My First Harvard MIT Math Tournament Problem

If you like this Harvard-MIT Math tournament (HMMT) problem and want to learn more about problem-solving, then check out Brilliant brilliant.org/blackpenredpen/ (this link provides a 20% off discount).
I really like this problem because I like how nested radicals can be simplified to some very clean results. This is the cube root question that I showed in the video: • believe in the math, n...
About HMMT: “Founded in 1998, HMMT is one of the largest and most prestigious high school competitions in the world. Each tournament draws close to 1000 students from around the globe, including top scorers at national and international olympiads. HMMT is entirely student-organized, by students at Harvard, MIT, and nearby schools, many of whom are HMMT alumni themselves.” This description is from their site, www.hmmt.org
Here's the HMMT test: hmmt-archive.s3.amazonaws.com...
#Harvard #MIT #HMMT #MathCompetition
0:00 Harvard MIT Math Competition
0:27 typo the ans shouldd be 2
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7:25 Bonus question
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[@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?

[@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.

[@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

[@irinaignatova1799]

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

[@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

[@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?

[@pkmath12345]

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

[@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

[@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

[@danielemelocchi5745]

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

[@danielemelocchi5745]

@@gscreationss yep!!

[@gscreationss]

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

[@jameslin2423]

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

[@ps4m319]

Really nice initiative

[@blackpenredpen]

Thanks.

[@zerospeed6412]

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

[@evanbee5669]

omg your 1 hand marker transitions are so cold

[@protostar2007]

His cursor is bigger than my self-esteem

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@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.

[@tambuwalmathsclass]

So fun though very intuitive 💪

[@ABCD-bm2hs]

kzbin.info/www/bejne/fn6rfX9qjqaLmsU

[@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=

[@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.

[@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.

[@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.

[@NonTwinBrothers]

Yay captions!

[@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

[@mrajsatyam]

Love you sir

[@jorgetrevino325]

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

[@vol230]

Happy teachers' day sir😇

[@blackpenredpen]

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

[@Yamdoot29]

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

[@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

[@user-ik2kd9mb5t]

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)

[@KanishkMalkan]

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

[@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).

[@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)

[@jimmykitty]

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

[@cringelord7542]

bruh

[@gigagrzybiarz]

bruh

[@aashsyed1277]

yes i aggree

[@jimmykitty]

🤩🤩🦒

[@aashsyed1277]

@@jimmykitty giraffe love grass

[@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

[@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?

[@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.

[@yoyoyogames9527]

I GOT THIS ONE LETS GO

[@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

[@MonsterIsABlock]

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

[@tamimplayz]

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

[@namantenguriya]

Much much n much interesting❤❤❤❤

[@namantenguriya]

+1 again 😆🙂

[@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.

[@coreymonsta7505]

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

[@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.

[@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

[@anshumaangupta7398]

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

[@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.

[@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.)

[@HacksMathscience]

Can you explain the Beimar formula for calculating prime numbers?

[@C__YashKumar]

Happy teacher's day

[@smhaceofspades2813]

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

[@richardbraakman7469]

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!)

[@richardbraakman7469]

@@alexmcdonough4973 Thanks! That's good to know

[@vijaykulhari_IITB]

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

[@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.)

[@jagmarz]

On the bonus question, does biggest mean greatest or |greatest| ?

[@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

[@TranceGate92]

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

[@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

[@02adityajain2]

This goat is the smartest

[@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.

[@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]

[@MrColbrot]

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

[@uzma342]

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

[@derwolf7810]

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

[@SrisailamNavuluri]

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

[@bobby4976]

2:39 🤣🤣🤣 I like this part..

[@blackpenredpen]

😆

[@thealphazero_]

Me too.😂

[@user-er8ro4hv1e]

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

[@_AadiDevSharma]

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

[@MF-qb7ns]

Did I see blue pen ??

[@jaydono]

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

[@messibruh]

Me to

[@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

[@yosefouad2116]

can you put all the other questions' answers please I tried to solve them but unsure of it's correct

[@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

[@maxhenderson1890]

So were you allowed a calculator for this problem, and if not then how would you know thats the correct solution when you plug 6156 into the initial equation?

[@saveerjain6833]

yay got it

[@adityaekbote8498]

Nice 🦒 will be proud

[@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.

[@thegamerfromjuipiter7545]

9,216 was as close as I got myself, with only a few minutes of trying

[@juandelacruz9027]

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

[@yhamainjohn4157]

2 [6156, 9216]

[@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

[@michaeledwardharris]

Surprisingly tricky

[@k.m.junayedahmed3748]

You should solve problems more often

[@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?

[@Donkeyhead]

How do we now that is the smallest?

[@sugamnema9103]

what if we rationalise in first place?

[@pearmonkey9304]

Fun that n=44^2=1936 is the smallest answer. But the final result is 12+8, higher than 18

[@tambuwalmathsclass]

For the negative integer 'n' I think we should look for the biggest n to ease the task 👍

[@nickpham6526]

I get lost by the part 10000-n=99 square and where is the part said k=18

[@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

[@justanonverifiedyoutubechannel]

Lol I got n=9216 and total answer as 16 I solved this under 2 minutes Then watched the video and understood how wrong I was

[@fr3shxonl1ne]

SAME

[@RohitKumar-on5xd]

9216 is correct answer, which gives 16 in equation 1

[@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]

😆

[@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.

[@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 ?

[@bitcoinbuy]

I really went crazy after seeing this video, envying also lol

[@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

[@sathvikmalgikar2842]

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