।। maths olympiad question ।। An algebraic exponential problem
HTML-код
- Опубликовано: 11 июн 2024
- ।। A Beautiful exponential equation ।। find the value of a,b and c
#mathsolympiad
#algebra
#mathslearning
#mathtricks
#mathematics
#olympiadquestion
Math Olympiod Question
Math Olympiod
Maths Olympiod Question
Math problem
Algebra
Olympiod Mathematics
Olympic Math
Math Olympiod l A Nice Algebra problem
A Nice Math olympiod Algebra Question
A nice Exponential problem
A Nice Exponents math problem
Olympiod Algebra
Solving A Quadratic Equations
International Math Olympiod Problem
Olympiod Mathematics
International Olympiod Maths
Olympiod
International Math Olympiod
Math Olympiod Question
Math Olympiod Preparation
American Math Olympiod Question
Math Olympiod Questions
Math Olympiod Problem
Math Olympiod Algebra Problem
France Math Olympiod Preparation
Can you solve this? Math Olympiod
Olympiod Maths Problem
Beautiful Algebra problem
Nice Algebra Math simplification,find value of a and b
Russian Math Olympiod Question
Nice Math olympiod Problem
A Nice Chinese Olympiod Algebra Problem
Just convert 148 to binary system 148 = 94 hexadecimal = 10010100 binary. Take the positions of ones, counting from right to left zero-based. They are 2, 4 and 7. So 148= 2² + 2⁴ + 2⁷.
READ: The problem says that it be solved using ALGEBRA!
@@motordvd148=128+16+4, it's algebra? 😂
Chả logic chả đại số????
Really the answer is that a, b and c belong to the exclusive set (2, 4, 7) because all permutations work I.e. a=7, b=2, c=4 still = 148, etc.
It is straight forward as base is 2 for solution. 2⁸=256 So 148= 2⁷ +20 Where 2⁷=128. Next is easy 20=16+4=2⁴ +2² Thus 2⁷+2⁴+2²=148
(A,B,C)::(7,4,2) every permutation is fine A,B,C::7,4,2 : 7,2,4: 4,2,7: 4,7,2: 2,7,4 or 2,4,7
U r also correct bur your solution method is not scientific
Easier (and more correct) way to do it with mental arithmetic:
apparently, we are talking about a binary number.
So 148 can be written as 128+16+4.
So it is easy to see that we are talking about 2^7 + 2^4 + 2^2.
So the solutions(!) are:
a=7,b=4,c=2
a=7,b=2,c=4
a=4,b=7,c=2
a=4,b=2,c=7
a=2,b=7,c=4
a=2,b=4,c=7
same pinch friend! ...🎉this is what I did few seconds... !
अपने प्रश्न को अच्छी तरह से हल किया धन्यवाद
Thank you sir the method of teaching is most encouraging those for newly learners
Like done 👍 Jai mata di 🙏 super video Bhai very nice jankari 👍👌👌❤❤
Like done 👍
jai shree krishana 🙏
radhe radhe bhaiya ji
awesome 🥰👌👌👍👍🙏🙏
Very nice sharing friend thanks 👍
Your final answer is incomplete, Sir. Since any permutation of {2, 4, 7} satisfies the equation. Thus (a, b, c) = (7, 2, 4) is also a solution
Thank u manish sir.
Like 👍 Jai mata di 🙏 super video bhaiya very nice jankari 👍👌👍👍❤❤
On the R hs you should also show( 148x2a,)/2a
Bahut badhiya prastuti 👌🥳
As per distribution law of addition you can give c=2 ,b=7,and c=4. In this case also you will get answer 148.hence attributing values can not be correct.value of a+b+c only can be correct for this question.
36.2
👌👌👌👌👌
Manakanakku a,b,c, can be 2,4,7in any combination
😂when an equation is multiped divided subtracted or added on one side then same on the both sides 😂
2,4,7
Very good. Thank you Sir
Look at powers of 2..128+16+4.No manipulation. Once again ,observation wins!!The symmetry allows for values to to be permuted.
Excellent working although there are shortcuts.
I am also from same branch,sir has made no shortcut.
