Math Olympiad | A Nice Exponential Problem 😊
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7, 4, 2 - brute force in my head. 128+16+4
Not elegant, but quick.
Same here...
em... 他们的数学确实不行@@christopherdean1326
I did the same. Easy
same bro
But there is not compulsory that value of a will be 2 it also will be 7 or 4
Just write 148 in binary notation can do the job very easily.
Came here to say this. Easy if you are a programmer.
Bilikul
you are not matematician, please shut up
The first that I've thought.
You can represent any number in binary => as sum of pows of 2.
Can u pls explain how u do this?
You are partially correct..
i.e =
Now the triplets can form in following manner :
1.a=7,b=4,c=2
2.a=7,b=2,c=4
3.a=2,b=7,c=4
4.a=2,b=4,c=7
5.a=4,b=7,c=2
6.a=4,b=2,c=7
In each case :
2⁷+2⁴+2²=128+16+4=148
And you need to impose the condition that a,b,c are positive integers
2,4,7. You can do it in your head if you are familiar with common numbers in binary, as most computing hardware people are :)
This equation has 6 different solutions if there is not any restriction which is biger of a. b, c.
Yes
Absolutely. Any combination of (2, 4, 7) will be a solution: (2, 7, 4), (4, 2, 7), (4, 7, 2), etc.
There are an infinite number of solutions because there's no restriction on using integers only. I can rewrite the problem as finding the sum of any 3 numbers to be 148: a' + b' + c' = 148 . I can chose two random values for a' and b' , and then compute c' as c' = 148 - a' - b' . This has an infinite number of solutions. Then, I can cast any such solution to the original problem by taking the log base 2 of a' , b' , and c' to get a, b, and c. The numbers can be complex values, too. [Remember that any complex number can be converted to r*e^(i*θ) form, and it's easy to get the log base 2 of that, even if it gets messy.]
For fun, let's use 2^e and 2^π as two of the three terms to sum up to 148 . We have:
2^e + 2^π + 2^c = 148
2^c = 148 - 2^e - 2^π
c = ln(148 - 2^e - 2^π)/ln(2) ≈ 7.0508731651623102433373853534896 .
Thus, a = e , b = π , and c = ln(148 - 2^e - 2^π)/ln(2) ≈ 7.0508731651623102433373853534896 .
We can have more fun with 148 = 128 + 32 - 12 = 2^7 + 2^5 + 2^c , with a = 7 and b = 5 .
2^c = -12 = 12*(-1) = 12*e^(i*π*(1+2*k)) , k any integer
c = ln[12*e^(i*π*(1+2*k))]/ln(2)
c = ln(12)/ln(2) + i*π*(1+2*k)/ln(2)
c = 3 + ln(1.5)/ln(2) + i*π*(1+2*k)/ln(2) , k any integer
You re right! And don not forget if a=b=c.
One needs the restriction that a,b,c are integers. Otherwise there are solutions such as a=pi, b=pi, c=7.026247...
Since the base common 2 , therefore , under the conditions of a
yes, but to prove that this is the only solution you need to then calculate for c = 6. while when you reach to 2^a*(1+..) = 2^n*(odd number) a is n and only n and this is a theorem.
@@user-ok1mi3pw5w only tried to suggest an alternative and quicker solution in order to save time, taking given condition (a
@@user-ok1mi3pw5wyou can take the next power and see it has no solution. Smaller power after that pointless as failed higher power is needed for a solution to work. QED
Me encantan estos ejercicios!!! Se resuelven fácilmente con cálculo mental. Sigo la misma lógica que usted. Son potencias de 2.
Creo que lo importante que aporta este video es el método para resolverlo cuando no es tan fácil.
Самое прикольное решение - записать 148 в двоичной системе. 148 = 10010100. Разряды с единичкой 7, 4 и 2)
Тсс, не пали контору 😂
Осталось не забыть про перестановки и доказать единственность )
If you grew up with The Book of Change, aka I Ching 易經, then you realize it is Yin Yang 陰陽 and realize that writing the numbers in binary gives you the answers right away.
I like recursive procedures. It really builds skill.
