Firstly, your videos are always entertaining, informative and educational. I would also like to congratulate you on the clarity of your hand writing. I've seen videos of mathematical content creators whose handwriting bears more similarity to ancient egyptian hieroglyphics! Should you ever wish to take sponsorship for your channel, consider hats!
Great video! I solved it without directly using the formula, but in a similar way to how the formula would work: If you let 2^x = 10, you can easily tell the number of digits of a power of 10 by doing (power of 10) + 1. so if 2^x = 10, then xlog(2) = log(10) = 1, so x = 1/log(2). 2^(1/log(2)) = 10, and using laws of exponents to make this equal to 2^100 we need to do 100/1/log(2), and this simplifies to 100log(2), which is roughly 30. Therefore (2^(1/log(2)))^30 = 10^30, which has 31 digits as this is the (power of 10) + 1.
Prior to watching the video: 2^100 = 4^50 or 16^25 = (1.6 x 10)^25 = 10^25 * 1.6^25 26 + [pow of 10 of(M = 1.6^25)] M = (2 - .4)^25 = 2^25 - 2^24*.4 + 2^23*.4^2+ … -.4^25 Yeah idk. maybe just calculating it directly is easier, if you restrict yourself to not be able to use logs. 2^100 = (2^10)^10 = 1024^10 = (10^3 + 24)^10 = 10^30 = 30 + 1 digits = 31. 24 is 2.4% of 1000, and 1024 is thereby 1.024 of 1000. Luckily I remember from doing percentage growth over time, yearly compounding, that 7% extra is nearly doubling every 10 years: 1.07^10 is below 2. So the 1000 is not even close to doubled with only 2.4%, meanwhile it must be multiplied by at least 10 to reach the next digit. So the +24 indeed does not affect the digit amt.
2^(10 k) > 10^(3 k) is an approximation that is off by a factor of 1.024 If you try to find the smallest power of 2 that satisfies 2^(10 k - 1) > 10^(3 k) you get k = 30 or 2^299 > 10^90 which is super close when comparing ratios. So an even better approximation that can be used when dealing with larger powers of 2 is 2^(299 k) > 10^(90 k)
Here is a method with no calculator and no memorized numbers, just logic: 2^100 = 1024^10 = 10^(log_10(1024^10)) = 10^(10*log_10(1024)) log_10(1024) = 3 + log_10(1.024) = 3 + log_10(1 + k) where k = 0.024 We know that 0 < log_10(1 + x) < x for all x > 0 So, choosing x = k = 0.024 0 < log_10(1 + 0.024) < 0.024 Now just reverse calculate 3 < 3 + log_10(1.024) < 3.024 3 < log_10(1024) < 3.024 30 < 10 * log_10(1024) < 30.24 30 < log_10(1024^10) < 30.24 10^30 < 10^(log_10(1024^10)) < 10^(30.24) 10^30 < 1024^10 < 10^(30.24) 10^30 < 2^100 < 10^(30.24) The number of digits in 10^x is floor(x)+1 by definition of the decimal notation. So, you get 31-digit number < 2^100 < 31-digit number and *hence 2^100 must also have exactly 31 digits.*
Maybe try to somehow leverage 2^x * 7^(2 x) = 98^x < 100^x = 10^(2 x) That's where I would start, since we already know how to compute digits from some powers of 2 to some powers of 10 ? So this should in theory give you some upper bound on the digits. So 2*61 * 7^122 * 7 < 7 * 10^122 2^60 * 2 * 7^123 < 7 * 10^122 (10^3)^6 * 2 * 7^123 ≈ 7 * 10^122 10^18 * 2 * 7^123 ≈ 7 * 10^122 2 * 7^123 ≈ 7 * 10^104 7^123 ≈ (7/2) * 10^104 7^123 ≈ 3.5 * 10^104 7^123 ≈ 105 digit number ? This is an estimate that should be close, but is not guaranteed to be correct.
