I mean, you don't even need to be that familiar with binary. You just need to know powers of 2 up to 7. It's pretty intuitive that the only combination of 3 variables that will add up to 148 are 2, 4, 7, even if you brute force it.
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.
@@КирилАнастасовyou 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.
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
If you're dealing with natural numbers, such as 148, then it's best to limit the powers to nonnegative integers. By including 0, it's possible to represent odd numbers, such as 7: 2² + 2¹ + 2⁰ . The lowest sum of 3 nonnegative integer powers of 2 is 3. It's possible to get down to sums of 2 or 1 using negative integer powers. Any sum below that requires complex numbers.
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
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
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.
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.
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)😊
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)
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).
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
Simple thinking gets me the highest exponent of 2 that is lower than 148. This number is 7 for 2^7=128. Thus 148-128=20. Now, which exponent of 2 is lower than 20. The answer is 4 for 2^4=16. Thus 20-14=4. And, of course, 2^2=4. The answer is 7, 4 and 2. So, since there are not any requirements for the values of a, b and c, you can have these combinations showing their relative values in increasing (or decreasing) order : abc, acb, bac, bca, cab, cba. Then, there are 6 solutions. All this suppose that a, b and c must be integers. Otherwise, there can be infinite possibilities.
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.
@@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
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
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 .
I do not know why you went through this prolonged and exhaustive manipulation. However, you can write 148 as a sum of numbers in N each value with a base of 2 and a positive exponent in N, such as: 2^7 + 2^4 + 2^2 = 128 + 16 + 4 = 148. Any of the exponents of 2^a + 2^b + 2^c = 148 could assume any of the numerical values of 2, 4, and 7. Therefore, there can never be any unique integer solution for a, b, and c, since any of these exponents can assume any of the values of 2, 4, and 7.
Problemita de 1 de secundaria diria yo , solo usando lógica llegas al relsultado y matematicamente sigue siendo fácil , no se porque le ponen math olympiad Aunque tampoco quiero ser hater asi que si solo pusiera ecuación exponencial me hubiera encantado verlo, aunque pudo ser para llamar más la atención que si lo logró hacer, pues me molesto que pusiera math olympiad pero bueno , almenos hizo cada procedimiento para los nuevos 😃
Le di me gusta pero no suscribi solo por que no sé inglés intermedio y no me gustaria tener videos solo en inglés el único canal que me hizo hacer estudiar inglés fue "Prime Newtons"
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)
The sum of all three exponents must equal 148. Since 2^8=256, we know the highest exponent can only be to the 7th power. 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 2^6=64 2^7=124 Pick the three above exponents that sum to 148 2^2=4 2^4=16 2^7=124 =148 2,4,7
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
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
You've 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
You wrote "1, 4, 7". The answer is any of 6 permutations of "2, 4, 7" if only integers are allowed. Actually, (a, b, c) aren't restricted to whole numbers, so there are an infinite number of solutions: (0, 0, log2(146)) { 1+1+146 } (-1, -1, log2(147)) { 1/2 + 1/2 + 147 } (i*π/ln(2), 0, log2(148)) { -1 + 1 + 148 } (i*2*π/3/ln(2), -i*2*π/3/ln(2), log2(149))
Всё на много проще: • перевести число 148 в двоичную систему : 10010100 • так как у нас 3 единицы в числе, то они соответствуют степеням двойки : 2, 4, 7 Значит, 148 = 2^2 + 2^4 + 2^7 Ответ: (2,4,7), (2,7,4), (4,2,7), (4,7,2), (7,2,4), (7,4,2)
There is a lot of simpler ways to resolve this problem, I propose this one. Powers of 2 end in 2,4,6 or 8. 148 ends in 8 so the possible combinations are 2,2,4 or 4,6,8. The powers involved are up to 128, using 2,2,4 It is not possible to get to 148, at most with 32,32,64 you get to 128, and Is really simple and immediate, analysing the possible combinations, find 4, 16 and 128.
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.
@Change_Verification: You also can get 144 with 128 + 32 - 16 using exponents of 7, 5, and 4+i*π*(1+2*k)/ln(2) with k being any integer. The original problem doesn't have a restriction on the powers, so complex numbers are allowed.
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?
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
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 :)
I mean, you don't even need to be that familiar with binary. You just need to know powers of 2 up to 7. It's pretty intuitive that the only combination of 3 variables that will add up to 148 are 2, 4, 7, even if you brute force it.
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.
@@КирилАнастасов only tried to suggest an alternative and quicker solution in order to save time, taking given condition (a
@@КирилАнастасовyou 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.
Nice method to solve equation
Very good solution.
its too effective solution
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
I used complex numbers, just for fun, by using 148 + 1 - 1 = 148:
a = ln(148)/ln(2)
b = 0
c = i*π*(1+2*k)/ln(2), k integer
If you're dealing with natural numbers, such as 148, then it's best to limit the powers to nonnegative integers. By including 0, it's possible to represent odd numbers, such as 7: 2² + 2¹ + 2⁰ . The lowest sum of 3 nonnegative integer powers of 2 is 3. It's possible to get down to sums of 2 or 1 using negative integer powers. Any sum below that requires complex numbers.