Very nice shering bhai 👌🏽 👍🏽 Thank-you for sharing 👌🏽 🙏 😀 13:58
Pol
🎉🎉🎉🎉🎉
🙏🙏🙏🙏🙏🙏🙏😊
Решение неполное - для поиска всех корней (a,b,c): нет проверки на a=1, b=(1,2,3):
Don't check by a=1, b=(1,2,3).
a= 2 b=7 c=4
Because this is SMALL NUMBER, just do TRIAL ERROR ...
148 = 128 + 16 + 4
= 2^7 + 2^4 + 2^2 ...
So a, b and c are 7, 4 and 2 ...
As I said, it because this is SMALL NUMBER....
No need more minutes to solve...
Very long method. Itna time nahi hota competitive exam mai.
Very nice 😊good morning bhaiya ji🙏
Very nice sharing. 13. 58
a=2
b=4
c=7
Big lk usefull video
Ha, when you are a low level or embedded programmer you spot it right away. It's 128+16+4 makes 7, 4, 2. Programmers tend to remember all powers of 2 up to 2 ^16. Of course whole numbers. I think your method is very good when dealing with fractions.
👍🤗
Answer may be ,a :b:c. 2:4:7,7:4:2,7:2:4,More of three by the process of statistics or permutation.
Radhey radhey bhaiya 🙏💐
148÷2÷2=37 ... therefore a=2 and 2^2=4
148 -4= 144
144÷2÷2÷2÷2=9 b = 4 and 2^4= 16
148 -4 -16 =128
128 = 2^7 therefore c= 7.
A simpler method following numerical logic.
a=2, b=4 and c=7
Nice sharing
The nearest number to 148 is 128=2^7 now substruct 148_128=20 , 20=16+4 =2^4+2^2
a=7 , b=4 and c=2
a=7 , b=4 , c= 2
👍👍👍👏👏🙏🙏🙏 sorry bhaiya ji Mere Aankh Mein problem hai 13:58
Bravissimo prof.
a=2, b=4, c=7
Soluția e Ok dar și Permutare mai Adaugă inca cinci solutii valabile
2^a + 2^b + 2^c = 148
2^a + 2^b + 2^c = 2²*37
2^a + 2^b + 2^c = 2²(2^0+2²+2^5)
2^a + 2^b + 2^c = 2² + 2⁴ + 2^7
a = 2,4,7
b = 4,7,2
c = 7,2,4
Full combinations :
2,4,7 || 2,7,4 || 4,2,7 || 4,7,2 || 7,2,4 || 7,4,2
à 20, b 24, ç 30
a = 2, b = 4, c = 7.
a=2 ; b= 4 ; c= 7
When u multiply by 2a on L 8:12 h.S you are not multyply by 2a on the R h s pl clarify
A=2,b=4,c=7
Vậy sao không thử từ ban đầu. Một phương pháp không hay ho và không chứng minh được tính duy nhất nghiệm. Nếu chỉ tìm một bộ ba giá trị thì chỉ cần vài bước thử sẽ tìm ra câu trả lời
You are a good teacher
Excellent brain storming
Like done watching bhai
Keep dividing 148 by 2 until you arrive at an odd number. Deduce 148 = 2^2 x 37
Powers of 2 smaller than 37 are 1,2,4,8,16,32.
37 = 32 +5 = 32 + 4 +1 = 2^5 + 2^2 + 2^0
Therefore 148 = 2^2 (2^5 + 2^2 + 2^0) = 2^7 + 2^4 +2^2
{a,b,c} = {7,4,2} ie a,b,c could be in any order
Very nice sharing bahut badhiya ❤❤❤❤❤
7, 4, 2
Accha hai but too slow
FIRST FIND THAT POWER OF 2 THAT'LL BE WITHIN 148.THAT IS 7 AND ASSIGN C THAT VALUE AND GET 128.THE REMAINING 2O MAY BE ASSUMED TO BE SUM TOTAL OF 2 TO THE POWER 2 AND 4 .ASSIGN 2 FOR A AND 4 FOR B.
A IS 2 B IS 4 C IS 7.
