Math Olympiad | A Nice Exponential Problem | VIJAY Maths

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[@shaankundu7249]

The easiest approach is to start with the largest power of 2 that is less than 148, which is (2^7 = 128). Then, find the remaining difference (148 - 128 = 20) and express it as a sum of powers of 2. In this case, (20 = 2^4 + 2^2), so (a = 2), (b = 4), and (c = 7).

[@piggypiggy9743]

คิดเหมือนกัน

[@manfredbottchergo]

For this I needed 1 minute.

[@amoswittenbergsmusings]

Why not simply convert 148 to binary? The solution can then simply been read off by inspection.

[@dorienjames5276]

Exactly this. 10 seconds mental arithmetic. How the hell does he spend 11:45 solving it? I'm spending the day downvoting convoluted solutions to clickbait problems in the hope that the algorithm will serve up more interesting challenges.

[@amoswittenbergsmusings]

@@dorienjames5276 I am not much of a crusader. I like problems even when the morally righteous way of solving them is missed by the poster. Paul Erdos called such proofs and solutions "book proofs" as he, a principled atheist, postulated that God has a book with all the best, most elegant, most enlightening proofs of all (provable) theorems. I am always looking for insights from that book. Please indulge me. I will give an example. The other day I came across this problem, on the Prime Newtons channel: How many non-integer solutions for x when x^7 - 1 + x^3(x-1) = 0? It isn't hard to do just with high school algebra and some factorisation but the beauty of the problem only becomes visible when one finds the factorisation Here is the comment I posted: QUOTE First: such a beautiful exposition! Prime Newton's love for his subject shines through his videos. Second: look at the factorisation of the expression! It is (x-1) . (x+1)^2 . (x*2+1) . (x^2-x+1), The last factor shouts TRIANGLE if you recognise a cyclotomic polynomial. If you now take one of the factors (x+1) and multiply it into the last factor and also multiply the remaining factors, ypu get a beautiful and revealing form of the polynomial! (x^4 - 1) . (x^3 + 1). The roots of the first factor are obviously +/- 1 and +/- i, the x (or y!) coordinates of the vertices of a square. Similarly the second factor for an equilateral trangle. The solution now stares you in the face: exactly three integer roots. I have not done the work but I am pretty sure that the polynomial can be written as a trigonimetric identity and with help of Euler and de Moivre as some beautiful function of a complex variable. Were the framers of this problem inspired by geometry? It certainly looks like it... All this is not to suggest a better solution but merely to have a closer look at the beauty of the problem. This is just one of the ways in which the universe sings. UNQUOTE Link: kzbin.info/www/bejne/oHTLmplsoJiSjtU Do I guess right if I think you watch 3blue1brown, Numberphile and Mathologer?

[@thomasw.eggers4303]

Exactly again! Perhaps the instructor isn't overly familiar with binary. If the solution presented was the general solution to more complicated problems, then I could understand 11:45 to solve it.

[@breadfor_bobross4055]

I just thought WHT was the highest exponent under 148, then I saw 2^4 and 2^2 must be the other solutions

[@MrEliseoD]

This was exactly the approach I took… if we’re dealing with a sum of powers of two, binary conversion is the ultimate way to go… 148 in binary is 10010100 and then read backwards for a, b and c starting at 2^0. a = 2, b = 4, c =7. Time taken: 5 seconds Why complicate it?

[@mikezilberbrand1663]

Shorter. 2^a is the GCF. Factorization of 148= 2X2X37. Then a=2. 148-4=144. The highest power of 2 in 144 is 4. Then b=4. 144-16=128. Then c=7.