I like recursive procedures. It really builds skill.
Thank you very much my dear friend,
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.
Simply Amazing!!! 🎉
Love the explanation
Very nice, thankyou
Excelente videom
Amazing 😊😊
Your video is very informative videos mam
Perfect for students
Very nice handwriting.
Thank you 🙏
Good job
Amazing
Nice 👍🏿
बहुत बढ़िया
Clumzy way to solve
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 !
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
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 )
Это единственное правильное решение. Как бы она решала если бы неизвестных было к примеру 20 скажем для числа 2 в 20 минус 1?
148 = 94h = 10010100b
Excelente...!!!!!! Tú explicación es perfecta.
So much Laborious
Nice video mam
Mind = blown
Thanks for good explanation.
In comment section, some are getting cocky but no one presented any systematic method
FANTASTIC!!!
Excellent 👏👏👏
Very nice
Thank you soo much mam for the explanation
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
This is quite fun
thank you great
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
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.
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)😊
Actually, (a, b, c) aren't restricted to whole numbers, so there are an infinite number of solutions:
(0, 0, log2(146)) { 1+1+146 }
(-1, -1, log2(147)) { 1/2 + 1/2 + 147 }
(i*π/ln(2), 0, log2(148)) { -1 + 1 + 148 }
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)
Nice to learn these maths
Solução maravilhosa
your video is realy very helpful for us
Answer us obvious:
a=b=c=ln(148/3)/ln(2)
These solutions for (a, b, c) are easy, too:
(7, 1/log(2), 1/log(2)) { 128 + 10 + 10 }
(2/log(2), 5, 4) { 100 + 32 + 16 }
(7, 5, 3 + ln(1.5)/ln(2) + i*π/ln(2)) { 128 + 32 + -12 }
(log2(148), 0, i*π/ln(2)) { 148 + 1 + -1 }
(log2(150), 1/2+i*3*π/ln(16), 1/2-i*3*π/ln(16)) { 150 + (-1+i) + (-1-i) }
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).
Perfect teaching
One needs the restriction that a,b,c are integers. Otherwise there are solutions such as a=pi, b=pi, c=7.026247...
You can use complex numbers, too.
Treat this like converting to base 2.
2^a = 128
2^b = 16
2^c = 4
Therefore a=7, b=4, c=2
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
🎉
thanks
Simple thinking gets me the highest exponent of 2 that is lower than 148. This number is 7 for 2^7=128. Thus 148-128=20. Now, which exponent of 2 is lower than 20. The answer is 4 for 2^4=16. Thus 20-14=4. And, of course, 2^2=4. The answer is 7, 4 and 2. So, since there are not any requirements for the values of a, b and c, you can have these combinations showing their relative values in increasing (or decreasing) order : abc, acb, bac, bca, cab, cba. Then, there are 6 solutions. All this suppose that a, b and c must be integers. Otherwise, there can be infinite possibilities.
148 - 20 = 128 (наибольший квадрат от двух) то есть одно из чисел 2^7
А 20 можно получить только из двух 16 и 4, значит 2^2, и 2^4 остаётся
Me gusta lo concreta y ordenada que es tu resolución. Saludos desde Córdoba en Argentina.
I like it
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.
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.
Just for fun, I used complex numbers with 148 + 1 - 1 = 148:
a = ln(148)/ln(2)
b = 0
c = i*π*(1+2*k)/ln(2), k integer
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.
Excellent. Brasil.
Divide left and right by 4 and the response appears immediately.
hit nd trial krlete madam ji
Your answer is incomplete. There are several more answers: a,b,c=2,7,4 a,b,c=4,2,7 a,b,c=4,7,2 a,b,c=7,2,4 a,b,c=7,4,2
Oh please
It's because all unknown terms composed of similar base which is 2.
If you want to be pedantic, your answer is incomplete too. The question isn't limited to integers, and there are an infinite number of solutions.
@@jakefromstatefarm6969 This is the entire answer, there is no other answer.
@@Huaxiaviewpoint华夏观点 what about a=0, b=0, c=log2 146.
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.
Bravo bombai!
Great 👍🏻.
Shortcut approach
148 = 128+16+4
Value of a,b,c can be 7,4,2
Its a simple one but you make it complicated. :) :) :)
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
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
In fact there are 3C3 ie ( 3 combination 3 )number of solutions to a , b and c
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 .
I do not know why you went through this prolonged and exhaustive manipulation.
However, you can write 148 as a sum of numbers in N each value with a base of 2 and a positive exponent in N, such as:
2^7 + 2^4 + 2^2 = 128 + 16 + 4 = 148.
