My First Harvard MIT Math Tournament Problem

[@blackpenredpen]

0:31 if you did pause 👇

[@mathmathician8250]

Let k=√(100+√n) +√(100-√n) k²=200+2√(10000-n) Since k is an integer, k² (perfect square) is also an integer. And since 200 is also an integer, one way to make this true is to make √(10000-n) an integer, which makes 10000-n a perfect square. Hence, set m²=10000-n, mEz+. So we have: k²=200+2m (1) and n=10000-m² (2). Notice that in (1), the RHS are sum of two even numbers, that means LHS is an even perfect square. From(2), notice that minimize the value of n will maximize the value of m and k and since n>=1 => 10000-m²>=1, then 0=

[@anshumanagrawal346]

@@mathmathician8250 You copied my solution :P

[@MathSolvingChannel]

Bonus Q answer: n=-10164 solution steps: kzbin.info/www/bejne/qIq3kHp7hJWIhdk

[@mathmathician8250]

@@anshumanagrawal346 Nah I didn't it took me 30 minutes to type in on my phone :)

[@mathevengers1131]

Don't see in 2x speed at 0:46

[@ilickcatnip]

The solution is simple: *define 0 to be a positive integer*

[@blackpenredpen]

😆

[@namantenguriya]

🤨🤨🤭😜😅🤣🤣🤣🤦🏻‍♂️clever

[@benedictchoong]

This person is going places

[@particleonazock2246]

Lmao

[@goodplacetostop2973]

@@benedictchoong Which places?

[@Harlequin_3141]

Oops I was assuming things had to stay as integers throughout, so I was assuming the sqrt(n) was also an integer. If you go that way, I came up with n=9,216 which sums to 16 in the original equation.

[@blackpenredpen]

That is cool! It would have been more fun that way 😆

[@virmontisfbg]

Did the same thing!

[@javiernasser3574]

I had the same error, thought that sqrt(n) was an integer and had n=9216, glad to see that i wasnt the only one 😅

[@Harlequin_3141]

@@blackpenredpen Yeah I guess I was just assuming the left term and the right term were both integers already. Because, what are the odds that they would be non-integers and then exactly cancel out to make an integer once you add them up? And yet that's exactly the solution you presented. How cool!

[@joaquingutierrez3072]

I did the same

[@wowZhenek]

Solved the bonus problem by doing a substitution: n = -t, which then turned, considering that i^2 = -1, the final equation into a k^2 = 200 + sqrt(100^2 + t) where I have to find the smallest non-negative t. Using the same method I found 22 as the first smallest potential k (after 20) and, respectively, t = 10164 => n = -10164

[@ipcheng8022]

It is very weird that the squre root of 6156 is not even an integer, but when it was plugged into this equation, the result is an integer.

[@normal981]

The magic of square root

[@normal981]

Also I'm confused

[@seedmole]

It's a perfect square less than 10,000. Namely, 62 squared = 3844, and 3844 + 6156 = 10000. Now why it's 62 squared of all values, I couldn't say.

[@samueldeandrade8535]

That shouldn't be weird. If you take any number of the form x = a+b√n you can rewrite it as x² = a²+nb²+2ab√n x = √(a²+nb²+2ab√n) So, any integer can be written as sum of radicals of this weird form. For example, 4 = √(15+4√11)+√(15-4√11) which I obtained just calculating (2±√11)² Such numbers and similar expressions became popular after Cardano's formula to solve cubics. Indeed, a third degree and complex version of the previous example is the solution of the classic x³=15x+4 The Cardano formula for depressed/reduced cubics x³=3px+2q is x = ³√R(+) + ³√R(-) with R(±) = q±√(q²-p³) You can easily check that such x is ONE solution for the cubic. For the cubic x³=15x+4 we have p=5, q=2 so R(±) = 2±√(2²-5³) = 2±√-121 = 2±11i Such numbers have as one of their cubic roots 2±i, since (2±i)² = 4-1±2*2i = 3±4i (2±i)³ = 6-4±(2*4+3)i = 2±11i making x = ³√R(+) + ³√R(-) = 2+i + 2-i = 4 which satisfies 4³ = 64 = 15*4+4 By the way, this is also the actual origin of complex numbers. I hope I didn't make any mistake.

[@rayg5445]

Wow I have not seen your videos since I was in college and your channel was so small. Now you have hundreds of thousands of subscribers. This is amazing. Congratulations

[@blackpenredpen]

Thanks. I am happy and grateful about it.

[@The-Cosmos]

Maybe correct the grammar?