[@RodgersHolmes-nf8vb]

I‘m at the same point👍

[@lechaiku]

There is much much shorter solution, you can do it almost immediately. But, of course, only for the real solution and only when a, b, c are integers. All what you need to solve it almost immediately is only a logic analysis of RHS. 148 = 2 * 2 * 37 148 = 2 ^2 * (36 + 1) 148 = 2 ^2 * (32 + 4 + 1) 148 = 2 ^2 * (2^5 + 2^2 + 2^0) 148 = 2^7 + 2^4 + 2^2 2^a + 2^b + 2^c = 2^2 + 2^4 + 2^7 a = 2; b = 4; c = 7 ---------------> for a < b < c

[@slacklineapuntacorvo]

One way is to convert in binary: 148d = 10010100 so the exponent are 2,4,7

[@sauronbadeye]

Wrong fella! No zero needed, you can't count in binary mode if you say that!

[@user-it6fh7hy6t]

​@@sauronbadeyeЯ ошибся. 10010100 - правильный ответ.

[@luiscostacarlos]

Soma de potências de 2, é óbvio que começando pela maior 2⁷ = 128, restando 20= 4 + 16, então: 2⁷ + 2⁴ + 2² ✓ ♦️Mas valeu pela resolução algébrica.

[@Hans_Magnusson]

Every computer engineer should be able to solve this in their sleep… It is simply GF(2⁸) or a byte … 2⁷ =128 2⁶ = 64 2⁵ = 32 2⁴ = 16 2³ = 8 2² = 4 2¹ = 2 2⁰ = 1 Now just start at the highest number, that is less than the number you are looking for ie 128 as you are looking at 148 So 148 -128 → 7 -16 → 4 -4 → 2 And Bob’s your uncle!

[@joseemanuelmonteiro]

Muito mais prático, simples e consequente. Para quê dar a volta ao mundo?

[@joseemanuelmonteiro]

Rápido e rasteiro. Parabéns.

[@ezzatabdo5027]

Excellent, nice and wonderful illustration, thanks sir.

[@andrewlayton9760]

A quicker way is to represent 148 as a binary number since all of the bases are 2 and look for the digits that are 1. 148 --> 10010100 in binary with 1s located at exponents 7, 4, and 2. We know this is the only solution because the sum of all preceding powers of 2 = (2^n)-1.

[@rathinaveluthiruvenkatam6203]

Right, Sir . partition as sum of powers of 2. 148 = 4+16+128

[@davidbrisbane7206]

148 = 10010100 base 2. There are only three '1's and so we only have three powers of 2 . 2^c = 10000000, so c = 7 2^b = 00010000, so b = 4 2^a = 00000100, so a = 2

[@Sharonjrainey]

How did you do that? This way is really unknown to me. I’m quite obsolete 😂. May I ask you for Please be kind and explain🙏🏻🤗

[@davidbrisbane7206]

@@Sharonjrainey You can look up how to convert a number (base 10), which is an ordinary number such as 13 (say), to (base 2), i.e, a binary number. Essentially, you divide 13 by 2 and look at the remainder (it is always 0 or 1) and keep dividing until there is a 0 with a remainder Example: 13 divided by 2 is 6 remainder *1* 6 divided by 2 is 3 remainder *0* 3 divided by 2 is 1 remainder *1* 1 divided by 2 is 0 remainder *1* So, reading *up* the remainders, we find that 13 (base 10) is .. 1101 (base 2), which is ... 1x2³ + 1x2² + 0x2¹ + 1x2⁰ = 13 (base 2). Try this algorithm on 148 (bases 10).

[@manojkantsamal4945]

a=2, b=4, c=7

[@vijaymaths5483]

Nice👍☺

[@deema1974]

2^a + 2^b + 2^c = 2*2*37. Поделим обе части на 2^2, тогда 2^(a-2) + 2^(b-2) + 2^(c-2) = 37, но нечетное число возможно только если одно из слагаемых равно 1 (2^0), соответственно, а=2. Тогда 1 + 2^(b-2) + 2^(c-2) = 37. Отсюда 2^(b-2) + 2^(c-2) = 36 = 2*2*9. Снова поделим обе части на 2^2, тогда 2^(b-4) + 2^(c-4) = 9. И снова нечетное число возможно только тогда, когда одно из слагаемых равно 2^0. Значит b=4, а 1 + 2^(c-4) = 9 => 2^(c-4) = 2^3 => c-4=3 => c=7.

