If you found 3, you should give evidence that no other solution exists. But this is easy as the power function is strictly monotonous. The solution given is otherwise tooooo complicated and wrong by the way as no one mentioned that we are searching for integer solutions as 1*35 or 5*7.
You can do this more easily by lettling f(x) = 2^x + 3^x. We know f'(x)=2^x * log(2) + 3^x * log(3) > 0, thus f is injective. We know by trial and error that f(3)=35. But because f is injective, this is the only value of x for which f(x)=35. So n=3 is the only solution to our original problem.
If you're going to do trial and error, it's a lot simpler to just plug integer exponents into the original equation. 3^4 = 81, so you know immeditately that n is 2 or 3, which doesn't leave you with a whole lot of trials. Unless you're asked to _derive_ the solution and show your work, this is one where you solve by inspection, in ~15 seconds, and move on.
yes, but the point of this exercise is how to calculate n. empyrical calculations are not based on assumptions but on logic operations based on the given data.
as many don't seem to understand there is a difference between assuming and testing a result and instead calculating a result, i want instead to point out something. this calculation still is not-so-empyrical. we're assuming n is an integer value, along with 35 factors. there are actually infinite possible values of n (before calculating its value, i mean), not just integer values, along with the factors resulting in 35. you can multiply 5/2 times 14 and the result is still 35 but this exercise is assuming (based on what?) that the 2 parenthesis are equal to integer values, along with the fact that consequently the left parenthesis is lower than the right one. if x and y are both 1/2, the left parenthesis is bigger than the right one (1 > 1/4). also, even if they are both 1, it results in 2 > 1... so i don't see why the left parenthesis should be "granted" to be < than the right one.
@@ThembaNzama-q7c And why a cubic sum instead of a quintic sum ? I didn't follow the video to the end but my guess is 3 works simply because 3 is the solution. And if you realize that you don't need the cubic sum to conclude.
3 is a solution. 2^n + 3^n is a strictly increasing function so that solution is unique. No need for the added complexity in the video to prove it properly, I just did it in one sentence.
"Well", I thought, "the number n can't be that big, let's ballpark it!" First I put n=1, then I tried n=2, then I tried N=3 and I was done in less than a minute.
You look to the expression mod 3 and you will find (-1)^n=-1, so that means n is odd. Then observe that 2^n=35-3^n. The left hand side is positive and so is the right hand side. Thus, n
The hit and trial solution was very intuitive, but since some equations might have extra solutions, so always solving completely helps in understanding the functions and solutions properly.
Three quarters of the way through, you suddenly assumed that you were multiplying two integers without any justification for this step. If it was stated that x is an integer to start with then this could have been found very easily by simple inspection with no work necessary.
Seriously!why you have to go through so complicated algebra! By inspection, and by graph, the left side function only intersect y=35 at one point, and by inspection n=3
This solution is very very wrong - you can not assume that x and y are integers !!! They are only when n is multiple of 3. So you indirectly assumes this in your "general" solution, so the result is only correct through coincidence.
Somehow I think the point of this particular question is: did you solve it in 30 seconds (or less) with a simple trial and error, or did you spend 5 minutes on the rigorous solution? I knew the answer before he started writing. This makes me wonder if there was more to the actual presentation of this problem.
Step by step every equation & logically flow to another. sometimes you solve it in your head faster than this but you made it clear how our brain solved it without realising 😅😅 in a very detailed way
Never seen a worse explanation than this. A much simpler solution is 1) n has to be greater than 0 else 3^n + 2^n can't be 35 2) n can't be >= 4 as 3^4 > 35, 3) so n is eithert 1 or 2 or 3. 4) n= 3.
A very thorough approach, and I do wonder about forcing the lhs into the sum of cubes. The alternative solution most likely uses modulo arithmetic. It is enough to guess and check, which is easy. Then demonstrate that the lhs is an increasing function, therefore just one solution.
The title of the video is misleading. Why title it a Harvard entrance exam problem, but in the actual introduction omit the name Harvard? Wholly unnecessary.