Answer us obvious:
a=b=c=ln(148/3)/ln(2)
Thank you very much my dear friend,
Its a simple one but you make it complicated. :) :) :)
Mind = blown
Thanks for good explanation.
In comment section, some are getting cocky but no one presented any systematic method
الطريقة جميلة
Divide left and right by 4 and the response appears immediately.
If you count powers of 2, you cannot guo beyond 2^7. as 2^8 will exceed 148. Therefore let us assume a=7, that makes 2^7=128
Balance is 148-128=20
If we assume b next, it cannot exceed 4 so b=4
Now the total is 128+16 =144
Balance is 4 means 2^2. Therefore c=2
Therefore a,b,c=7,4,2
They can be in any more combination
Vì 148 < 2^8 nên a, b, c < 8.
Trường hợp 1:
a = 7 thì 2^a = 2^7 = 128 => 2^b + 2^c = 20.
Lại vì 2^5 = 32 > 20 nên b, c < 5
Xét b = 4 thì 2^b = 2^4 = 16 => 2^c = 4 => c = 2
Vậy ta tìm được một bộ nghiệm (a, b, c) = (7, 4, 2) và các hoán vị của nó.
Xét b = 3 thì 2^b = 2^3 = 8 => 2^c = 12 , loại vì 12 không phải số chính phương.
Xét b = 2 thì c = 4
Xét b = 1 thì 2^b = 2 => 2^c = 18, loại vì 18 không phải số chính phương.
Xét b = 0 thì 2^b = 2^0 = 1 > 2^c = 19, loại vì 19 không phải số chính phương.
Trường hợp 2
a = 6 thì 2^a = 2^6 = 32 => 2^b + 2^c = 112.
Lại vì 2^7 = 128 > 112 nên b, c < 7
Xét b = 6 thì 2^6 = 64 => 2^c = 58. Loại vì 58 không phải số chính phương.
Xét b = 5 thì 2^b = 2^5 = 32 => 2^c = 80. Loại vì 80 không phải số chính phương.
Do vai trò của a, b, c là như nhau. Nên bài toán chỉ có một bộ nghiệm (a, b, c) = (7, 4 , 2) và các hoán vị của nó: (7, 4, 2); (7, 2, 4); (4, 2, 7); (4, 7, 2); (2, 7, 4) và (2, 4, 7).
Divide both sides by 4. Let x = a-2, y = b-2, z = c-2 then 2^x + 2^y + 2^z = 37. Since 37 is odd, one of the terms must be odd on the left. Let z=0 so 2^x + 2^y = 36. Divide both sides by 4. Let j = x-2, k = y-2 then 2^j + 2^k = 9. Since 9 is odd, one of the terms must be odd on the left. Let k=0 so 2^j = 8. 2^j = 2^3 so j=3. Now we plug in our solutions to the previous substitutions. j = 3 = x - 2 = a - 2 - 2 so a = 7. k = 0 = y - 2 = b - 2 - 2 so b = 4. z = 0 = c - 2, so c = 2. a = 7, b = 4, c = 2 is the solution, or more accurately 7, 4 , and 2 are the combination of solutions for a, b, and c that lead to all of the solutions.
Treat this like converting to base 2.
2^a = 128
2^b = 16
2^c = 4
Therefore a=7, b=4, c=2
The easy solution in 5 sec in my head: 2^7+2^4+2^2=128+16+4=148
So (2,4,7) is certainly a solution set. But maybe there are more solutions? (at least you can have various combinations, of which is a, b or c, if it is unclear that eg a>b>c is one condition)
Excelente...!!!!!! Tú explicación es perfecta.
आपने बिल्कुल् सही तरीके से समझाया है लोगो की कमेंट पे मत जाए आप सबका कहना सही हो सकता है madam ने पहले a की वैल्यू निकाली इस लिये a ki value 2 आयी यदि madam ने b या c पहले लिया hota to b या c की वैल्यू भी 2 ही आती
Let's assume a = 2 and b=4, then applying the assumption in the equation,
2 power c = 148-20 = 128
2 power c = 2 power 7
Hence, a = 2, b = 4 and c = 7
its too effective solution
Divide both sides to 4. and 37 is 32 + 4 + 1 which are powers of 2. a-2=5, b-2=2, c-2=0. So a=7, b=4, c=2.