[@hydraslair4723]

I found the solution a different way! It is possible to rewrite √(100 + √n) = √a + √b, where a and b are integers and a>b. If you square both sides you get that whatever n is, it can be written as 4a(100-a). Writing the sum with our new numbers we get that the sum of the two terms is just 2√a. For this to be an integer, since a

[@kromydas5063]

for the common types of problems involving sqrt[a+sqrt(b)]+sqrt[a-sqrt(b)], i usually find out first if i can get something in the form of sqrt[x+2 sqrt(y)], then figure out if i can factor it using (a+b)^2, and if i can i can write it as sqrt x_2 + sqrt y_2, and similarily, i can do the same thing with the other part of the equation such that the y_2's cancel out giving us 2 sqrt x_2

[@melocchidaniele]

I found the same result and i want to explain my solution. Let me call the expression f(n) and substitute n with x belonging to R so that f(x) is a real CONTINUOUS function. Existence conditions implie that x must belong to the bounded interval [0, 10000]. f(0) = 20 whereas f(10000) = sqrt(200) that is similar to 14.14 Now, f(x) is monotonous decreasing ( HINT: study the positivity of the derivative function) and this means that 20 is his maximum value and 14.14 is his minimum value. Thanks to Darboux's theorem ( or the intermediate value theorem) we can claim that f(x) attains every integer value from 20 to 15. From 20 to 15 x will increase from 0 to 10000 since f(x) is monotonous decreasing. So compute the counter image of K from 19 to 15 and if x is an integer we stop the algorithm, else the result is impossible. To compute x knowing K we write the inverse function: x= 10000-((K^2-200)/2)^2. f^(-1)(19) is not an integer, we pass to 18. f^(-1)(18) is an integer and we find that x=6156. This implies that x=6156 is the smallest positive integer which makes f(x) an integer. The proof is now concluded. I hopefully you appreciate my demonstration and if I made mistakes please don't hold it against me, I'm only an italian mathematician that wanted to share his solution with the global community ^.^ All the best, D.M.

[@petarscekic3898]

👌

[@basil1400]

That’s beautiful

[@gscreationss]

Are you a mathematician

[@melocchidaniele]

@@gscreationss yep!!

[@gscreationss]

Please give me some tips for a future mathematician (😜it's me) what I want to learn well

[@drpkmath12345]

I have covered the exact same problem a couple of months ago in my channel. Glad you covered the same problem too!

[@Min-cv7nt]

Their explanation is basically almost identical. good thing

[@javiernasser3574]

Other way to solve its calling sqrt(100+sqrt(10))= b then b+sqrt(200-b²)=N then by using the quadratic formula we get (N+sqrt(400-N²))/2 then just search for the lowest N value with b= some natural number +sqrt(natural number) as we want b to be the least start with 19 and then move on to 18, and thats it with 18 you get 9+sqrt(19)

[@blackpenredpen]

Ah! Nice one.

[@muhammadridho7680]

The fact that I can't understand it, it's the proof that I can't get into Harvard

[@iceIceCold666]

I did the same method lol and i found that expression gives values that are less or equal to 20 and when the expression is equal to 18 then it gives the lowest N ...and N=

[@VSN1001]

Quite a stander math Olympiad number theory question. Great video as always

[@blackpenredpen]

Thanks.

[@sharpnova2]

I'd say this is far too easy for an Olympiad problem

[@VSN1001]

Junior Olympiad

[@richardfeynman7332]

srsly??? olympiad???😂, go check out olympiad kiddo

[@irinaignatova1799]

2:38 my confidence, 30 seconds before the exam ends

[@anandk9220]

I think I've actually solved this by little logic and trial error method. Since [ 100 - sqrt (n) ] must be positive perfect square, sqrt (n) < 100 Trial error values of 100 +- sqrt (n) fails for squares of 11, 12 and 13, more specifically 100 - 21, 100 - 44 and 100 - 69 (as these differences are not perfect squares). But it succeeds for sqrt (n) = 96, as both 100 + 96 and 100 - 96 are perfect squares. Hence n = 96^2 = 9216 EDIT 1 : I tried this orally without using any calculator as I thankfully remember perfect squares till 111. EDIT 2 : I checked the solution and was sad to know my answer is not correct here. No issues. It would have been correct for a rational square root result. Anyway. I enjoyed trying this one.

[@enoua5222]

This is the same method and result I used/got

[@VSN1001]

I also got your answer initially and assumed each of the roots are positive integers

[@sreeharie821]

Yeah I used the same method

[@mojtabasaleh8842]

I did the same,but why this answer is wrong?

[@anandk9220]

@@mojtabasaleh8842 It's wrong because square root of 100 +- sqrt (n) can be irrational, which on addition, result in an integer. That's something which we didn't think of.