You could use a simpler method: Let N=2¹⁰⁰ Then logN=100log2= 0.30×100=30 Hence N=10³⁰ If we notice a general trend, 10¹=10= 2 digits 10²=100= 3 digits Hence 10³⁰ has 31 digits
2^100 = (2^10)^10 = 1024^10. Multiply and divide by 1000^10: (1024/1000)^10 * 1000^10 This is the same as (1.024)^10 * 1000^10 = (1.024)^10 * 10^30 (1+x)^n ~ 1+nx (for small x) So, (1.024)^10 * 10^30 = 1+0.24 * 10^30 = 1.24 * 10^30 and this number has 31 digits.
Someone may know that 2^10 = 1024 ≈ 10^3 (if you learned about computer science, 1GB = 1000MB but it's actually 1024MB.) so, 2^100 = (2^10)^10 ≈ (10^3)^10 = 10^30. and 10^33 = (10^3)^11 ≈ (2^10)^11 = 2^110. so digits of 2^100 is 30~33. That's all what I can guess
Your logic is incorrect, since what you proved is that 2^100 > 10^30 and 2^110 > 10^33 that does not really mean that 2^100 < 10^33 Both the ≈ signs should have been > signs since 1024 > 1000 Also, 10^30 is the smallest 31 digit number and 10^33 is the smallest 34 digit number. So even applying your logic, which is incorrect, 2^100 has ~ 31 - 33 digits.
well the formular is interesting yes because then we can use even more log properties and turn the multiplication into addition and then use a slide rule, which no one teaches or uses any more. The point though is that we have ways to handle that question without having to resort to memorization or using a calculator.
Very nice video. I think about the sequence of 2 up to power 9 and there is pattern, from 1-3 (3 integers) one digit, 4-6 (3 integers) two digits, 7-9 (3 integers) three digits. Still this is not linear. To increase the number of digits from 4 to 5 you will have to go from 10 to 13, which 4 integers. So, I assumed that for every 13 integers the number grows by 4 digits. Then 100/13*4=30.769. I was really close but I don't know if my method could work in other combinations. Thank you!!
It's fairly easy to see. Smallest number with 3 digits is 100, log10(100) = 2, Smallest number of 4 digits is 1000, log10(1000) = 3, so any number whose log10 is between 2 inclusive and 3 exclusive (i.e. the floor of the log = 2) will have 3 digits, because log is an always increasing function. This can be generalized to any size. b*log10(a) is just another way to write log10(a^b), or the log base 10 of the original number.
The formula you use is a special case of the generic formula: the number of digits of N is floor(log10(N))+1, it just so happens that log10(a^b) = b log10(a)
2^100 = 2^(log2 10 * 100/log2 10) = (2^log2 10)^(100/log2 10) [Laws of Exponent] = 10^(100/log2 10) = 10^(100 * log10 2) Approximately =10^(100*0.3010) =10^30.1 =10^30 We know 10^0 = 1 which has one digit, 10^1 = 10 which has 2 digits, similarly 10^x has x + 1 digits. 2^100 ~ 10^30 Has 31 digits.
My first thought was, What base ? For octal, 100/3 or 34, hex 100/4 = 25. For decimal, 2^10 is 1024 and 2^20 is approximately million 1,000,000 giving 4 decimal digits for the first 10, and 3 more for subsequent 10, giving 31. (1.02^10 is less 9)
I have estimated out of my head: 2^(10*10) = 1024^10 As the effect of 24 is rather small, only 2.4 %, per multiplication of 1024 you and 3 more digits: 1024 has 4 digits plus 9*3 digits = 31 digits.
Tried to write 2 in the power of 10. So what is x if 10^x = 2 x = log 2 So the problem becomes how many digits does (10^log10)^100 have. Which is equal to 10 to the power of some 100×0.3010 …. At which point it should be obvious.
