@@letsthinkcritically but why the hell would anyone rhino of subtracting those two digits anyway?? It comes outnof nowhere..surely there's a waybtonsolve without doing that..wouldnt most ppl be more likely to divide the denominator 8118 by the numerstor 2244 and divide some of the other terms?? I would think so..hope you can respond
@@leif1075 It’s not really out of nowhere. It’s an Olympiad problem, so you’d expect there to be some kind of trick with solving it. The repetition on the bottom should pop out to anyone. On the top, we *almost* have the same repetition, except only in the middle 3 sections. So we try making the first 4 sections involve that same kind of repetition by taking out the first 4 digits from each of them, and it turns out that’s the right way to go about it. It’s not obvious, but I would argue that it is natural once you think of it.
@@Deathranger999 but just becaise it repeats wouldn't make one think of subtracting..wouldn't a lotnof ppl think of dividing those 2 terms like I did..can't you solve it by dividing them instead of subtracting at all?
@@leif1075 I mean all we’re really doing is dividing, just in a clever way that allows us to get to the actual answer. You could certainly try dividing by small factors one after the other, but the problem is that large factor equal to the sum of powers of 10. It’s gonna be really hard to pull that out later if you don’t do it at the beginning. And it may not seem obvious, but it’s not unreasonable. What I’m trying to say is that we have this repetition on the bottom, and something close-ish on the top - how can we make the top look more similar to the bottom? Well, since the top starts with 2244, what would happen if we tried to isolate a term of 2244224422442244000 on the top, since that’s what the term on the bottom is like? And then you’d try that, this would lead to the subtraction, you’d notice the other repeated term, and boom, you’re on your way to solving the problem. So it’s less of just arbitrarily deciding to subtract things, and more trying to isolate similar terms that ends up leading you naturally to doing that subtraction.
There is a mistake in the thumbnail. I mistyped the numerator, and that number should be 2244 8514 8514 8514 627. I accidentally swapped the 1 and 5. Apologies for that!
I have just wasted 9 minutes of my life trying to work out what on Earth you were talking about. Then I read the comments. If you made such a fundamental mistake in writing the number you started with, why not just delete the video and start it again and do it properly?
He doesn't know that. The point was that since the denominator has a 41 as one of its factors, it would be good to check the numerator if it has 41 as well. You can perform long division to see if 41 goes into it easily. And again, the motivation to do that is the fact that we already have it in the denominator
@@karolakkolo123 The Vedic divisibility rule for 41 uses a multiplier m of -4, which is quite manageable here for a quick test: 1) Calculate multiplier m as: m = (1 + 41*9) / 10 - 41 = 370 / 10 = 37 - 41 = -4. 2) Test divisibility of 22673 by 41 as: 41 | 22673 41 | (2267 + (-4 * 3) = 2255 41 | (225 + (-4 * 5) = 205 41 | (20 + (-4 * 5) = 0 true. The Vedic divisibility rules are available for all numbers coprime to 10; so all numbers with 1, 3, 7, or 9 as a ones digit. The multiplier used to calculate m as above varies with that ones digit of the divisor d being tested : divisors d with a ones digit of 1 or 9 use the other while those with a ones digit of 3 or 7 each use themselves. The negative multiplier is typically more convenient when the divisor d being tested ends with 1 or 7; and the positive multiplier typically more convenient for divisors d ending in 3 or 9.
John Cochran (previous comment) is right: there is no need to find the prime factors; the Euclidean algorithm is the most efficient way to solve this problem. Also, the numerator is odd and not divisible by 5, so it is obvious that 2000 remains as a factor after any factors in common with the numerator have been cancelled. However, spotting the common factor of 10^12+10^8+10^4+1 at the start cuts out a lot of work.
If you spot the big common factor then at 4:28 in the video the fraction is n/d with n=2244627, d=8118000. If you know the "rule of 11" (sum of digits with alternating +/- signs is a multiple of 11) then you can spot that 11 is a common factor of n and d; and also if you know the "rule of 9" you can see that 9 is another common factor. But supposing you don't know either of these things, you can just do division with remainder to get d=3n+r_0, with r_0=1384119, and similarly r_j=r_{j+1}+r_{j+2} for j=0,1,2,3,4, with r_1=860508, r_2=523611, r_3=336897, r_4=186714, r_5=150283, r_6=36431. Then r_5=4r_6+r_7, and finally r_6=9r_7 with r_7=4059=9x11x41=g being gcd(n,d). This is the Euclidean algorithm for finding the gcd, and cancelling out this big factor gives the fraction 553/2000 in lowest terms: you can avoid big divisions/multiplications by running the algorithm backwards to find r_5=37g, r_4=46g, etc., and finally n=553g, d=2000g. Even if you don't spot the big common factor at the start, the algorithm works in exactly the same way with the original pair of 19-digit numbers.