There is a MUCH faster analytical way to solve this in 3 easy steps using modular math: STEP 1 3^n + 2^n = 35 => n ≠ 0 (mod 2) 1 ≡ 1 (mod 3) [1,2,...] ≡ 2 => n = 2 m + 1 for m>=0 STEP 2 3 9^m + 2 4^m = 35 (mod 4) 3 ≡ 3 (mod 9) [2,8,5,...] ≡ 8 => m = 3 a + 1 STEP 3 27 9^3a + 8 4^3a = 35 If a>0, left side > 35 => a=0 => m = 1 => n = 3 Also works with much larger numbers that can't be guessed with trial and error ;)
Notes for those not familiar with modular math: In step 1: - apply modulus operator to both sides - modulus gives remainder after division by a divisor, so 5 mod 3 = 2 Or 5 % 3 == 2 common code syntax Or 5 ≡ 3 (mod 3) common math notation - 2^n mod 2 = 0 for all n>0, since all powers of 2 are multiples of 2 - 3^n mod 2 = 1 for all n>=0, since 3^n=(2+1)^n and all binomial terms containing 2^k are eliminated, leaving only last binomial term of 1 - (3^n + 2^n) mod 2 = (3^n mod 2 + 2^n mod 2) mod 2 = (1 + 0) mod 2 = 1 - 3^n mod 3 = 0 for all n>0 - 2^i mod 3 = [1,2,1,2...] - meaning 2^n mod 3 repeats w/ period 2, starting with n=0 - we're looking for a remainder value equal to 2 so we have a match on phase 1 - so n must have a period of 2 and phase of 1, so n=2i+1 In Step 2: - mod repeats w/ period 3 and a match on index 1
Méthode très longue et ennuyeuse. 3^n est multiple de 3 et impair, 2^n est multiple de 2 et pair.3^n>2^n. 35=30+5=3+32=9+26=27+8.27=3^3 et 8=2^3 .3^3=27et 2^3=8 d'où n=3.
Master of Disaster! 3^x + 2^x is an increasing function as sum of 2 increasing functions therefore a unique solution equals 35 which is obvious 3. Harvard requires SAT or ACT.
Finding 3 in your head takes less time. Like a lot less time than this.
Yeah anyone who's used to juggling numbers can see that.
Exactly
If you found 3, you should give evidence that no other solution exists. But this is easy as the power function is strictly monotonous.
The solution given is otherwise tooooo complicated and wrong by the way as no one mentioned that we are searching for integer solutions as 1*35 or 5*7.
Agreed, and I flunked math in school 55+ years ago. Took less than a minute to work through it.
I needed 6 seconds, then started watching convinced that there was a second solution: why make this video otherwise?
You can do this more easily by lettling f(x) = 2^x + 3^x. We know f'(x)=2^x * log(2) + 3^x * log(3) > 0, thus f is injective. We know by trial and error that f(3)=35. But because f is injective, this is the only value of x for which f(x)=35. So n=3 is the only solution to our original problem.
If you're going to do trial and error, it's a lot simpler to just plug integer exponents into the original equation. 3^4 = 81, so you know immeditately that n is 2 or 3, which doesn't leave you with a whole lot of trials. Unless you're asked to _derive_ the solution and show your work, this is one where you solve by inspection, in ~15 seconds, and move on.
@ Well okay, but is this the only solution? Prove it. It is one thing to find one solution, but have you found all of then?
35 = 27 + 8 = 2^3 + 3^3
Found the same in just within seconds.
Simple et efficace et sans baratin!
yes, but the point of this exercise is how to calculate n. empyrical calculations are not based on assumptions but on logic operations based on the given data.
@sunbrogilgamesh yes, but why 3^n=(3^n)^(3/3) , 2^n=...=(2^n)^(3/3) ?!!!
3³ + 2³ = 27 + 8 = 35
What the hell is he doing all the time ..!😂😂😂😂
I think he teaches us how to make simple problem into complex one.,.......
Yes
Trial and error is much easier than this
for a iquality smallest than 35, yes.
But there are no working steps
as many don't seem to understand there is a difference between assuming and testing a result and instead calculating a result, i want instead to point out something. this calculation still is not-so-empyrical. we're assuming n is an integer value, along with 35 factors. there are actually infinite possible values of n (before calculating its value, i mean), not just integer values, along with the factors resulting in 35. you can multiply 5/2 times 14 and the result is still 35 but this exercise is assuming (based on what?) that the 2 parenthesis are equal to integer values, along with the fact that consequently the left parenthesis is lower than the right one. if x and y are both 1/2, the left parenthesis is bigger than the right one (1 > 1/4). also, even if they are both 1, it results in 2 > 1... so i don't see why the left parenthesis should be "granted" to be < than the right one.
But there is only one value of n.
Good Analysis Bro
If you see 'n' in a problem it usually means you're looking for an integer.
Why choose 3 in the first case rather anything else?
Because you worked it out in your head like most. This video is BS.
He chose 3 because he wanted to create a cubic sum.