In physical chemistry exercises (55 years ago) we memorized log(2) = .30103 as a cartoon. One round like a head, and then 3 for the ear, 0 for the eye, 1 for the nose, 0 for the other eye and 3 for the other ear. At that time, it was a rule that the base of the logarithm (if not indicated) was assumed to be 10. If the base was e, then we used the label ln.😎
Hi, great video😊 I got a suggestion for a video, I saw this math problem and I really can’t solve it, though it is proof-based so I am not sure if it would fit your video style, anyway here it is: Prove that the equation 7272772272 + n = n^5 doesn’t have an integer solution. Please 🙏, it would really help if you could solve it🙏🙏🙏
Btw, this is the first/easiest question on a super difficult olympiad test, If you are interested, I will send you the whole thing on email @PrimeNewtons👍
This is an immediate corollary of Fermat's (little) Theorem. If n is an integer thrn 5 | n^5-n, but 7272772272 is not a multiple of 5. Therefore the equation does nor have an integer solution.
Firstly, your videos are always entertaining, informative and educational.
I would also like to congratulate you on the clarity of your hand writing. I've seen videos of mathematical content creators whose handwriting bears more similarity to ancient egyptian hieroglyphics!
Should you ever wish to take sponsorship for your channel, consider hats!
@@robertsandy3794
indeed, the man sure can write neatly.
Great video! I solved it without directly using the formula, but in a similar way to how the formula would work:
If you let 2^x = 10, you can easily tell the number of digits of a power of 10 by doing (power of 10) + 1.
so if 2^x = 10, then xlog(2) = log(10) = 1, so x = 1/log(2).
2^(1/log(2)) = 10, and using laws of exponents to make this equal to 2^100 we need to do 100/1/log(2), and this simplifies to 100log(2), which is roughly 30. Therefore (2^(1/log(2)))^30 = 10^30, which has 31 digits as this is the (power of 10) + 1.
It's easier to read with commas: 1,267,650,600,228,229,401,496,703,205,376
Prior to watching the video:
2^100 = 4^50 or 16^25 =
(1.6 x 10)^25
= 10^25 * 1.6^25
26 + [pow of 10 of(M = 1.6^25)]
M = (2 - .4)^25
= 2^25 - 2^24*.4 + 2^23*.4^2+ … -.4^25
Yeah idk. maybe just calculating it directly is easier, if you restrict yourself to not be able to use logs. 2^100 = (2^10)^10
= 1024^10
= (10^3 + 24)^10
= 10^30 = 30 + 1 digits = 31.
24 is 2.4% of 1000, and 1024 is thereby 1.024 of 1000. Luckily I remember from doing percentage growth over time, yearly compounding, that 7% extra is nearly doubling every 10 years: 1.07^10 is below 2. So the 1000 is not even close to doubled with only 2.4%, meanwhile it must be multiplied by at least 10 to reach the next digit. So the +24 indeed does not affect the digit amt.
2^(10 k) > 10^(3 k)
is an approximation that is off by a factor of 1.024
If you try to find the smallest power of 2 that satisfies
2^(10 k - 1) > 10^(3 k)
you get k = 30 or
2^299 > 10^90
which is super close when comparing ratios.
So an even better approximation that can be used when dealing with larger powers of 2 is
2^(299 k) > 10^(90 k)
Simple, Short and Sweet. Love it.
Here is a method with no calculator and no memorized numbers, just logic:
2^100 = 1024^10 = 10^(log_10(1024^10)) = 10^(10*log_10(1024))
log_10(1024) = 3 + log_10(1.024) = 3 + log_10(1 + k)
where k = 0.024
We know that 0 < log_10(1 + x) < x for all x > 0
So, choosing x = k = 0.024
0 < log_10(1 + 0.024) < 0.024
Now just reverse calculate
3 < 3 + log_10(1.024) < 3.024
3 < log_10(1024) < 3.024
30 < 10 * log_10(1024) < 30.24
30 < log_10(1024^10) < 30.24
10^30 < 10^(log_10(1024^10)) < 10^(30.24)
10^30 < 1024^10 < 10^(30.24)
10^30 < 2^100 < 10^(30.24)
The number of digits in 10^x is floor(x)+1 by definition of the decimal notation. So, you get
31-digit number < 2^100 < 31-digit number
and *hence 2^100 must also have exactly 31 digits.*
Cool. I'll see how to adapt it for 7^123
Maybe try to somehow leverage
2^x * 7^(2 x) = 98^x < 100^x = 10^(2 x)
That's where I would start, since we already know how to compute digits from some powers of 2 to some powers of 10 ? So this should in theory give you some upper bound on the digits.