[@anshumanagrawal346]

My answer before watching the video is 96^2 My approach is to let the expression equal to x and then make x^2 a perfect square, I got in the form 2(100 +√(100^2-n) so I concluded 50 + √(50^2 - n/4) must be a perfect square. Now since the √ is always +ve, I checked for the first perfect square after 50, which is 64 and that gave an answer for n, which was 4(50-14)(50+14), then I simplified a little and wrote in terms of power of primes (luckily it only contained powers of 2 and 3) to solve for √n, and I got it's value 96, which I verified that it makes both square roots perfect square so answer is integer without simplifying But the value of n I got is pretty large, so I suspect I may have made a mistake somewhere and missed a smaller solution, but I'm almost sure my answer is correct (Unless you remove the positive integers and change it to non-negative :P)

[@anshumanagrawal346]

Looks like he's taking a similar approach, I thought I was clever to find this easy way :P

[@anshumanagrawal346]

Wow, turns out he didn't realise he had to make that expression a perfect square and mistakenly just solved for it to be an integer. So, he didn't get it right on his first try. Yay!

[@anshumanagrawal346]

Never mind, I just realised my mistake halfway through the video, the answer is actually 4(50-31)(50+31) or ((18)^2)(19) which gives after simplifying gives (9+√19)+(9-√19) = 18 Earlier, I actually mistakenly solved for the largest value of n, because I solved for smallest value of x, in fact I have to solve for biggest value of x to get smallest n

[@qo7052]

well done

[@caesaraugustus6231]

Other solution: If we have sqrt{a+\sqrt{b}}=sqrt{x}+sqrt{y} then sqrt{a-sqrt{b}}=sqrt{x}-sqrt{y} With x=frac{a+\sqrt{a^2-b}}{2} and y=frac{a-sqrt{a^2-b}}{2} [the proof is easy, try it] So, in this problem we have that the sum is sqrt{100+sqrt{n}}+sqrt{100-\sqrt{n}}=2(sqrt{x})=sqrt{4x} in Z With x=frac{100+\sqrt{10000-n}}{2} and 4x=200+2\sqrt{10000-n} and how do we know that $4x=k^2\inZ$, we continue as bprp.

[@Viollatie]

lets consider n=4k than (root/100-,+//n)=(/100-,+2//k) we know that if a+b=100 and a*b=k we can write that as /a -,+/b a is bigger than b so we can get positive number(for the negative one) if we do all these we get (/a+/b)+ (/a-/b)=2/a so a must be x^2 and a=81 and b=19 perfectly fits into this so 2/81 =18 n=4*81*19=6156 ''/'' = square root

[@jameslin2423]

I was the problem czar for this tournament, during my freshman fall! I hope you enjoyed the problems!

[@sanyalox01]

Solution to the question at the end: With the same approach as in the video we get k² = 200 + 2sqrt(100² - n), and for the sake of (my own) better understanding, let -n = m; in which case we're looking for the smallest positive integer m. so k² = 200 + 2sqrt(100² + m), if m was equal to zero, k² would be 400, so we're looking for the perfect squares bigger than 400 but as small as possible 21² = 441 doesn't do the job, as k² is also even. take 22² = 484 then 484 = 200 + 2sqrt(100² + m) 284 = 2sqrt(100² + m) 142 = sqrt(100² + m) 142² = 100² + m m = 142² - 100² = (142-100)(142+100) = 42*242 = 484*21 = 9680 + 484 = 10164; therefore n = -10164 plug this into any programm able to work with complex numbers, you get an integer in the original expression.

[@forcelifeforce]

Regarding the bonus part, the biggest negative integer is the negative integer closest to zero. If there are no other possible solutions between the claimed -10164 and -1, inclusive, then -10164 would be the biggest (assuming it works).

[@jgsh8062]

Yayy did I get it right with n=-10164 then? Not used to getting BPRP problems right lol

[@latermyfriend8934]

A quick python script to verify the answer : from math import sqrt j = 0 while True: j += 1 f = sqrt(j) s = sqrt(100+f)+sqrt(100-f) if s == int(s): break print(j)

[@adriancarpio7536]

Wow, I tried solving this and ended up making the exact same mistake as you did! It must have stumped many more students during the tournament

[@bryantg8749]

The problem with the tournaments is that their purpose is to trip people up on mental calculations (for the most part) rather than any rigorous type of math involved. Seeing or scratch writing 10000, might as well be shorthand to 10k or 1kk, or any other moniker.

[@ps4m319]

Really nice initiative

[@blackpenredpen]

Thanks.