What do you use to make videos? I think it would be nice if you linked software and amazon affiliate link to tools, that way you profit from affiliate link and interested lectureres profit from the equipment. Thank you for the videos as always!
I did it differently. A combination of my version after your "power of ten" extraction would have been quickest. (Find the 1000100010001 then long divide). GCF: divide greater by smaller until remainder = 0 (so old remainder becomes new divisor) the last divisor is the GCF Clearly a factor of 1000 can be taken out of the denominator 𝟸𝟸𝟺𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟼𝟸𝟽 𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 ×𝟸𝟶𝟶 𝟷𝟼𝟸𝟹𝟽𝟼𝟸𝟹𝟽𝟼𝟸𝟹𝟽𝟼𝟸𝟹𝟼𝟶𝟶 - 𝟼𝟸𝟷𝟶𝟾𝟿𝟷𝟶𝟾𝟿𝟷𝟶𝟾𝟿𝟷𝟶𝟸𝟽 𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 ×𝟽𝟶 𝟻𝟼𝟾𝟹𝟷𝟼𝟾𝟹𝟷𝟼𝟾𝟹𝟷𝟼𝟾𝟸𝟼𝟶 - 𝟻𝟸𝟽𝟽𝟸𝟸𝟽𝟽𝟸𝟸𝟽𝟽𝟸𝟸𝟽𝟼𝟽 𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 ×𝟼 𝟺𝟾𝟽𝟷𝟸𝟾𝟽𝟷𝟸𝟾𝟽𝟷𝟸𝟾𝟽𝟶𝟾 - 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟸 𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 - 0 𝚐𝚌𝚏 = 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚜𝚘 𝚌𝚊𝚗𝚌𝚎𝚕 𝚝𝚘𝚙 𝚊𝚗𝚍 𝚋𝚘𝚝𝚝𝚘𝚖 denominator = 2𝚡1000 numerator : 𝟸𝟸𝟺𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟼𝟸𝟽 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟻𝟶𝟶 𝟸𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟻𝟶𝟶 - 𝟸𝟷𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟸𝟽 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟻𝟶 𝟸𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟻𝟶 - 𝟷𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟽 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟹 𝟷𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟽 - 0 𝟻𝟻𝟹/𝟸𝟶𝟶𝟶
On the contrary, spotting the large common factor at the start (despite the typo in the video) does involve critical thinking: finding a strategy to solve the problem quickly is the whole point, even if the rest of the solution given here is not the most efficient one.
Lmao, just divide the numerator and denominator by 10^18 u get 2.2/8.1 (since the digits are near I'm not rounding off) u will get .27 as the answer matching the final answer.
8154 - 2244 = 5910 is it not? I think you swapped some digits in the question.
Yes! Sorry for the typo. The original number at the numerator should be 2244 8514 8514 8514 627. Thanks for pointing that out!
@@letsthinkcritically but why the hell would anyone rhino of subtracting those two digits anyway?? It comes outnof nowhere..surely there's a waybtonsolve without doing that..wouldnt most ppl be more likely to divide the denominator 8118 by the numerstor 2244 and divide some of the other terms?? I would think so..hope you can respond
@@leif1075 It’s not really out of nowhere. It’s an Olympiad problem, so you’d expect there to be some kind of trick with solving it. The repetition on the bottom should pop out to anyone. On the top, we *almost* have the same repetition, except only in the middle 3 sections. So we try making the first 4 sections involve that same kind of repetition by taking out the first 4 digits from each of them, and it turns out that’s the right way to go about it. It’s not obvious, but I would argue that it is natural once you think of it.
@@Deathranger999 but just becaise it repeats wouldn't make one think of subtracting..wouldn't a lotnof ppl think of dividing those 2 terms like I did..can't you solve it by dividing them instead of subtracting at all?
@@leif1075 I mean all we’re really doing is dividing, just in a clever way that allows us to get to the actual answer. You could certainly try dividing by small factors one after the other, but the problem is that large factor equal to the sum of powers of 10. It’s gonna be really hard to pull that out later if you don’t do it at the beginning. And it may not seem obvious, but it’s not unreasonable. What I’m trying to say is that we have this repetition on the bottom, and something close-ish on the top - how can we make the top look more similar to the bottom? Well, since the top starts with 2244, what would happen if we tried to isolate a term of 2244224422442244000 on the top, since that’s what the term on the bottom is like? And then you’d try that, this would lead to the subtraction, you’d notice the other repeated term, and boom, you’re on your way to solving the problem. So it’s less of just arbitrarily deciding to subtract things, and more trying to isolate similar terms that ends up leading you naturally to doing that subtraction.