Like done watching
It will be solved by just estimation. The maximum multiple below 148 is 2×2×2×2×2×2×2 = 128. So, if we take 128 + 16 + 4, then it will come to 148. It means 147 is 2×2×2×2×2×2×2 + 2×2×2×2 + 2×2, i.e. a, b and c will be 7, 4 & 2. Ok.
Easier by mental work!
Not enough;it is 3 answers
Radhey radhey bhaiya 🙏🙋♀️🚩
Radhe radhe sister 🙏🏻
Veri nice sharing🎉🎉🎉🎉❤❤❤ 13:58
2 üzeri 7 2 üzeri 4 2 üzeri 2 = 148😉
13:58 nice sharing
Very nice sharing ❤ 13:58
Where can find ,i searched everywhere but I couldn't found
Quite boarding and wrong😂
Very nice sharing 👍
Thanks for visiting
There are 6 solutions. You only get 1/6 point for it.
Re watch 🤗
Nyc sharing ram bhai🎉nidhi
Very nice shering bhai 13:58
128+16+4=148
Good example of how to make solutions overly complicated. Plus, your solution is incorrect. Found values are intermittent...
Unless it is stated that a, b, & c are integers, there is an infinite number of solutions!
How?
Very slow
People who watch this are clearly maths students who have a greater knowledge of maths than the average person
You should not have to go through every little detail too show the solution
a=2,b=4&c=7
Right
a, b and c ∈ {7, 4, 2} with a # b # c
The mathmatician ,s scenes is very poor. He is doing it as a goods train.
2 .2.37 wrong
Sir didn't explain how 2.2.37 come😢
Prime factor of 148 = 2×2×37
Math Olympiad Question: 2^a + 2^b + 2^c = 148; a, b, c = ?
2^a + 2^b + 2^c = 148 = 4(37) = 4(32 + 4 + 1) = 2²(2⁵ + 2² + 1) = 2⁷ + 2⁴ + 2²
= 2⁷ + 2² + 2⁴ = 2⁴ + 2² + 2⁷ = 2⁴ + 2⁷ + 2² = 2² + 2⁷ + 2⁴ = 2² + 2⁴ + 2⁷
a = 7, b = 4, c = 2; a = 7, b = 2, c = 4;
a = 4, b = 2, c = 7; a = 4, b = 7, c = 2;
a = 2, b = 7, c = 4 or a = 2, b = 4, c = 7
Answer check:
2^a + 2^b + 2^c = 148; Confirmed as shown
Final answer:
a = 7, b = 4, c = 2; a = 7, b = 2, c = 4;
a = 4, b = 2, c = 7; a = 4, b = 7, c = 2;
a = 2, b = 7, c = 4 or a = 2, b = 4, c = 7
Note:
If the Question is specified; a < b < c
The only answer is, a = 2, b = 4, c = 7
(a=2;b=4;c=7);(a=2;b=7;c=4);(a=4;b=2;c=7);(a=4;b=7;c=2);(c=7;a=2;c=4);(c=7;a=4;b=2)❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤😂
Totally wrong😂😂
Wrong
Why are complicating a very simple math problem??? You cant spend 5 minutes explaining to this😂😂😂😂
Very nice solution guru ji 🙏
आपका पढाया हुआ समझ से परे है तरीका सही नही है
❤❤❤❤
Why spend 13+ minutes solving something that one can see just by looking at it: a=2, b=4, c=7: 2^2+2^4+2^7 = 4+16+128 = 148.
Because the process can be used for more complex situations, that's why!
READ: The problem says that it be solved using ALGEBRA!
Who has not accepted the Lord Jesus yet, pray like this
Lord God, I come to You, as a sinner that I am, in the name
of Jesus asking you for forgiveness for my sins. Forgive me, Lord,
my sins.
Help me to improve, I can't do it alone.
Free me from all evil, because I want to walk in your presence
seeking to do good.
Make all things new in my life that you don't like
I ask you to come into my life so I can have eternal life
that comes after this life that the Lord has given me. and have the
salvation not to suffer eternally.
Amen to the God who made me and loves me.
a=2 , b=4 , c=7
a=2,b=4,c=7