@@ThembaNzama-q7c And why a cubic sum instead of a quintic sum ? I didn't follow the video to the end but my guess is 3 works simply because 3 is the solution. And if you realize that you don't need the cubic sum to conclude.
For anyone saying that the problem was a lot easier ,he wanted to do it properly ,to be SURE there aren't any other solutions
3 is a solution. 2^n + 3^n is a strictly increasing function so that solution is unique. No need for the added complexity in the video to prove it properly, I just did it in one sentence.
"Well", I thought, "the number n can't be that big, let's ballpark it!" First I put n=1, then I tried n=2, then I tried N=3 and I was done in less than a minute.
you earned yourself a new subscriber, I really enjoyed your content as an olympiad participant
You look to the expression mod 3 and you will find (-1)^n=-1, so that means n is odd. Then observe that 2^n=35-3^n. The left hand side is positive and so is the right hand side. Thus, n
Its simple hit and trial
With the remark that functions f_1(n)=2^n+3^n and f_2(n)=35 have obviously only one common point (one solution for n).
The hit and trial solution was very intuitive, but since some equations might have extra solutions, so always solving completely helps in understanding the functions and solutions properly.
Well done,Sir,you have done all the necessary steps.We want mathematical analysis,not mental maths.
It appears that you made a mistake in case 1. You were adding the two equations not subtracting.
Three quarters of the way through, you suddenly assumed that you were multiplying two integers without any justification for this step. If it was stated that x is an integer to start with then this could have been found very easily by simple inspection with no work necessary.
Так как 3^n < 35, то n=3.
27+8 = 35.
Данная функция не является параболой, поэтому прямая у=35 пересекает её в одной точке. Ответ один.
Seriously!why you have to go through so complicated algebra! By inspection, and by graph, the left side function only intersect y=35 at one point, and by inspection n=3
This solution is very very wrong - you can not assume that x and y are integers !!!
They are only when n is multiple of 3. So you indirectly assumes this in your "general" solution, so the result is only correct through coincidence.
Somehow I think the point of this particular question is: did you solve it in 30 seconds (or less) with a simple trial and error, or did you spend 5 minutes on the rigorous solution? I knew the answer before he started writing.
This makes me wonder if there was more to the actual presentation of this problem.
Step by step every equation & logically flow to another. sometimes you solve it in your head faster than this but you made it clear how our brain solved it without realising 😅😅 in a very detailed way
Does anyone have any proof that x+y
Why all the dicking around? Throwing 3 in there takes a couple of seconds.
More easy.... you can do 27^n/3 + 8^n /3=35 ; 35^n/3 = 35 then n=3
Never seen a worse explanation than this. A much simpler solution is 1) n has to be greater than 0 else 3^n + 2^n can't be 35 2) n can't be >= 4 as 3^4 > 35, 3) so n is eithert 1 or 2 or 3. 4) n= 3.
This approach is better than pure guess and check, yet simple.
the author didn't prove that the equation has only 1 solution
Can you resolve next? 3^x - 2^x = 6
{11.2+17.5}=28.7 11.2 8^9.5 5^6.28^3^2.2^3 5^3^3.22^3^1^1.1^1 2^3^1^1.1 1^1^1 2^3 (n ➖ 3n+2).
What's the connection between Hawkins and Harvard?
Immediately saw it’s 3, because 27 (=3^3) + 8 (=2^3) = 35.
Harvard class
How ddi you know to raise the exponatials to the power of 3/3 and not for example 2/2?
A very thorough approach, and I do wonder about forcing the lhs into the sum of cubes. The alternative solution most likely uses modulo arithmetic.
It is enough to guess and check, which is easy. Then demonstrate that the lhs is an increasing function, therefore just one solution.
Didn’t say that x and y are integers
After u get xy=6 u can directly put
(6)n/3. And solved that right?
Ans. n=3. 3^3=27+2^3=8. Now 27+8=35.
Did it in my head. n equals cubed 3 cubed= 27 2 cubed =8 27+8=35
Harvard University Admission Entrance Exam: 3ⁿ + 2ⁿ = 35; n =?
35 > 3ⁿ > 2ⁿ > 0; n ϵ ℕ, 4 > n > 2: 3ⁿ + 2ⁿ = 35 = 27 + 8 = 3³ + 2³; n = 3
Answer check:
n = 3: 3ⁿ + 2ⁿ = 35; Confirmed as shown
Final answer:
n = 3
5:07
When X=0 and Y=1, X+Y = XX - XY + YY
What do I miss?
Also why do we know that X + Y is a natural number?
It's XX-2XY+YY.
Also idk why it is a natural number (,,•᷄ࡇ•᷅ ,,)?
sir could you help with integration problems?