[@vol230]

Happy teachers' day sir😇

[@blackpenredpen]

Thanks. I don’t think we actually have that here but I appreciate it! 😃

[@Yuvrxj29]

@@blackpenredpen I think you have teachers day at 5 October we are from India 🇮🇳

[@vishalmishra3046]

*Related Problem* Prove that sum of [ C(4k-2, 2k) / (2k-1) x (n/20^4)^k ] from k = 1 to k = Infinity equals 1/20 where C(n,r) is binomial co-efficient function. You will find that n = 6156

[@zerospeed6412]

I thought bprp went mad when setting 10,000-n to 9801 😂

[@bobzarnke1706]

More generally, given a and any k such that a/2 ≤ k² ≤ a, then n = 4k²(a-k²) makes √(a+√n) + √(a-√n) an integer, namely, 2k. In the case above, a = 100; so 100/2 ≤ k² ≤ 100 implies k = 8, 9 or 10, making n = 9216, 6156 or 0 and √(a+√n) + √(a-√n) = 16, 18 or 20. n = 0 is explicitly excluded, making n = 6156 the minimum. (9216 = 96², in which case √n is an integer, 96, whereas 6156 = 4·9²·19 makes √n = 18√19.)

[@tanmayshukla5330]

The answer to your challenge question:- (-6156) because let n = -m, where m is positive int. when we find for m, for smallest, it is similar to solving for -n for biggest. If I am right, pls give a heart!!

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@protostar2007]

His cursor is bigger than my self-esteem

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@BlockStah]

Can You Solve This Harvard MIT Math Tournament Problem? *no*

[@L开门见山]

4:49 let k=2p, square both sides and you will get n=f(p)

[@amirthya]

There is a related approach as well - consider 100 + \sqrt(n) = ( a + \sqrt(b) )^2 for some a,b. Then, 100 - \sqrt(n) = ( a - \sqrt(b) )^2. It follows that 100 = a^2 + b while n = 4 a^2 b = 4 a^2 (100 - a^2). However, there are two conditions for this to hold: 1) a > \sqrt(b) else the \sqrt(100 - \sqrt(n)) = \sqrt(b) - a and the result is no longer an integer. This also means a^2 > b = 100 - a^2 => a^2 > 50. 2) b > 0 (This is to ensue n>0, and for the bonus can we changed to b < 0). This gives a^2 < 100. Given n is a decreasing function in 50 < a^2 < 100, choose a^2 = 81 and correspondingly n = 4 x 81 x 19 = 6516. The solution for bonus is similar with b < 0. which corresponds to a^2 > 100. Choose a = 11 to get n = - 4 x 121 x 21 = 10164. (Assuming the square root with positive real part is considered) In fact, the set of all admissible n corresponds to all a^2 >= 50 or a >= 8.

[@jaymercer4692]

There are only two integer solutions to this problem. Obviously one for n=6156 but the other is n=9216.

[@Harlequin_3141]

This was how I approached it by looking at cases where n was a square number. Turns out that isnt the best solution, but I kinda just assumed it would be the case. n=9,216 is nice and neat in that things are integers at every step.

[@jaymercer4692]

@@Harlequin_3141 Yeah when I was in the gym I wasn’t able to square it all in my head and work it out that way so at first I was just looking for numbers that fit and found that. But later when I got home I managed to reduce the equation and get an inequality that showed the only possible solutions were 6156 and 9216. Also n=9216 makes the equation it’s smallest possible integer solutions so I prefer it.

[@seedmole]

I missed that it's about finding a positive integer for n, so I stopped once I found the trivial n=0 case. But the method I used easily spat out 6126 (which interestingly enough when subtracted from 10,000 gives exactly 62 squared).

[@andy-kg5fb]

What's wrong in: let k=√(100-√n)+√(100+√n) Squaring both sides(and skiping a step or two) K²=100-√n+2√(100²-n)+100+√n K²=2(100+√(10,000-n)) Rearranging to get: (K²/2)-100=√(10,000-n) Squaring both sides to get (K⁴/4)-100k²+10,000=10,000-n Rearranging to get 100k²-(k⁴/4)=n You can try finding integer solutions from here but it doesn't work out. I don't know where the mistake is.

[@andy-kg5fb]

Ok I got the right answer after seeing your working out.(also pretty surprised seeing that everyone's first step is the same. I thought I did something horribly wrong in the beginning but turns out it's the same as everyone.)

[@SebastienPatriote]

It's weird how all the videos in english say that zero is not positive nor negative, while in french we learned it was actually both. In a question like this it would say n is strictly positive, if it only said positive then zero would be an acceptable answer.

[@rohitchaoji]

My solution was different, but incorrect, because of an incorrect assumption I made. Moreover, I did this algorithmically rather than mathematically. I made a couple of assumptions based on the fact that our result, k is an integer and n is a positive integer: 1. n squared must be less than 100 2. each term of k individually must be an integer (an incorrect assumption based on the given facts) Then, based on this, I introduced a new term, p = sqrt(n) Now my objective was to find an integer, p, at an equal distance from 100 in both directions, landing on a perfect square. This gave me p = 96, or n = 9216, which is a solution, but not the answer to the question, as it is not the smallest value of n. This as all based on the incorrect assumption that p would be an integer, stemming from the assumption that each term of k would be an integer. This method gives me k = 16, however.