[@erapesmobile4216]

О русскоязычный, решил тоже так

[@manojkantsamal4945]

^=raed as "to the power" According to the question a

[@robisonluiz5826]

Parabéns pela excelente solução

[@prabhushettysangame6601]

ತುಂಬಾ ಸುಂದರವಾಗಿ ಬಿಡಿಸಿ ತಿಳಿಸಲಾಗಿದೆ 🎉ಎಷ್ಟು ವರ್ಣೀಸಿದರೂ ಸಾಲದು ❤

[@kiranjoshiacc]

Max power of 2 tending to 148 is 2^7 (= 128). Now, 148-128 = 20. This has to be the addition of powers of 2. Simply, 2^4 = 16 and 2^2 = 4. Now, arrange them according to the condition = 2^2 + 2^4 + 2^7 = 148. So simple.

[@erapesmobile4216]

можно заметить что что 148 можно представить виде 4 × 37 и можно заметить что левое выражения делится на 4.значит сделал замену два в степени m=a-2 соответственно и аналогично со всеми остальными то заметим что 3 × 2 в степени m меньше чем 37. Но чтобы было целое число близкое и меньше 37 это 36. Получаем два степени m меньше 12 получаем удовлетворяющие значения m выходит это значение ноль, один, два, три. Посмотрим значение ноль. Тогда 2^0 +2^ n +2^ L будет = 37. Тогда 2^ n +2^ L= 36 можно не разбирать случаи больше нуля так как сумма двух четных не должно давать нам нечетные. Методом подбора очевидный ответ это числа 2 и 5.Плюсую два будет такой ответ: 2,4,7

[@davisnganga6266]

Wow. Fantastic solution.

[@vijaymaths5483]

Thank you

[@grzegorzkondracki4630]

2, 4, 7 - this is natural intuitive solution - 148 = 128+20 = 128+16+4 = 2^7 + 2^4 + 2^2. But we want to find a general solution. This is my proposition: 1. Suppouse a=b=c. I know, a

[@jorgepinonesjauch8023]

Está muy bien tu análisis aritmético, y también lo analice así mismo, pero me parece mejor la herramienta del álgebra para analizar este problema ya que involucra una serie de otros aspectos de análisis y despeje

[@grzegorzkondracki4630]

@@jorgepinonesjauch8023 Thx 🙂

[@l.muriel598]

Así es,por la "cuenta la vieja"

[@sergey_r]

it can be noted that the solution exists only at c = 7, because if c =6, then max(2^a+2^b+2^c)=(2^4+2^5+2^6)=(16+32+64)=112

[@urmilas5356]

Great problem thank for solution😊

[@vijaymaths5483]

Most welcome 😊

[@ezzatabdo5027]

Sure is excellent and nice, thanks sir.

[@vijaymaths5483]

Thanks a lot 🎈

[@dkajjuaxz]

в условии к заданию вроде бы не было сказано про то, что ответы должны быть целыми числами по этому когда мы уже нашли "а" и преобразовали уравнение в 2^b(1+2^(c-b))=4*4*9, то далее правую часть можно было бы расписать как 48 * 3, и по факту у нас первое чётное, а второе нечётное. тогда "b" равно log(2, 48), и при вычислении "c" получился бы log(2,96). Подставляем в изначальное уравнение 2^(2)+2^(log(2, 48))+2^(log(2,96)) = 148 => 4 + 48 + 96 = 148 => 148 = 148. Второе решение тоже верное А вообще здесь приблизительно бесконечно решений, так как если представлять степени "а", "b", "c" в виде логарифмов log(2, Xa), log(2, Xb), log(2, Xc), вместо Xa, Xb, Xc подойдут любые вещественные числа, которые в сумме дают 148, где 0

[@zorg6020]

Решение автора видео неверное, ибо он угадывает числа. Метод подбора - не решение.