8154 = 2244 + 5910
There is a mistake
It must be 8514, not 8154
Yes I made a mistake in the thumbnail. Thank you for pointing that out!
No wonder I couldn't do it in my head
@@letsthinkcritically The mistake is not just in the thumbnail, it’s in the video.
@@letsthinkcritically no wonder it was impossible to solve. Disappointing!
There is a mistake in the thumbnail. I mistyped the numerator, and that number should be 2244 8514 8514 8514 627. I accidentally swapped the 1 and 5. Apologies for that!
I understand it’s impossible to change the video, but please at least swap the thumbnail
let's just refilm this before view rate goes too high. the confusion no good for math lovers :)
Hi. You can make a "comment" on the video and cover the video during a few seconds to let people know that there's error.
you can change thumbnails
I think you owe it to your viewers to simplify the fraction actually given in the thumbnail too!
Why are you finding the prime factors?
Once you've simplified things, use Euclid's algorithm to find the GCF and then simply divide.
I have just wasted 9 minutes of my life trying to work out what on Earth you were talking about. Then I read the comments. If you made such a fundamental mistake in writing the number you started with, why not just delete the video and start it again and do it properly?
damn, hard criticize
The video isn't wrong, just the thumbnail. Thumbnail is easy to swap out
@@karolakkolo123 No, the video is also wrong
I honestly don't know how you knew that 22673 has 41 as a factor. That's the only factor (41) I didn't cancel.
He doesn't know that. The point was that since the denominator has a 41 as one of its factors, it would be good to check the numerator if it has 41 as well. You can perform long division to see if 41 goes into it easily. And again, the motivation to do that is the fact that we already have it in the denominator
@@karolakkolo123 The Vedic divisibility rule for 41 uses a multiplier m of -4, which is quite manageable here for a quick test:
1) Calculate multiplier m as:
m = (1 + 41*9) / 10 - 41 = 370 / 10 = 37 - 41 = -4.
2) Test divisibility of 22673 by 41 as:
41 | 22673
41 | (2267 + (-4 * 3) = 2255
41 | (225 + (-4 * 5) = 205
41 | (20 + (-4 * 5) = 0
true.
The Vedic divisibility rules are available for all numbers coprime to 10; so all numbers with 1, 3, 7, or 9 as a ones digit. The multiplier used to calculate m as above varies with that ones digit of the divisor d being tested : divisors d with a ones digit of 1 or 9 use the other while those with a ones digit of 3 or 7 each use themselves. The negative multiplier is typically more convenient when the divisor d being tested ends with 1 or 7; and the positive multiplier typically more convenient for divisors d ending in 3 or 9.
This took me like half an hour😑
I think people put too much weight on the thumbnail. Solution to the problem solved in the video is correct. Amazing solution.
John Cochran (previous comment) is right: there is no need to find the prime factors; the Euclidean algorithm is the most efficient way to solve this problem. Also, the numerator is odd and not divisible by 5, so it is obvious that 2000 remains as a factor after any factors in common with the numerator have been cancelled. However, spotting the common factor of 10^12+10^8+10^4+1 at the start cuts out a lot of work.
If you spot the big common factor then at 4:28 in the video the fraction is n/d with n=2244627, d=8118000. If you know the "rule of 11" (sum of digits with alternating +/- signs is a multiple of 11) then you can spot that 11 is a common factor of n and d; and also if you know the "rule of 9" you can see that 9 is another common factor. But supposing you don't know either of these things, you can just do division with remainder to get d=3n+r_0, with r_0=1384119, and similarly r_j=r_{j+1}+r_{j+2} for j=0,1,2,3,4, with r_1=860508, r_2=523611, r_3=336897, r_4=186714, r_5=150283, r_6=36431. Then r_5=4r_6+r_7, and finally r_6=9r_7 with r_7=4059=9x11x41=g being gcd(n,d). This is the Euclidean algorithm for finding the gcd, and cancelling out this big factor gives the fraction 553/2000 in lowest terms: you can avoid big divisions/multiplications by running the algorithm backwards to find r_5=37g, r_4=46g, etc., and finally n=553g, d=2000g. Even if you don't spot the big common factor at the start, the algorithm works in exactly the same way with the original pair of 19-digit numbers.