Any of the exponents of 2^a + 2^b + 2^c = 148 could assume any of the numerical values of 2, 4, and 7. Therefore, there can never be any unique integer solution for a, b, and c, since any of these exponents can assume any of the values of 2, 4, and 7.
Pensei nas potências de 2 e rapidamente combinei os números que davam a resposta. Quando são inteiros, é certeiro.
Problemita de 1 de secundaria diria yo , solo usando lógica llegas al relsultado y matematicamente sigue siendo fácil , no se porque le ponen math olympiad
Aunque tampoco quiero ser hater asi que si solo pusiera ecuación exponencial me hubiera encantado verlo, aunque pudo ser para llamar más la atención que si lo logró hacer, pues me molesto que pusiera math olympiad pero bueno , almenos hizo cada procedimiento para los nuevos 😃
Dije primero de secundaria por que yo estoy en primero quise decir primaria pero seria muy arrogante 😁
Le di me gusta pero no suscribi solo por que no sé inglés intermedio y no me gustaria tener videos solo en inglés el único canal que me hizo hacer estudiar inglés fue "Prime Newtons"
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.
Depois que esta resolvido todo mundo acha fácil ..rs..queo ver na hora
Thank's
الطريقة جميلة
You need the condition that a, b, and c are integers.
Convert 148 into the binary number. U will find three 1. That's the answer.
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)
Actually, (a, b, c) isn't restricted to whole numbers, so there's an infinite number of solutions, where 2^a + 2^b + 2^c = 148 .
128 + 10 + 10:
(7, 1/log(2), 1/log(2))
100 + 32 + 16:
(2/log(2), 5, 4)
128 + 32 + -12:
(7, 5, 3 + ln(1.5)/ln(2) + i*π/ln(2))
146 + 1 + 1:
(log2(146), 0, 0)
147 + 1/2 + 1/2:
(log2(147), -1, -1)
148 + 1 + -1:
(log2(148), 0, i*π/ln(2))
149 + (-1+i*√3)/2 + (-1-i*√3)/2:
(log2(149), i*2*π/ln(8), -i*2*π/ln(8))
150 + (-1+i) + (-1-i):
(log2(150), 1/2+i*3*π/ln(16), 1/2-i*3*π/ln(16))
Thanks for easy solution.
The sum of all three exponents must equal 148.
Since 2^8=256, we know the highest exponent can only be to the 7th power.
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=124
Pick the three above exponents that sum to 148
2^2=4
2^4=16
2^7=124
=148
2,4,7
2^7=128
The exponents are 2, 4, and 7. Their sum is 13, not 148.
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
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
You've 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
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
2, 4, 7
You wrote "1, 4, 7". The answer is any of 6 permutations of "2, 4, 7" if only integers are allowed.
Actually, (a, b, c) aren't restricted to whole numbers, so there are an infinite number of solutions:
(0, 0, log2(146)) { 1+1+146 }
(-1, -1, log2(147)) { 1/2 + 1/2 + 147 }
(i*π/ln(2), 0, log2(148)) { -1 + 1 + 148 }
(i*2*π/3/ln(2), -i*2*π/3/ln(2), log2(149))
Why are you making simple mathematics so complex?
Correct
Всё на много проще:
• перевести число 148 в двоичную систему : 10010100
• так как у нас 3 единицы в числе, то они соответствуют степеням двойки : 2, 4, 7
Значит, 148 = 2^2 + 2^4 + 2^7
Ответ: (2,4,7), (2,7,4), (4,2,7), (4,7,2), (7,2,4), (7,4,2)
Выписать степени 2 от 0 до 7 и подобрать те,которые в сумме дадут 148 это и будут целые корни
решается в уме с налета, менее чем за полминуты. 129+16+4
Além do exercício gostei da sua voz!❤
Thank you 🙏
I was searching for this only from half and hour😅😅
HOW MUCH TIME WILL BE ALLOTED FOR A SUMS ?
There is a lot of simpler ways to resolve this problem, I propose this one. Powers of 2 end in 2,4,6 or 8. 148 ends in 8 so the possible combinations are 2,2,4 or 4,6,8. The powers involved are up to 128, using 2,2,4 It is not possible to get to 148, at most with 32,32,64 you get to 128, and Is really simple and immediate, analysing the possible combinations, find 4, 16 and 128.
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.
@Change_Verification: You also can get 144 with 128 + 32 - 16 using exponents of 7, 5, and 4+i*π*(1+2*k)/ln(2) with k being any integer. The original problem doesn't have a restriction on the powers, so complex numbers are allowed.
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
Задачку можно решить поэтапным делением на 2, но можно и по стандарту, как на видео. Если это и олимпиада, то школьный этап, не более.
Да, по сути так. А как Вам маникюрчик?...
Может маневр отвлекающий...
@@elenalivshits3514 О мертвых ногтях либо хорошо, либо ничего..решил промолчать
2, 4, 7
148 = 128 + 16 + 4
Solved it instantly
147
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