।। maths olympiad question ।। An algebraic exponential problem
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- čas přidán 11. 06. 2024
- ।। A Beautiful exponential equation ।। find the value of a,b and c
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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.
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... !
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
Like done 👍 Jai mata di 🙏 super video Bhai very nice jankari 👍👌👌❤❤
अपने प्रश्न को अच्छी तरह से हल किया धन्यवाद
Like done 👍
jai shree krishana 🙏
radhe radhe bhaiya ji
awesome 🥰👌👌👍👍🙏🙏
Thank you sir the method of teaching is most encouraging those for newly learners
Very nice sharing friend thanks 👍
a= 2 b=7 c=4
Bahut badhiya prastuti 👌🥳
Like 👍 Jai mata di 🙏 super video bhaiya very nice jankari 👍👌👍👍❤❤
Valde Valde facilis! Hic est a= 7,b= 4 et c=2.Responsi.
a=2
b=4
c=7
Thank u manish sir.
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
Manakanakku a,b,c, can be 2,4,7in any combination
Big lk usefull video
Very nice sharing. 13. 58
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.
😂when an equation is multiped divided subtracted or added on one side then same on the both sides 😂
Excellent working although there are shortcuts.
I am also from same branch,sir has made no shortcut.
Very nice 😊good morning bhaiya ji🙏
Very nice shering bhai 👌🏽 👍🏽 Thank-you for sharing 👌🏽 🙏 😀 13:58
Pol
Very good. Thank you Sir
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
2,4,7
Very nice sharing bahut badhiya ❤❤❤❤❤
Radhey radhey bhaiya 🙏💐
On the R hs you should also show( 148x2a,)/2a
Look at powers of 2..128+16+4.No manipulation. Once again ,observation wins!!The symmetry allows for values to to be permuted.
Nice sharing
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.
Veri nice sharing🎉🎉🎉🎉❤❤❤ 13:58
Very nice shering bhai 13:58
👌👌👌👌👌
Bravissimo prof.
Like done watching bhai
Where can find ,i searched everywhere but I couldn't found
Like done watching
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...
Radhey radhey bhaiya 🙏🙋♀️🚩
Radhe radhe sister 🙏🏻
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).
à 20, b 24, ç 30
Nyc sharing ram bhai🎉nidhi
A=2,b=4,c=7
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.
Soluția e Ok dar și Permutare mai Adaugă inca cinci solutii valabile
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
You are a good teacher
a=2, b=4 and c=7
Very nice sharing ❤ 13:58
👍👍👍👏👏🙏🙏🙏 sorry bhaiya ji Mere Aankh Mein problem hai 13:58
2 üzeri 7 2 üzeri 4 2 üzeri 2 = 148😉
🙏🙏🙏🙏🙏🙏🙏😊
a=2 , b=4 , c=7
a = 2, b = 4, c = 7.
a=7 , b=4 , c= 2
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.
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
Excellent brain storming
a=2 ; b= 4 ; c= 7
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
When u multiply by 2a on L 8:12 h.S you are not multyply by 2a on the R h s pl clarify
Easier by mental work!
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.
Very nice sharing 👍
Thanks for visiting
13:58 nice sharing
Accha hai but too slow
👍🤗
Quite boarding and wrong😂
7, 4, 2
🎉🎉🎉🎉🎉
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
Unless it is stated that a, b, & c are integers, there is an infinite number of solutions!
How?
36.2
Re watch 🤗
128+16+4=148
(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)❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤😂
a=2,b=4&c=7
Right
Very long method. Itna time nahi hota competitive exam mai.
Very slow
Good example of how to make solutions overly complicated. Plus, your solution is incorrect. Found values are intermittent...
There are 6 solutions. You only get 1/6 point for it.
Not enough;it is 3 answers
Sir didn't explain how 2.2.37 come😢
Prime factor of 148 = 2×2×37
The mathmatician ,s scenes is very poor. He is doing it as a goods train.
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
2 .2.37 wrong
a, b and c ∈ {7, 4, 2} with a # b # c
Very nice solution guru ji 🙏
Why are complicating a very simple math problem??? You cant spend 5 minutes explaining to this😂😂😂😂
The commentary is a disaster. I don’t understand the language the commentator is speaking.
Wrong
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!
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Free me from all evil, because I want to walk in your presence
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Totally wrong😂😂
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