I got this question in my olympiad exam and i didn't know how to solve it : [2^(x-3)] × [3^(2x-8) ]=36 ; find x
This test make your head going bold Mr hehe...try and error' the fastest solution
The title of the video is misleading. Why title it a Harvard entrance exam problem, but in the actual introduction omit the name Harvard? Wholly unnecessary.
One of the solution : n = 3
There is a MUCH faster analytical way to solve this in 3 easy steps using modular math:
STEP 1
3^n + 2^n = 35
=> n ≠ 0
(mod 2) 1 ≡ 1
(mod 3) [1,2,...] ≡ 2
=> n = 2 m + 1 for m>=0
STEP 2
3 9^m + 2 4^m = 35
(mod 4) 3 ≡ 3
(mod 9) [2,8,5,...] ≡ 8
=> m = 3 a + 1
STEP 3
27 9^3a + 8 4^3a = 35
If a>0, left side > 35
=> a=0 => m = 1 => n = 3
Also works with much larger numbers that can't be guessed with trial and error ;)
Notes for those not familiar with modular math:
In step 1:
- apply modulus operator to both sides
- modulus gives remainder after division by a divisor, so
5 mod 3 = 2
Or
5 % 3 == 2 common code syntax
Or
5 ≡ 3 (mod 3) common math notation
- 2^n mod 2 = 0 for all n>0, since all powers of 2 are multiples of 2
- 3^n mod 2 = 1 for all n>=0, since 3^n=(2+1)^n and all binomial terms containing 2^k are eliminated, leaving only last binomial term of 1
- (3^n + 2^n) mod 2 = (3^n mod 2 + 2^n mod 2) mod 2 = (1 + 0) mod 2 = 1
- 3^n mod 3 = 0 for all n>0
- 2^i mod 3 = [1,2,1,2...]
- meaning 2^n mod 3 repeats w/ period 2, starting with n=0
- we're looking for a remainder value equal to 2 so we have a match on phase 1
- so n must have a period of 2 and phase of 1, so n=2i+1
In Step 2:
- mod repeats w/ period 3 and a match on index 1
3ⁿ + 2ⁿ = 35 ⇒ 0< 3ⁿ < 35 ⇒ 0≦n≦3 ⇒ n=0 or 1 or 2 or 3
n=0 ⇒ 3ⁿ + 2ⁿ = 1 + 1 = 2, No
n=1 ⇒ 3ⁿ + 2ⁿ = 3 + 2 = 5, No
n=2 ⇒ 3ⁿ + 2ⁿ = 9 + 4 = 13, No
n=3 ⇒ 3ⁿ + 2ⁿ = 27 + 8 = 35, Bingo!
Put me in Harvard 😤
I aolve this within a second
Good solution
Isn't it obvious that n=3? a "tricky question"? really?
Answer 3
n=3ans
did this in my brain in 2 seconds lmao
N=3
he is making things complicated
THIS IS SO EASY😊
n=3
You make easy thing loook hard
word for word replicate of other videos...
Bhai ek second me 3 answer de diya maine😂
Méthode très longue et ennuyeuse. 3^n est multiple de 3 et impair, 2^n est multiple de 2 et pair.3^n>2^n. 35=30+5=3+32=9+26=27+8.27=3^3 et 8=2^3 .3^3=27et 2^3=8 d'où n=3.
Master of Disaster! 3^x + 2^x is an increasing function as sum of 2 increasing functions therefore a unique solution equals 35 which is obvious 3. Harvard requires SAT or ACT.
Very nice presentation, really entertaining, keep up the good work😃
Better is 2/2
Bravo
What a waste of paper and ink!
3 piece of cake
3^x +2^x=35
3^x +2^x=45-10
3^x -45=-2^x -10
3^2(3^x-2 -5)=-2(2^x-1+5)
3^x-2 -5=-2 et
2^x-1+5=9
3^(x-2)=3
X-2=1
X=3
2^(x-1)=9-5
X-1=2
X=3
X appartient a Z
I really need to find u 😂
3.
th th th th th :D
Deslike!
Nul
Very bad
3
Vous nous faites chier avec vos pseudo théories que vous avez créé vous même !!!!!!!!!!
De plus, 2700 vues et 100likes en 8h!
Ça motive les charlatans comme lui.
2840 vues !
😬
(3^3)=27+(2^3)=8
27+8=35
so n=3
I don´t understand you why not?: N(lo3+log2)=log35, N=log35/(log3+log2)=1,984277
👍
N=3
Bravo
3
3
3