[@VSN1001]

I made the same mistake and got the same answer as you. It seems like the usual technique so it’s the first time I encountered such a question!

[@sanjitgera1160]

rohitchaoji, i did the same mistake

[@anonim5926]

We did same mistake.I found it 96 too

[@ВасилийТёркин-к8х]

There is quiet a few natural numbers that makes the expression a positive integer. 6156 give integer because (9±sqrt(19))^2 = 100±18sqrt(19)=100±sqrt(6156)

[@tambuwalmathsclass]

So fun though very intuitive 💪

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@The1RandomFool]

I decided to take this a step further and created a small Python script to find all possible values. The value n must be greater than 0, but less than or equal to 10,000. There are only two values, the one he found, and 9216.

[@SlipperyTeeth]

This is because k^2 is at least 200, so k is at least 15. Since k is even and less than 20, that only leaves solutions for k=16 and k=18.

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@jimmykitty]

Elegantly explained Boss!! Awww.... 🦒❤😯😍😍 It's really adorable 😅😅

[@cringelord7542]

bruh

[@gigagrzybiarz]

bruh

[@aashsyed1277]

yes i aggree

[@jimmykitty]

🤩🤩🦒

[@aashsyed1277]

@@jimmykitty giraffe love grass

[@derwolf7810]

0:28 "that expression gave us an integer" ("= - 2" pops up) The minus sign is wrong (just nitpicking).

[@evanbee5669]

omg your 1 hand marker transitions are so cold

[@danielcopeland3544]

I don't pause the video and try it when you say, because I've usually already had a go from the thumbnail and now I want to know the answer.

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@UzziWallendorf]

100+n/100-n->ⁿ defined as an postive integer ->10000n->200+2×->->18

[@amtep]

I think before concluding "k is even" you also have to prove that you can't choose an n so that the square root comes out to a half number (like 80.5) which when multiplied by two comes out as an odd integer

[@alexmcdonough4973]

Good catch, but it turns out this is always impossible. The square root of a positive integer is either an integer or irrational (look up the proof of the irrationality of 2, it's a classic!)

[@amtep]

@@alexmcdonough4973 Thanks! That's good to know

[@ittesafyeasir3438]

What I'd probably do is at first, deduce that n must be a square number and 100-sqrt(n) AND 100+sqrt(n) must be a square number. then plug in sqrt(n)=1,2,3... Since sqrt(n) cannot be equal to 10, it will be one of the values from 1-9

[@tamimplayz]

you know things getting complicated when he pull out the blue pen

[@yogeshkhatri1618]

I tried to find any value of n that will yield the integer in the expression and I found 396. sqrt(100+sqrt(n)), putting n=396 sqrt(100+sqrt(396)) = sqrt(100+2sqrt(99)) = sqrt(1+2.1.sqrt(99)+(sqrt(99))^2) = sqrt((1+sqrt(99))^2) = 1+sqrt(99) In similar fashion, sqrt(100-sqrt(396)) becomes 1-sqrt(99) now , the given expression becomes 1+sqrt(99)+1-sqrt(99) =2 , which is an integer. I do know that second term can be sqrt(99)-1, in that case the solution won't work but I don't see any reason why cant we put 1-sqrt(99). Is it because the term becomes negative? I'm really confused.

[@RMV6]

I'm only in 9th grade, and I do these equations so I can get ahead in math, I want to go to MIT, thankfully you have helped me expand my math knowledge. These problems are difficult but honestly I understand them a little, I was off to a good start but then got lost about halfway through trying it. Watching you do it made it make more sense though.

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@jorgetrevino325]

The last time I watched was the 100 integrals video. Now he is a wizard 😳. Math God

[@Setiny]

It’s pretty easy if you notice that it is simply 9 + sqt(19) + 9 - sqt(19), then it goes 81*19*4=6156

[@moeisthis1490]

where did you get the numbers from?

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@KanishkMalkan]

successfully solved this! and i furthermore challenged myself to do it all in my mind! great video as always

[@aluiziofjr]

I have a general question: is it ok to say naturals instead of positive integers?

[@blackpenredpen]

Some people count 0 as a natural number. So it’s better to just say pos integers.

[@sharpnova2]

@@blackpenredpen typically computer scientists. and i have no idea why they ever fumbled into that obnoxious definition. of course the naturals should start with 1. 0 is hardly "natural". didn't it take life something like 4.5 billion years to even conceptualize it? whereas the positive integers took merely 4.49999 billion.