Simply Amazing!!! 🎉
So much Laborious
Nice method to solve equation
Very nice, thankyou
Your video is very informative videos mam
Very good solution.
128 + 16 + 4
a, b, and c can be any in order
There are 3 combinations of the answer. Therefore there are 6 possible answers for a, b, and c
1, 4, 7
Correct
Clumzy way to solve
2, 4, 7
148 = 128 + 16 + 4
Solved it instantly
thanks
Love the explanation
(4; 2; 7). Et on fait toutes les permutations sur ce triplet pour avoir 6 solutions.
Le problème on peut le poser de la manière suivante:
On cherche dans l'espace de repere (O ; x ; y ; z ) les points M( a; b ; c ) dont les coordonnées vérifient l'équation 2^a + 2^b + 2^c = 148.
L'enssmble des solutions, s'il n'est pas vide, contiendra des triplets différents ou égaux.
Par la nature de l'équation les nombres a, b, et c sont permutables ce qui donne 3! = 6 triplets égaux ou différents.
Et comme les nombres obtenus sont deux à deux distincts alors on a 6 triplets solutions donc 6 points .
Very nice handwriting.
Thank you 🙏
Excelente videom
Perfect for students
Excellent 👏👏👏
Good job
You need the condition that a, b, and c are integers.
同除2次2後右邊出現單數 可以推測此時左邊有一項變成1 再同減1 繼續除2 除了2次後右邊又出現單數 代表左邊又出現1 可以推出a為2 b為4 再推c就不難了 用心算就解開了
Перевода не знаю, но А ,В и С могут быть равными и 2 и 4 и 7 или 4, 2, 7 и в других комбинациях.
Amazing
your video is realy very helpful for us
Well, 148 = 128 + 16 + 4 = 2^7 + 2^4 + 2^2.
I expect guys in my profession (digital circuit desing) will eat this question for lunch - we know our powers of two up one side and down the other. The thing to note here is that if you just write 148 in binary, then only the bits corresponding to those three powers will be set.
Now, if you'd instead used 149, or really any number that's not the sum of three powers of two, then it becomes a MATH problem, and it would be much harder. But when it's a simple case like this we just sort of "see it" without thinking much at all.
I was searching for this only from half and hour😅😅
Thank you soo much mam for the explanation
Nice 👍🏿
решается в уме с налета, менее чем за полминуты. 129+16+4
🎉
Very nice
So complicated)). Just divide both sides of given equation by 2^a and that’s it.! Right side will be equal to odd number and only possible number on the right will be 37, a=2 comes right away, that’s a key. Further just technic
बहुत बढ़िया
Ese problema tiene una solución extremadamente facil, la base se trataria de entender, que el 2 elevado a cualquier potencia tiene un comportamiento al sistema binario en las computadoras
Nice video mam
Excellent. Brasil.
FANTASTIC!!!
148 попробуем разложить на сумму трех чисел с основанием 2. Одно из слагаемых 128, так как следующие степени числа 2 в сумме дают 64+32=96, что при вычитании из 148 дает 52, а это число не степень 2., далее 148-128=20, 20 это 16 и 4 однозначно, следовательно 148=2⁷+2⁴+2², a,b,c(7,4,2)
Можно просто выписать степени числа 2. Это 2, 4, 8, 16, 32, 64, 128. Очевидно, что одно из чисел это 128, не хватает 20, а это 16 +4. Т.е. 7, 4 2.
Solução maravilhosa
Mathematicians sure do like to play with themselves. Filling two pages with equations, when all you have to do is decompose 148 into its three largest powers of 2 (128, 16, & 4) and then find the exponents for 2.
,😊👍👏
Thanks for easy solution.
I started by taking the biggest exponent out first. That being 128 or 2^7. That leaves 20. I took out 16, or 2^4 and 4 or 2^2. My answer is 2^2+2^4+3^7=148
Выписать степени 2 от 0 до 7 и подобрать те,которые в сумме дадут 148 это и будут целые корни
Bravo bombai!