So
2*61 * 7^122 * 7 < 7 * 10^122
2^60 * 2 * 7^123 < 7 * 10^122
(10^3)^6 * 2 * 7^123 ≈ 7 * 10^122
10^18 * 2 * 7^123 ≈ 7 * 10^122
2 * 7^123 ≈ 7 * 10^104
7^123 ≈ (7/2) * 10^104
7^123 ≈ 3.5 * 10^104
7^123 ≈ 105 digit number ?
This is an estimate that should be close, but is not guaranteed to be correct.
You can leverage the approximations
10^3 = 1000 < 1024 = 2^10
2^93 = 9.9035 * 10^27 < 10^28
So
10^3 < 2^10
2^93 < 10^28
Trying to combine
10^27 * 2^3 < 2^90 * 2^3 < 10^28
8 * 10^27 < 2^93 < 10^28
*28-digit number* (that is 80% of the way to being a 29-digit number) < 2^93 < 29-digit number *(smallest 29-digit number)*
2 * 2^61 = 2^62 = (2^93)^(2/3)
(8 * 10^27)^(2/3) < 2^62 < (10^28)^(2/3)
4 * 10^18 < 2^62 < 10^18 * 10^(2/3)
2 * 10^18 < 2^61 < 10^(2/3) / 2 * 10^18
2 * 10^18 < 2^61 < 100^(1/3) / 8^(1/3) * 10^18
2 * 10^18 < 2^61 < 12.5^(1/3) * 10^18
2 * 10^18 < 2^61 < 4 * 10^18
Since
2^x * 7^(2 x) = 98^x < 100^x = 10^(2 x)
2^61 * 7^122 < 10^122
7 * 2^61 * 7^122 < 7 * 10^122
(2^61) * 7^123 < 7 * 10^122
(2 * 10^18) * 7^123 < (2^61) * 7^123 < 7 * 10^122
10^18 * 7^123 < 3.5 * 10^122
7^123 < 3.5 * 10^104
num_digits(7^123) (2^14)^24 * 7^3 = 2^336 * 343
7^123 > 2^336 * 343
7^123 > 2^330 * 64 * 343
7^123 > (2^10)^33 * 21952
7^123 > (10^3)^33 * 21952
7^123 > 10^99 * 2.1952 * 10^4
7^123 > 10^99 * 2.1952 * 10^4
7^123 > 2.1952 * 10^103
num_digits(7^123) >= num_digits(2.1952 * 10^103)
num_digits(7^123) >= 104
So best that I could estimate is *104
You could use a simpler method:
Let N=2¹⁰⁰
Then logN=100log2= 0.30×100=30
Hence N=10³⁰
If we notice a general trend, 10¹=10= 2 digits
10²=100= 3 digits
Hence 10³⁰ has 31 digits
@@PrimeNewtons By the method i did in reply section, let N=7¹²³
logN=124log7 =124×0.84 =104.79...
Hence N=10¹⁰⁴... So number of digits is 105.
2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰
(1024)¹⁰ = 10³⁰(1.024)¹⁰
=> 31 digits
curiosity
(1.024)⁹⁷ < 10 < (1.024)⁹⁸
(1.024)^10=((1+0.024)^(1/0.024))^0.24 which is approximately e^0.24. To be technically correct, (1+x)^(1/x)
@@johnlv12 thank you so much! I will save your comment. (1 + a/x)ˣ ≤ eᵃ is a great trick!!
