^100.

@gabrieldcf 19 дней назад

100*log10(3) = 47.7121... I knew there was a catch with both 47 and 48 being options in the test! Before watching the video, my guess is that the correct answer is 47. The remainder, 0.7121, probably has to do with the mantissa of the result in scientific notation... so 3^100 ≃ 5.15 * 10^47? EDIT AFTER WATCHING: okay, I forgot to account that a number to the power of 47 in scientific notation will have 48 digits, just like log(1000) = 3 and it still has 4 digits!

@Guishan_Lingyou 23 дня назад

I skipped school the week we learned logs in 10th grade. Now, 30 something years later, I'm enjoying this video about how to use them to solve a problem I don't need to know the answer to ;-)

@kmyc89 26 дней назад

Well, I was like: 3^(2*50) = 9^50. Since 9^50 is less them 10^50, I have already an upper bond of 51 digits (since 10^n have n+1 digits) . From here (in best case still no calculator), how to confirm or cancel aswer A).

@neilmccoy9390 19 дней назад

Without calculator I observe that 3^100=9^50 < 10*50. So # digits is less than 51. So now I have 50% chance of guessing answer instead of 25% chance.

@NoosaHeads 18 дней назад

Brilliant! I'd have been totally lost, but you explained this simply, cleverly and clearly. Thank you so much.

@shivnathbanerjee5868 25 дней назад

Conclusion : Number of digits in a number N is given by: Floor(log base 10( N )) + 1

@tobybartels8426 25 дней назад

If x≥1, then the number of digits of x before the decimal point is [log(x)]+1.

@christhefella 18 дней назад

It's B because: log10(3) = ~0.4771212... x = log10(3) * 100 x = 47,71212... 10^47,71212 has 48 digits, therefore B

@geralynpinto5971 9 дней назад

Great! Thanks so much. It's so nice and fine to learn something new every day.

@JubilantJerry 25 дней назад

I stupidly thought that I'm supposed to solve this without a calculator, and was trying hard to figure out how to choose between 47 and 48 (since I don't know log_10(3) to 2 decimal places)

@InPursuitOfCuriosity 25 дней назад

Nice kitty cat. I did not know where to start for this math question. Thanks for explaining.

@stvp68 16 дней назад

Amazingly simple!

@naygoats955 25 дней назад

3^100=10^x Log10(3^100)=x 100log10(3)=x X=47

@TerjeMathisen 21 день назад

Hmmm... since 3 is close to but less than sqrt(10), we get a bit less than an additional decimal digit for every two powers, so 47 or 48 are the only possible answers among the alternatives given here. I can't immediately see any easy way to pick between them, will I have to watch the video? I do remember that log10(2) is about 0.301 and log(3) is about 0.477, so that would indicate that the final answer would have 47 digits and start with a somewhat large digit, maybe 5-8?

@yurenchu 19 дней назад

10^47 has 48 digits (it's a 1 , followed by 47 zeroes) , not 47 digits. Hence 10^(47.7) also has 48 digits.

@yurenchu 25 дней назад

0:08 Poll: A : 21% B : 36% C : 21% D : 22% So only 15% actually knew the correct answer, 84% just made a random guess, and 1% was drunk and choose D.

@yurenchu 25 дней назад

*chose

@MasterReady12 25 дней назад

How I solved it in my mind was like: The order of magnitude is 100, so digits must be 100. But then I remembered it would be if it was 3 × 10^100. So I thought 10^100 has 100 digits, and 3^2 = 9, which is close to 10, which means I can approximate (10^0.5)^100 to 3^100. Which means 3^100 has order of magnitude of 50, and 48 was closest to 50 so that was my answer. This sounds complicated but it took 20 seconds in my mind.

@excelmaster2496 25 дней назад

3¹ and 3² have 1 digit, so 3¹⁰⁰ has less than 100 digits. 48 is divisible by 3, so it's the nicer answer out of the remaining 2.

@bushwalker6214 19 дней назад

I calculated it on a piece of paper for 2 minutes, w/o a calculator and uncommon knowledge like the ln(3) value. Pretty straightforward using the properties of "e".

@yurenchu 19 дней назад

Not ln(3), but log(3), which (in this video) is the _base ten_ logarithm of 3 . You did 3^100 = 9^50 = (1 - 1/10)^50 * 10^50 ≈ (1/e)^5 * 10^50 = (10^50)/(e^5) = 10^[50 - 5/ln(10)] , didn't you? How did you quickly evaluate e^5 or ln(10) ?