Muito mais simples assim:
Compor tudo com base 2 elevado a "n":
2 4 8 16 32 64 128
148 - 128 = 20
20 = 4 + 16
Pronto: 148 = 4 + 16 + 128
a=2 b=4 c=7
Olimpíada requer rapidez !
Thank's
Произвольно назначаем значение "а" И "в" И вычисляем"с" Легко и просто т. к другой зависимости между "а" " в" и "с" нет
Please stop calling every basic math problem “a nice Olympiad one”. There is nothing Olympiad about this. It’s trivial and is solvable in 10 seconds without a pen and paper.
yes as a 7th grade i literally thought of 128+16+4 which is 7,4,2
Depois que esta resolvido todo mundo acha fácil ..rs..queo ver na hora
Me gusta lo concreta y ordenada que es tu resolución. Saludos desde Córdoba en Argentina.
I like it
As we have practice, just by looking at the problem comes the solution. In this case, as there is no great order of the exponents, we can have six different combinations of solutions.
There is a trick that simplifies the resolution:
2^a + a^b + 2^c = 2² * 37
Step 2² to the other side and divide, then it's easy and even an elementary school child can find the solution:
2^a-2 + 2^b-2 + 2^c-2 = 37
It has to be: 1 + 4 + 32
So one of the six solutions is:
a = 2
b = 4
C = 7
Proof: 4 + 16 + 128 = 148
You're been sloppy in your writing:
2^a-2 + 2^b-2 + 2^c-2 = 37
You're missing parentheses to group the exponents to handle proper precedence rules, so you have:
2^a + 2^b + 2^c = 43
You should've written:
2^(a-2) + 2^(b-2) + 2^(c-2) = 37
Mr up why all this in wanted rubbish nearest 2 power of 148 is 128,+20, 20 in 2 summation powers is 4+16, very easy, lengthy unwanted is not required at all
Mukund
Very good ❤❤❤❤❤❤❤
This is a trivial problem for anyone used to converting from decimal to binary.
148 = 10010100 = 128+16+4 = 2^7+2^4+2^2
If in the original example we replace 148 with 144=10010000=(2^6+2^6+2^4 or 2^7+2^3+2^3) this method will not work.
Since you didn't indicate that you meant "10010100" to be binary, your answer is wrong because 148 ≠ 10,010,100.
Also, the problem asked for the values of a, b, and c, which you didn't provide; you didn't answer the question.
This is like having a task to get 3 items from the store, so you go there to fetch them from the shelves, but go home, leaving them in the cart.
@@Change_Verification: FYI, 144 ≠ 10,010,000 . You're being sloppy with your work.
Also, using your example of 144, once it's decomposed to the sum of two powers of 2, namely 2^7 and 2^4, we still need the sum of three powers of 2. That's accomplished by splitting either term into the sum of two halves; i.e., substitute 2^7 with 2^6 + 2^6, or 2^4 with 2^3 + 2^3 in 144 = 2^7 + 2^4 . This is just an added step after converting to binary.
@@oahuhawaii2141 144 ≠ 10,010,000 ? Indeed! 😂😂😂 "This is just an added step" - this essentially means another solution that differs from the original one.
I used binary (base 2) combinations to get 7, 4, 2
The equation has infinite solutions because it is one equation and three unknowns. One of them is
a=b=c=(ln(148/3))/ln2
she tacitly assumed a, b, c to be natural numbers. ((Diophantine equation))
You can't see so closely.
2-4-7 just by logic🎉
2^7=128 so 148 is equivalent to a 7+2=8 digit binary number.
2^7+2^6+2^5+2^4+2^3+2^2+2^1+2^0 will be equal to 148 if 2^6, 2^5, 2^3, 2^1 and 2^0 is equal to 0. So, the binary equivalent number of 148 is 10010100.
Is it logical to multiply and divide LHS with the same value. mathematically operation on LHS should also hold for RHS.
a=0 b=0 c=7,21 is a possible answer too.
Turn it into binary and the digit positions give the answer - I did it in less than a minute in my head that way.
Same here, I solved it before clicking on the thumbnail image.