2^100 = (2^10)^10 = 1024^10.
Multiply and divide by 1000^10: (1024/1000)^10 * 1000^10
This is the same as (1.024)^10 * 1000^10 = (1.024)^10 * 10^30
(1+x)^n ~ 1+nx (for small x)
So, (1.024)^10 * 10^30 = 1+0.24 * 10^30 = 1.24 * 10^30 and this number has 31 digits.
Someone may know that 2^10 = 1024 ≈ 10^3 (if you learned about computer science, 1GB = 1000MB but it's actually 1024MB.)
so, 2^100 = (2^10)^10 ≈ (10^3)^10 = 10^30.
and 10^33 = (10^3)^11 ≈ (2^10)^11 = 2^110.
so digits of 2^100 is 30~33. That's all what I can guess
2¹⁰² = 2²*2¹⁰⁰ ≅ 4*10³⁰, which has 31 digits.
Omg
it's 31~34, not 30~33. 10^n is n+1 digits I missed it.
Goal: rewrite 2¹⁰⁰ as 10^x
2^x = 10
x = log_2(10)
x= 1/log(2) ≈ 1/0.3 ≈ 3.33
(2^3.33)^x=10^x
(2^3.33)^(100/3.33)=10^x
10^(100/3.33)=10^x
100/3.33 = x
x ≈ 30
x+1 = 31
Also, 1GB isnt 1024MB right? That's a GiB
Your logic is incorrect, since what you proved is that
2^100 > 10^30
and
2^110 > 10^33
that does not really mean that 2^100 < 10^33
Both the ≈ signs should have been > signs since 1024 > 1000
Also, 10^30 is the smallest 31 digit number and 10^33 is the smallest 34 digit number. So even applying your logic, which is incorrect, 2^100 has ~ 31 - 33 digits.
Need the number theory one. Also a video on proof for the formula
Very easy
well the formular is interesting yes because then we can use even more log properties and turn the multiplication into addition and then use a slide rule, which no one teaches or uses any more. The point though is that we have ways to handle that question without having to resort to memorization or using a calculator.
Let N=2¹⁰⁰
Then logN=100log2= 0.30×100=30
Hence N=10³⁰
If we notice a general trend, 10¹=10= 2 digits
10²=100= 3 digits
Hence 10³⁰ has 31 digits
2^10 has 3 digit, so 2^100=(2^10)^10 has more than 3x10=30 digit. 😊
2^10 has 4 digits
Very nice video. I think about the sequence of 2 up to power 9 and there is pattern, from 1-3 (3 integers) one digit, 4-6 (3 integers) two digits, 7-9 (3 integers) three digits. Still this is not linear. To increase the number of digits from 4 to 5 you will have to go from 10 to 13, which 4 integers. So, I assumed that for every 13 integers the number grows by 4 digits. Then 100/13*4=30.769. I was really close but I don't know if my method could work in other combinations. Thank you!!
I thought you would prove the formula first and then use the formula
It's fairly easy to see. Smallest number with 3 digits is 100, log10(100) = 2, Smallest number of 4 digits is 1000, log10(1000) = 3, so any number whose log10 is between 2 inclusive and 3 exclusive (i.e. the floor of the log = 2) will have 3 digits, because log is an always increasing function. This can be generalized to any size.
b*log10(a) is just another way to write log10(a^b), or the log base 10 of the original number.
@@MadaraUchihaSecondRikudo -- Don't write it that way. Try log_10(100) or log[base 10](100), for example.
@@robertveith6383 I think my meaning was fairly obvious.
@@robertveith6383 if you knew what he meant, it doesn't matter
The formula you use is a special case of the generic formula: the number of digits of N is floor(log10(N))+1, it just so happens that log10(a^b) = b log10(a)
Yours is a "special case" too. Because you already know N.