[@Zane_Zaminsky]

Not the right way to do it, but if you’re a computer programmer and you’re used to working with 2 to various powers, you can easily do this in your head. 2^7 = 128, leaving 2^a + 2^b = 20. a = 2 and b = 4 follows easily. So 2, 4, 7. Of course, if you don’t get credit without a mathematical derivation, then you’re sunk. 😉

[@doanhnguyenquoc4707]

Phản logic!!!!!

[@helioxghost]

This. Took me less than a minute to figure out a,b,c just in my head.

[@harrydonnison5343]

Write it as a binary number and work out the value of the 1s in the number in base 10.

[@rabinarayansahoo166]

Excellent

[@vijaymaths5483]

Thank you

[@A.V.N.SRIRAM]

🙏 Sir, on the basis of which principle you equated the even and odd parts of both the expressions on the LHS and RHS equality symbols? Please explain.

[@ramanma9915]

Such an easy problem cannot be a math olympaid problem.

[@sirjuliusdeviscensus114]

muy bonito, aprendí buenas cosas!!!!

[@vijaymaths5483]

Thanks for watching!!

[@Rocio62154]

Excellent professor!!

[@vijaymaths5483]

Thank you

[@user-ux5uy3gs7j]

Оригинальный процесс решения!! Спасибо!!

[@jllaury75]

Por tanteo: a = 2; b = 4 y c = 7

[@prajnaparamitapatra9722]

Very nice question sir.

[@vijaymaths5483]

Thank you for feedback :) :)

[@manuelrivero124]

si pasas a base 2 el numero 148 en base decimal se obtiene el 2 , 4 y 7 como exponentes de 2.

[@quakers200]

I made a table of 2 to the power of one, two. Three etc. in three columns ten I picked a number from the third coumn, subtracted it from , 148. Then I looked at numbers in the first two rows that added up to that number. Took less than three minutes. I assume thanks t the answer was only positive integers.

[@HanhTran-tl8dm]

Very good solution.

[@vijaymaths5483]

Lot of thanks

[@jamalopez33]

Excelente, mil gracias

[@vijaymaths5483]

Thanks you !

[@Mordorian_Orque]

Just divide the expression by 4, substitute t=a-2, u=b-2, z=c-2 and deal with 37

[@manojkantsamal4945]

Respected Sir,, Pranam.. You tabled the solution through proper logical algebric method. 👍🏻👍🏻👍🏻

[@vijaymaths5483]

Thanks a ton :)

[@igorsidorin3585]

Very long and abstruse: Right part rewritten as 148=4+16+128 or 2∧2 +2∧4+2∧7 So 2∧a+2∧b+2∧c= 2∧2 +2∧4+2∧7/ From which it follows that a=2. b=4 and c=7

[@user-lc3er8bb8d]

There is so many(infinite) answers because a, b, c is not must be integer. (Ex) a=log_2_1 b=log_2_2 c=log_2_145)

[@joseagra7725]

Excellent teacher!

[@vijaymaths5483]

Thank you ⚘️

[@sauronbadeye]

Bad teacher!

[@ItsAnkush108]

Sir, I had done this calculation in my mind. Not exactly calculation but by putting power of 2 in mind.

[@vijaymaths5483]

Very good 👍👍 👍👍

[@veerakitchen1946]

👏👏👏

[@krishnannambiar5988]

Three powers of 2 is148.148has to get adding thtee powers of2=nearest power of 2 to148 is128.+16+4=powers of two to 2.4.&7.wecan take any of these threenumbers as value of a. b&c. Ans 2.4.7

[@rathinaveluthiruvenkatam6203]

Teach partition a powerful tool more so now , from number theory as you teach factorization

[@Sharonjrainey]

Nice & community comment interesting also🙏🏻🤔

[@vijaymaths5483]

Thanks for visiting ⚘️

[@williammarshal4043]

Put a number 2^a closest to 148 but not exceeding Subtract 148-2^a. 2^b closest to 148-2^a but not exceeding. Put 2^c closest to 148-2^a-2^b, hopefully it has 0 remainder. To put simply You got 128 + 20, then 128+ 16+4 You got 7, 4, 2

[@user-ph4iv9yd6t]

A = log2(148 - 2^b - 2^c) B is in range(log2(37);log2(74)) C is in range(log2(74);log2(148-2^b)). First you must find b, then C and a.