@StevenLubick
I only clicked the video to find out what the heck she could be doing for 12 minutes.
Can you write the solutio in your methode. Please
@@s.m.a9324 148 in binary = 1001010. The 2nd digit from the right indicates a value of 4, the 4th digit from the right indicates a value of 16 and the 7th digit indicates a value of 128. 128+16+4 =148. This is possible because all values are powers of 2.
Easy method using computer programming is to write it in binary which is 10010100 1 in 8th place, 5th place and 3rd place, 8-1=7, 5-1=4, 3-1=2@@s.m.a9324
a=2, b=4, c=6
Pensei nas potências de 2 e rapidamente combinei os números que davam a resposta. Quando são inteiros, é certeiro.
Too much descriptive and very helpful
Very lengthy. I did it just seeing it and the answer is c=7, b=4, a=2 . There are multiple solution with interchange of power of a, b and c.
148=10010100 then a=7, b=4, c=2.
First: an assumption that they are all integers.
Numbers that can be constituents: 1, 2, 4, 8, 16, 32, 64, 128
1 can only be used if there are two of them, because the answer is even. 148-2=146 which isn't an integer power of 2, so rule out 1. Therefore, exponents a,b,and c are not 0.
What remains is to find a combination of the numbers I've written to make 148, and repetition is allowed if needed.
4, 16, and 128, works, though there may be other combinations.
Exponents a, b, and c can be 2,4,and 7 in any order.
Although this looks long winded, I actually did it in my head much more quickly.
7,4,2
HOW MUCH TIME WILL BE ALLOTED FOR A SUMS ?
Easily, 148 = 128 + 16 + 4 = 2^7 + 2^4 + 2^2 so the answer is 7, 4, 2. Solved it in my head in 2 seconds.
Only one question with 3 variable not known , the, answer will be many. Need 2 more equestions to be exact solution.
2^a + 2^b + 2^c = 148 , and let a >= b >= c >= 0 (all integers).
I know the powers of 2 (up to 2^24), so I'll take out the biggest chunks first.
I see 2^7 = 128 is just below 148 .
So, let a = 7 , and simplify:
128 + 2^b + 2^c = 148
2^b + 2^c = 20
I see 2^4 = 16 is just below 20 .
So, let b = 4 , and simplify:
16 + 2^c = 20
2^c = 4
I know 2^2 = 4, so c = 2 .
Alternatively, we can convert to binary:
148/2 = 74 r 0
74/2 = 37 r 0
37/2 = 18 r 1
18/2 = 9 r 0
9/2 = 4 r 1
4/2 = 2 r 0
2/2 = 1 r 0
Thus, 148 = 10010100b = 2^7 + 2^4 + 2^2 .
Therefore, a = 7 , b = 4 , and c = 2 .
Another solution:
LHS is sum of powers of 2, and RHS has factors of 2, so factor them out.
148 = 4 * 37 = 2^2 * 37
Let a = 2 + a' , b = 2 + b' , and c = 2 + c' .
2^2 * (2^a' + 2^b' + 2^c') = 2^2 * 37
2^a' + 2^b' + 2^c' = 37
Let c' = 0 . Thus, c = 2 .
2^a' + 2^b' + 2^0 = 37
2^a' + 2^b' + 1 = 37
2^a' + 2^b' = 36
LHS is sum of powers of 2, and RHS has factors of 2, so factor them out.
36 = 4 * 9 = 2^2 * 9
Let a' = 2 + a" , and b' = 2 + b" .
2^2 * (2^a" + 2^b") = 2^2 * 9
2^a" + 2^b" = 9
Let b" = 0 . Thus, b = 2 + b' = 2 + (2 + b") = 4 .
2^a" + 2^0 = 9
2^a" + 1 = 9
2^a" = 8
2^a" = 2^3
a" = 3 . Thus, a = 2 + a' = 2 + (2 + a") = 7 .
Therefore, a = 7 , b = 4 , and c = 2 .
Actually, there are 6 possible answers for a, b, c because of the associative property of addition. So (a,b,c) = (2,4,7) or (2,7,4) or (4,2,7) or (4,7,2) or (7,2,4) or (7,4,2)😊