2^100
= 2^(log2 10 * 100/log2 10)
= (2^log2 10)^(100/log2 10) [Laws of Exponent]
= 10^(100/log2 10)
= 10^(100 * log10 2)
Approximately
=10^(100*0.3010)
=10^30.1
=10^30
We know 10^0 = 1 which has one digit, 10^1 = 10 which has 2 digits, similarly 10^x has x + 1 digits.
2^100 ~ 10^30
Has 31 digits.
My first thought was, What base ? For octal, 100/3 or 34, hex 100/4 = 25. For decimal, 2^10 is 1024 and 2^20 is approximately million 1,000,000 giving 4 decimal digits for the first 10, and 3 more for subsequent 10, giving 31.
(1.02^10 is less 9)
Because you want to write 2 two in the form of a power of 10. Then the problem becomes how many digits does 10^(1/3.010) ^ 100 have.
@@firstolasto1518
The first digit is units which explains why 4 digits is needed for 2^10 and 3 for each additional 2^10.
I have estimated out of my head:
2^(10*10) = 1024^10
As the effect of 24 is rather small, only 2.4 %, per multiplication of 1024 you and 3 more digits:
1024 has 4 digits plus 9*3 digits = 31 digits.
Thanks sir for such informative videos❤❤
First time I learn floor function ❤
Nice but I d like to know why this formula works correctly
2^100=1267650600228229401496703205376 31 digits
I was correct.❤🎉🎉❤
Without seeing the video, it seems very simple to me. 2¹⁰⁰ has 101 (binary) digits, 1 digit '1' followed by 100 digits '0'. 😉
31
1.9*10^30 kg is approximately mass of the Sun
and 2^{100} is a little bit more than half of it
Only works for people in this solar system.
How can we calculate the number of places in the factorial of a number?
I already did a video on that. But I'm planning to do something with Sterling approximation.
For logs are the terms 'characteristic' and 'mantissa' defunct?
HELLO TEACHER, PLEASE HOW TO SOLVE THIS EXERCISE? HOW MANY DIGITS ARE IN THE NUMBER 16^8 X 125^9 THANK YOU SO MUCH
Sir how is the formula derived ?
Hey can you explain whats the reasoning behind the formula?I am curious
Tried to write 2 in the power of 10. So what is x if 10^x = 2
x = log 2
So the problem becomes how many digits does (10^log10)^100 have. Which is equal to 10 to the power of some 100×0.3010 …. At which point it should be obvious.
In physical chemistry exercises (55 years ago) we memorized log(2) = .30103 as a cartoon. One round like a head, and then 3 for the ear, 0 for the eye, 1 for the nose, 0 for the other eye and 3 for the other ear. At that time, it was a rule that the base of the logarithm (if not indicated) was assumed to be 10. If the base was e, then we used the label ln.😎
Yeah, did you also pronounce ln as "log" like we did?
In just about 20 seconds I got the answer and that too without using calculator. #JEEASPIRANTS
This question is in my exam.
2on100 = 10onX then + 1
Number of digits in N is log(N)+1
Well 1 + the floor of log(N) because it will not be rational 90% of the time.
@@maxhagenauer24 Exactly
@@Harrykesh630 -- You should edit your first post by using the three dots by your comment.
Hi, great video😊 I got a suggestion for a video, I saw this math problem and I really can’t solve it, though it is proof-based so I am not sure if it would fit your video style, anyway here it is: Prove that the equation 7272772272 + n = n^5 doesn’t have an integer solution.
Please 🙏, it would really help if you could solve it🙏🙏🙏
Btw, this is the first/easiest question on a super difficult olympiad test, If you are interested, I will send you the whole thing on email @PrimeNewtons👍
This is an immediate corollary of Fermat's (little) Theorem. If n is an integer thrn 5 | n^5-n, but 7272772272 is not a multiple of 5. Therefore the equation does nor have an integer solution.
@@YAWTon thank you!