@bushwalker6214 18 дней назад

@@yurenchu Yep, misprinted regarding ln and lg. e^5 = (2.7*2.7)^2*2.7 = 7.3^2*2.7 = 140 approximately. Hence e^-5 = 7*10-3.

@yurenchu 18 дней назад

@@bushwalker6214 Ah, nice. Took me quite a while to evaluate (7.3)² * 2.7 = (49 + 4.2 + 0.09) * 2.7 = (53.29)*2.7 ≈ 53.3 * 2.7 = (28 + 25.3)(28 - 25.3) = 28² - 25.3² = 784 - (625 + 15 + 0.09) = 784 - 640.09 = 143.91 but you probably just did (7.3)² * 2.7 ≈ (7.5)² * 2.5 = 56.25 * 2.5 ≈ 14 * 10 = 140 .

@gamemakingkirb667 26 дней назад

I saw that 3^2 < 10 -> (3^2)^50 = 3^100 < 51 digits. Either way it’s hard without a calculator or writing out some arithmetic

@johnfox2483 24 дня назад

Some basis math, however ... how to find log(3) ? Of course, there were tables ...who of older viewers remember where they have one? In library far away. And where are our calculators? So we probably start with excel, calculators, or Wolfram or Google ... And can calculate 3^100 directly. So maybe next time ask how many digits 3^1000 have ...

@yoavmor9002 25 дней назад

10^30 < 2^100 < 3^100 < 4^100 = 2^200 < 1.7 × 10^60 That's as far as I could go lol

@michaelblankenau6598 17 дней назад

Dang … I was already up to 3 to the 17 th … felt confident I’d get there … eventually.

@zahirulhuq6233 26 дней назад

Can you take class on number theory

@CaptBackwards 25 дней назад

Technically just one digit 3^100 is just a single digit with a high power

@leventunal313 25 дней назад

what do you think for the solving of 1^100? its result is: 0 if you follow your method, is it true?

@8bit_pineapple 17 дней назад

Got this right but with weird logic. I said it has one hundred digits in base 3, and therefore fifty digits in base 9. A bit less in base 10, how much less? I guess it will be proportional to log10(3), calculator says thats 0.477 so every digit in base 3 is "worth" 0.477 dogits in base 10, so thats 47.7 but i have to round up so I have enough digits

@yurenchu 14 дней назад

Note: 3^100 has _one hundred and one_ digits in base 3 , and _fifty-one_ digits in base 9 .

@qubyy1714 25 дней назад

So the formula you would use based off the first part on logs would be 1+floor(log(n)) = number of digits when n>0 (estimate before he starts explanation) edit: realised it could be ceiling(log(n)) when n>0 lol

@yurenchu 25 дней назад

@@qubyy1714 ceiling(log(n)) doesn't work when n is a power of 10. The correct formula is floor(log(n))+1 , for integer n≥1 .

@qubyy1714 25 дней назад

@yurenchu it does work because when n is a power of 10 then it will be log(10^n) which is n and that is just how many zero's there are so you add 1 to get the total digits also I meant to say integers because then it doesn't work for non int but if you talk about integers then n>0 = n=>1 scince decimals aren't integers

@yurenchu 25 дней назад

@@qubyy1714 ceiling(log(n)) doesn't work when n is a power of 10. For examples: n=200 : ceiling(log(200)) = ceiling(2.301...) = 3 , which is correct because 200 has three digits, but n=1000 : ceiling(log(1000)) = ceiling(3) = 3 , which is incorrect because 1000 has *four* digits. The formula floor(log(n))+1 can also be used for non-integer (real) n to give the number of digits _before the decimal point_ , but only if n≥1 . That's why I didn't keep the "n>0".

@qubyy1714 25 дней назад

@yurenchu my bad I noticed that thanks for correcting

@yurenchu 25 дней назад

@@qubyy1714 no problem! :-)

@mmmhorsesteaks 25 дней назад

I did len(str(3**100)) in python. Which, I admit, is straight up cheating but was very fast to do and knowing your 'tools' can also be very useful.

@GuitarFR34K121 13 дней назад

So in theory you can also calculate how many bits would be needed to represent such a number by using a logarithm of base 2?

@maxmustermann3938 11 дней назад

You can also do that in praxis. 159 bits in this case if you use unsigned integer representation.

@LaussseTheCat 26 дней назад

"These are what digits are" Okay "Now, logarithms..."

@GeorgeAlexanderOz 15 дней назад

How is 47.7 digits 48 digits? Where inside the 48th digit do you exactly place the separator?

@GeorgeAlexanderOz 15 дней назад

Checked it, you are right.

@groovermctoober4508 13 часов назад

4. The question is how many digits does 3^100 have, not how many digits does the expanded value of 3^100 have. You need to express yourself more precisely!