[@cQunc]

@@sharpnova2 When I was a computer science major, I never heard anyone there talk about 0 as a natural number (I don't think they even used the term). From what I've heard, it's more of a geographical thing: 0 is often a natural number in Europe, but not in the U.S.

[@gravysnake78]

3:40, I feel like I kinda understand why but I can't really wrap my head around why that would make k even

[@brayden9458]

First examine all singular digits. Square every one. 0^2 = 0 1^2 = 1 2^2 = 4 3^2 = 9 4^2 = 16 5^2 = 25 6^2 = 36 7^2 = 49 8^2 = 64 9^2 = 81 We define an even number to be a number divisible by two with no remainders left and we define odd to be a number when divided by two has a remainder. Notice that every odd number produces an odd number when squared. This is why if we work backwards that we can conclude that ‘k’ is even. I would test bigger numbers to prove this to you further but it satisfies for every number because the only way you will get an even number is if your last digit is defined as even. i.e: 100^2 = 10000 and has no impact in the ones place. 102^2 = 10404. 102 has impact in the ones place because there exists and number in the ones place for 102.

[@anshumaangupta7398]

I was able to solve this question on my own ... and yea it made my day :P

[@smhaceofspades2813]

Yeah I'm just gonna admit Alexa just halfass solved it for me.

[@NonTwinBrothers]

Yay captions!

[@koibubbles3302]

I did this and got 9,216, which is wrong, but I’m still happy because it is the smallest value that also remains an integer when taken the square root. I did consider that maybe the answer would not necessarily abide by this rule but I didn’t really know how to solve the problem if that was the case.

[@SrisailamNavuluri]

Please try with k=16.Then n is a square number=36×256.

[@truejeffanderson]

Well it's not zero. I guess n is a square since the root of a root is irrational. For root of n, I use x. The value of x must be between 1 and 100 or else we obtain complex solutions from 100-x being negative under the root. The larger root will be 101 to 200. There are four roots in this range: 121, 144, 179, 196. Of these we search from smallest to find the first which is also a square at 200-x. It can't be 121, because 200-121 = 79 and that isn't a square of an integer. Likewise, we eliminate all but 196, because 200-196 = 4 .... an integer square as 2². So we now see that x is 96, because √(100+96) and √(100-96) are integers. Since x is the root of n, I think that n = 96²

[@i.jaydon]

Idk about y'all but I don't know how to do this math but I'm still watching 😭

[@messibruh]

Me to

[@bobby4976]

2:39 🤣🤣🤣 I like this part..

[@blackpenredpen]

😆

[@thealphazero_]

Me too.😂

[@Marek-db8wl]

The idea of the solution is quite easy but not making a numerical mistake along the way was impossible for me. I came up with a slightly different tactic from yours but I had to calculate the thing like three times until I eventually uncovered all the mistakes (things like 2x199=399 or whatever). I suppose simple 4-operation calculators aren`t allowed on the exam, right?

[@MyOneFiftiethOfADollar]

Thanks for this problem! I worked the dual on my channel, i.e. found largest natural number, n, such that sqrt(100+sqrt(n)) + sqrt(100-sqrt(n)) is positive integer. Got n=9216 which is in the domain=[0,10000]

[@SmartWorkingSmartWorker]

My try: let's say that the given thing equals k. k^2=200+2sqrt(10000-n). Since k is an integer, k^2 should be a square number. n is a positive integer, and k^2 must be smaller than 400. The largest square number smaller than 400 is 19^2=361, however this must be even and it is 18^2=324. This means, sqrt(10000-n)=62, 10000-n=3844, therefore n=6156. Uh, I didn't copy the video but I solved it in an identical way, I guess.

[@limew]

Sorry, can someone explain to me how does sqrt(100^2-n) become sqrt(99^2)?

[@AnkitThakur-rp6gp]

2:19 you are talking about he just tried a little troll or a showed a little little mistake but the question is why he said that so let me explain:- integer + integer is always integer but the square root of various integers are irrational (which is not integer) so if the square root part comes out to be integer we wont care about rest it will be integer which will be answer but the he said that there is square on answer which means we are not getting integer hence not the answer most importantly he knew it was closest perfect square (n should be integer) he set them both equal and square rooted both sides and equated(rest explaination is mentioned in this reply by me)

[@jagmarz]

Because we're looking for the smallest possible n>0, so we're looking for the largest perfect square less than 100^2 (to get an integer out of the radical), which would be 99^2.

[@limew]

Ohh ok, I understand now. Thanks guys 👍

[@MrColbrot]

It’s great to watch a 5 min video and confirm I had no business going to MIT. 😂

[@coreymonsta7505]

Probably couldn't do it when I was in highschool tbh but I can now

[@justinpark939]

Hello blackpenredpen, I would like to ask you if you accept user submissions with regards to question and papers and if you do, how we can contact you. I love the questions you put on your channel and allows me to practice my craft, thank you so much.