What do you use to make videos? I think it would be nice if you linked software and amazon affiliate link to tools, that way you profit from affiliate link and interested lectureres profit from the equipment. Thank you for the videos as always!
asnwer= 1
Great one
Beautiful
I did it differently. A combination of my version after your "power of ten" extraction would have been quickest. (Find the 1000100010001 then long divide).
GCF: divide greater by smaller until remainder = 0 (so old remainder becomes new divisor)
the last divisor is the GCF
Clearly a factor of 1000 can be taken out of the denominator
𝟸𝟸𝟺𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟼𝟸𝟽
𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 ×𝟸𝟶𝟶 𝟷𝟼𝟸𝟹𝟽𝟼𝟸𝟹𝟽𝟼𝟸𝟹𝟽𝟼𝟸𝟹𝟼𝟶𝟶 -
𝟼𝟸𝟷𝟶𝟾𝟿𝟷𝟶𝟾𝟿𝟷𝟶𝟾𝟿𝟷𝟶𝟸𝟽
𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 ×𝟽𝟶 𝟻𝟼𝟾𝟹𝟷𝟼𝟾𝟹𝟷𝟼𝟾𝟹𝟷𝟼𝟾𝟸𝟼𝟶 -
𝟻𝟸𝟽𝟽𝟸𝟸𝟽𝟽𝟸𝟸𝟽𝟽𝟸𝟸𝟽𝟼𝟽
𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 ×𝟼 𝟺𝟾𝟽𝟷𝟸𝟾𝟽𝟷𝟸𝟾𝟽𝟷𝟸𝟾𝟽𝟶𝟾 -
𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿
𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾
𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟸 𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾𝟾𝟷𝟷𝟾 -
0
𝚐𝚌𝚏 = 𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚜𝚘 𝚌𝚊𝚗𝚌𝚎𝚕 𝚝𝚘𝚙 𝚊𝚗𝚍 𝚋𝚘𝚝𝚝𝚘𝚖
denominator = 2𝚡1000
numerator :
𝟸𝟸𝟺𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟼𝟸𝟽
𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟻𝟶𝟶 𝟸𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟻𝟶𝟶 -
𝟸𝟷𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟺𝟾𝟻𝟷𝟸𝟽
𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟻𝟶 𝟸𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟽𝟶𝟸𝟿𝟻𝟶 -
𝟷𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟽
𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿𝟺𝟶𝟻𝟿 𝚡𝟹 𝟷𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟾𝟸𝟷𝟽𝟽 -
0
𝟻𝟻𝟹/𝟸𝟶𝟶𝟶
This video wasted my time! Due to the typo, I cannot enjoy the pleasure of finding the secret hiding in this problem.
First view and like
don't care + didn't ask
@@bruhyou197 i didn't reply for u so shut ur mouth
@@bruhyou197 Why you are so hostile
@@user-yg97f5hfvh Some people are just like that, we can forget about them
@@user-yg97f5hfvh cuz I don't like silly comments.
You should really correct the mistake in the video ...
Should be 8514 not 8154
Yes I made a mistake in the thumbnail. Thank you for pointing that out!
You have made great mistake! Check it again, please!
Well, you can multiply and divide by 3, no?
is it one number or are there invisible multiplication signs in between
Not understood anything.. you put 2244 4 times followed by three 0 a number. Similar to denominator., then how do you get 6270
error in the numerator/thumbnail. should be 2244 8514 8514 8514 627 and thats it
0.25 .. close enough for me
oi mate, a bit long fraction, innit?
Should be 8514 not 8154
Fabeeeee
This exercise has NOTHING whatsoever to do with “critical thinking”. How is this useful? Better watch Netflix!
On the contrary, spotting the large common factor at the start (despite the typo in the video) does involve critical thinking: finding a strategy to solve the problem quickly is the whole point, even if the rest of the solution given here is not the most efficient one.
Lmao, just divide the numerator and denominator by 10^18 u get 2.2/8.1 (since the digits are near I'm not rounding off) u will get .27 as the answer matching the final answer.
you seem to be unable to comprehend what no calculator means
@@24Kemistbro, that thing doesn't require calculator. If not what else do u mean??
0.2765 or just divide by 9/9 to get (249 423 942 394 239 403) / (902 090 209 020 902 000). You stated to simplify, not fully simplify.
It’s implied that simplify means fully simplify
@@shashanks626 Really? Where is that written down and agreed to?