@michelebrun613 20 дней назад

But since it is 47.7 it does not reach 48. I would say 48 (expecially after having checked in Mathematica .....)

@smithjohn383 24 дня назад

Interesting. Would've probably been more interesting if I had any clue what 'log' is.

@harshitrai5433 25 дней назад

Find the characteristic of the number floor it and add 1

@snowsearch 25 дней назад

seeing him shaved really made my day

@hikari1690 25 дней назад

Aaaaaargh, where cat?! I watched the whole thing waiting for cat to come back and only got knowledge!!! I was curious how to figure if it's 47 or 48 without a calculator though. Didn't expect the answer would be gitgud 😢

@tontonbeber4555 16 дней назад

How do you know log(3) = 0.477 if you don't know it by mind and have no calculator. Is it possible to calculate it by hand ?

@yurenchu 14 дней назад

Yes, it is possible; although this method below takes some effort and is not always going to work. 3^4 = 81 3^8 = 81*81 = 6400 + 160 + 1 = 6561 3^9 = 3*6561 = 19683 ≈ 20000 log₁₀(3^9) = log₁₀(20000) + log₁₀(19683/20000) *9*log₁₀(3)* = log₁₀(2*10^4) + log₁₀(0.98415) *9*log₁₀(3)* = log₁₀(10^4) + log₁₀(2) + log₁₀(1 - 0.01585) *9*log₁₀(3)* = log₁₀(10^4) + log₁₀(2) + ln(1 - 0.01585)/ln(10) ... note: for small x , ln(1-x) ≈ -x ; and 2 < ln(10) < 3 ... 4 + 0.301 + (-0.01585)/2 < *9*log₁₀(3)* < 4 + 0.302 + (-0.01585)/3 4 + 0.301 + (-0.007925) < *9*log₁₀(3)* < 4 + 0.302 + (-0.005283) 4.293 < *9*log₁₀(3)* < 4.297 (4.293)/9 < *log₁₀(3)* < (4.297)/9 0.4770 < *log₁₀(3)* < 0.4775 ==> *log₁₀(3)* ≈ 0.477 But now you're gonna ask: But wait, how do you know log(2) = 0.301 ? 2^10 = 1024 log₁₀(2^10) = log₁₀(1000) + log₁₀(1.024) 10*log₁₀(2) = 3 + ln(1.024)/ln(10) ... note: for small x , ln(1+x) ≈ x , and 2 < ln(10) < 3 ... 3 + (0.024)/3 < 10*log₁₀(2) < 3 + (0.024)/2 3.008 < 10*log₁₀(2) < 3.012 0.3008 < log₁₀(2) < 0.3012 ==> log₁₀(2) ≈ 0.301

@johnvonhorn2942 26 дней назад

Without watching the video my initial guess is: x^10 = 3 ^ 100 and then take log base 10 of x to find the number of digits?

@charcolew 23 дня назад

Does 3 to the 100th power equal 9 to the 50th power?

@MrAlony01 21 день назад

😂😂 so funny to watch people argue over remembering log value, bruh in my country its mendatory in 11th standard (around 16 years) to learn some of these values because it makes calculations fast and effective, even it is used in a lot of places in physical chemistry example for pH concentration, gibbs formula, much more tbh, its like people dont wanna learn 1/0 is infinity and wanna argue about finding its value using complex methods, maths is all about utilisation of pre made concepts and making complex things simpler... but damn these people really wanna show how they have the extraordinary ability of manipulating powers and going through each and every option just to end up wasting more time... in competitive exams and olympiads you gotta be fast and accurate, if you lack any one you fall behind and end up throwing everything on your fate but at the end it was on you not on your fate bro

@realdealsd 17 дней назад

It just wasn't something we had to memorize in the US. If we needed to know the approximate value of a log either we were provided a table of values or we were allowed to use a calculator. With the exception if pi, memorizing any approximation of an irrational number was deemed a waste of time. Even with pi, 3.14 or 22/7 were considered adequate enough.

@lovalmidas 22 дня назад

Try computing lg 3 (log10 3) without the use of a calculator. :D

@yurenchu 20 дней назад

3^4 = 81 3^8 = 81*81 = 6400 + 160 + 1 = 6561 3^9 = 3*6561 = 19683 ≈ 20000 log₁₀(3^9) ≈ log₁₀(20000) 9*log₁₀(3) ≈ log₁₀(10000) + log₁₀(2) 9*log₁₀(3) ≈ 4 + 0.301 log₁₀(3) ≈ (4.301)/9 log₁₀(3) ≈ 0.478