[@blackpenredpen]

Hi Justin. I am quite busy this semester. You can send the problem to my email [e-mail] but I am sorry that cannot promise if I can make a video on it soon.

[@justinpark939]

@@blackpenredpen no problem

[@bryantg8749]

@@blackpenredpen Stay safe and keep on bringing the videos! Old timers like me who haven't studied rigorous theoretical math or even basic algebra in the past 7 years still like to keep up with things. I ended up going the biochemistry route, but I still have a longing for that math itch from my 2nd degree in pure math.

[@namantenguriya]

Much much n much interesting❤❤❤❤

[@namantenguriya]

+1 again 😆🙂

[@privatejoshuadancer4370]

Why isnt 361 a vakid answer? Root of 361 can be +19 or -19. If you choose appropriate roots, you can get 18 as the final answer. Can somwone explain it to me please??

[@ZeroGravityDog]

n=361 looks good to me. I was working on the same trick but for some reason went upwards to 100+(21) and 100-(-21) making n=441

[@sathvikmalgikar2842]

The ball in his hand turned to Pokémon and he has grown a beard all in a matter of time.

[@johnjhonyk8976]

My lucky day , i just opened the phone to search this type of sqrt exercise

[@Xdetonando]

Sorry for my dumbness, but i didnt undestand why sqrt(10000 - n) can be set to sqrt(99^2)

[@sxz452]

10000 is 100^2, and n cannot be zero but we want to maximize sqrt(10000 - n). That way, the maximum possible answer is sqrt((100-1=99)^2).

[@tyjrthrtg]

I've watched two math videos and now youtube thinks i love math

[@blackpenredpen]

😆

[@mrajsatyam]

Love you sir

[@mainakgoswami21]

What if we found the maxima of the given expression at the first place and then put that in place of k to get n(of course less than 20).

[@PackSciences]

Both expressions k(n) and k²(n) are strictly decreasing and admit no local maximum (the maximum is n = 0). Which is more or less what BPRP did when he said k=20 and which is wrong. Then your method would find the closest n that gives k an integer, which would be again equivalent to what he said.

[@mainakgoswami21]

@@PackSciences yes i think you are right...the graph does not feature a local minima and maxima is at 20..so by my method it is impossible to find an integer. I guess hit and trial is the only way to solve it

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@Shashank_ssm]

Plot twist There is arceus in that pokeball

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@iceIceCold666]

Idk why i got 9216 Just say 100-sqrt(n)= k^2 Sqrt(n)= 100-k^2 So k

[@vladimirvolodarskya1904]

The wrong is to suppose 100-sqrt(n)=k^2 / It can be any real number

[@iceIceCold666]

@@vladimirvolodarskya1904 I worked in only perfect squares where's k is an integer...if i worked in any real number id get the smallest N and i think it's impossible for me lol

[@mrpersistent9086]

the biggest n is 9216 thanks it was easier when you explained it and predicted the value of k and the strategy was already given be careful if n exceeds 10000 the answer is complex *edit*: oh i thought the question on biggest positive integer, I'll work on the negative but I think I'll struggle if I couldn't focus on reading the question properly.

[@uzma342]

Sir how you write maths symbol in this video thumbnail, plz guide me

[@kimveranga]

I hope someone makes a general case for the cube root video. I’m quite interested that that thing just simplifies to 2, so I’m hoping to discover more of such.

[@MathSolvingChannel]

can you be more specific? what cubic root problem?

[@kimveranga]

@@MathSolvingChannel so here are two specific examples of the problem: kzbin.info/www/bejne/lWeznZKDg9WbnJY kzbin.info/www/bejne/a6HEYnSort6fpq8 These are specific examples. How do I generalize/generate other examples with other numbers inside the nested roots? Any leads would be highly appreciated. Thank you!

[@MathSolvingChannel]

@@kimveranga The general way to solve this type of problems is to use substitution. Let a=1st cubic root term, b=2nd cubic root term. so you will get a^3+b^3 = some integer and a*b=another integer, usually. then you use the complete cube formula to convert them into a cubic equation respect to a+b, then you solve this cubic equation. I also did two videos on how to apply this method to solving this type of problems. and here are the links: kzbin.info/www/bejne/ZoGZfWhuZqerq6s kzbin.info/www/bejne/i5OviIqEnNVgqac

[@kimveranga]

@@MathSolvingChannel I’m quite familiar on how to solve those things. My question is how do you even come up with the square roots inside cube roots problem like this. How do you generate these kinds of problems? Thank you!

[@MathSolvingChannel]

@@kimveranga Let me confirm if I understand your question correctly, do you mean: why they put this form of square root expression inside this cubic root? why not they put others such as 11+sqrt(97), or in general speaking, for what kind of square root expression this problem is solvable?

[@normalhuman512]

I'm a bit confused, granted I am well out of my range of knowledge. Doesn't √(100+√n)+√(100-√n) rearrange to 10+ √√n + 10 - √√n and then the √√n cancel out to just leave twenty?

[@normalhuman512]

Is there some imaginary reason this doesn't work maybe?

[@TranceGate92]

I tought that square roots were separate part of the formula.

[@vijaykulhari_IITB]

Sir please video on roots of 4 degree polynomial and sir I like your teaching thanks for open this channel

[@juandelacruz9027]

I get it. You are channeling that Pokemon ball to give you the answers. Good thinking.

[@adityaekbote8498]

Nice 🦒 will be proud

[@yoyoyogames9527]

I GOT THIS ONE LETS GO

[@GirishManjunathMusic]

What is the value of n that gives the smallest integer solution to sqrt(100 + sqrt(n)) - sqrt(100 - sqrt(n))?

[@SayXaNow]

smallest integer: 0 smallest positive integer: 396 other positive solutions: 1536 3276 5376 7500 9216 9996

[@02adityajain2]

This goat is the smartest

[@3nelad]

Biggest negative integer n = -10164 ?

[@MathSolvingChannel]

yes, correct, same result here :)

[@teslaaf5830]

How can sq root contain a negative integer?

[@MathSolvingChannel]

@@teslaaf5830 it becomes complex numbers :)

[@teslaaf5830]

@@MathSolvingChannel so u solved the problem in complex numbers? and how come a negative integer came as a result.. i tried but didnt get the ans

[@teslaaf5830]

@@MathSolvingChannel can you post that solution in ur youtube channel

[@C__YashKumar]

Happy teacher's day

[@hellNo116]

can you explain how those two irrationals make a rational. because it is not like adding φ and 1-φ. they are two different irrationals. what part of those numbers is the one canceling out? is that a stupid question?

[@ipcheng8022]

not at all. in fact, i believe this is the core to the debate whether 6156 is the correct answer or not.

[@hellNo116]

@@ipcheng8022 is there anywhere you recomend to look up for this one. I am an engineer so I am not familiar with where you go to discuss math. I haven't done math for the degree for 3 years now so yeah that didn't help either

[@ipcheng8022]

@@hellNo116 i am an engineer too, so i tried it on Python. when the two terms in the original equation are seperated, both of them have a lot of decimal places. First term: 13.358898943540673 Second: 4.641101056459326 Here is the thing that I cant figure out why: the decimal places were aligned too perfectly, so that all of the sum at the decimal places are nine. I believe this is not a mathematical issue but a technical issur, where computers automatically round up the 17.99999999999 to 18. On the other hand, I cant figure out why the result is so close to 18. There is no terms to should be cancelled due to the additional square roots applied to the two terms. It could just be coincidence, or this is some math theories that we dont know about. This is why I strongly support that 9216 should be the correct answer because it can be proven for sure that results of all terms are integers

[@hellNo116]

@@ipcheng8022 rounding errors are thing. they don't always round that way. it has to do with the printed versus the saved value of the number. also it is really depending on how the computer is programmed to deal with them. Also that is an approximation. My question isn't if the approximation add up, but how the irrational parts cancel out

[@ipcheng8022]

@@hellNo116 i cant find any other similar cases or related theories that can explain this phenomenon, which is why i can only explain it from a technical view.

[@fizixx]

Black pen friend of red pen, red pen friend of black pen.

[@MF-qb7ns]

Did I see blue pen ??

[@benoplustee]

how is it clear that 20 (n=0) is the biggest value of k? i don't see how k is strictly decreasing as n increases. i plotted the graph on desmos and i can see that it is decreasing, but how could one tell algebraically/analytically? EDIT: ah, i can see that k^2 is strictly decreasing as n increases. is that enough to determine that k is also strictly decreasing though? 2nd Edit: yes, because k is always positive, so a decreasing k^2 implies a decreasing k!

[@colereynolds2080]

because you cant have n>100^2, otherwise Sqrt(100-Sqrt(n)) is a complex number and this problem restricts us to the positive (real) integers.

[@benoplustee]

@@colereynolds2080 I figured it out in my edits lol thanks tho

[@THE_FIXOR]

we can just take n = 0 it will be inferior to 6156 and k will equal to 20 what are you saying about this please answer me ?

[@gobyg-major2057]

WolframAlpha gets it wrong sometimes…I got 361…nvm I got it wrong by using - in the first radical instead of +…..

[@daanwinne2596]

7:48 -10164, I think

[@_AadiDevSharma]

@blackpenredpen how can I send my q to you to solve..?