[@rurallifevy]

Thank you

[@joseangelocidreiradaconcei2505]

👏🏾👏🏼👏🏽👏👏🏿👏🏻

[@vijaymaths5483]

Thank you

[@juliovasquezdiaz2432]

Gracias. Me gustó ña solución. Bendiciones de nuestro Padre Celestial. Saludos desde Chiclayo Norte del Perú

[@DonaldKim33]

When I asked Chatgpt, I couldn't solve this math problem. I was so disappointed.

[@josejuanvallschafer3006]

Buenos dias , esta solución es slo una de las infinitas que existen, tenenmos un sistema compatible indeterminado, eso de que 2 elevado a a es igual a 4 en el caso de una multiplicación no es soluicion única, por favor sean correctos.

[@joseampueroacuna7662]

Por favor, traducirlo al español.y que los números sean más grandes para que se puedan visualizar. Gracias

[@toly1961]

a=2, b=4, c=7. Устная задачка на одну минуту.

[@superiorlyrics8326]

well explained ✅

[@vijaymaths5483]

Glad you liked it

[@user-nd7th3hy4l]

Intuition a=2 b=4 c=7 on peut faire des permutations sur ce triplet et obtenir six solutions.

[@user-it6fh7hy6t]

Не можем: а

[@betocoelho4863]

Sensacional

[@vijaymaths5483]

Thanks for watching !!

[@ToanPham-wr7xe]

😮

[@severinodearaujoferreira9996]

We can separete 148 = 4 + 144 Separete again and fatorate 148 = 4 + 16 + 128 148 = 2^2 + 2^4 + 2^7 So: a=2; b=4; c=7

[@user-rm4xj1ot3u]

이런 문제는 대입해서 조합 찾는게 가장 빠름

[@crazyindianvines1472]

Thanks 🙏

[@vijaymaths5483]

Welcome

[@razzp1155]

nice

[@vijaymaths5483]

Thanks a lot

[@Niyakun]

沒有a=2的過程啊，這跟我自己心算沒差別吧，沒證明會被扣分的 心算：2、4、8、16、32、64、128、20、16、4 7、4、2 填空10秒搞定，計算不行啊

[@KUNDANKUMAR-vv2qw]

A2, b4, c8

[@vijaymaths5483]

No, c=7

[@asifikh1736]

A=2,b=4,c=7

[@beyeu123]

It's too easy to solve. i need a more difficult exercise

[@oguzkaganatmaca9892]

Türkiye learning that in highschool

[@sauronbadeye]

No need to fill pages of formula! The solution is intuitive and quite easy for those who are familiar with binary numbers: 128 +16+4 It took me 1 minute! 😅😅😅😅

[@vijaymaths5483]

Very well 👏

[@Ditt_ty]

We know that largest power of 2 smaller then 148 is 2⁷ that is 128.... now 2⁷+2^a+2^b= 148 .... 2^a+2^b = 20.... now largest power of 2 smaller then 20 is 2⁴ ie 16.... now 12^a+2⁴=20 => 2^a = 4 => a=2 b=4 c=7.... a+b+c=13

[@DProgram-xb9pp]

128+16+4 a=7 b=4 C=2 Such a easy questions

[@mohammadhosseinian5986]

How do you know that a ,b,c are integer?

[@thomasw.eggers4303]

You are correct: it is not given that a,b,c are integer. But the form of the problem strongly suggests binary, so try that. In this case it worked.

[@HimuBora-me8nh]

,a=1. b=2c=4

[@banitasethy2323]

Long time 😮

[@bingwang7889]

148=4x37, 37=1+4+32. Done.

[@ezeldintayel6247]

Binary expansion: Am I a joke to you??

[@vijaymaths5483]

😂

[@abhirampanda5462]

It takes much time to solve

[@i_am_a_gugugu]

7,4,2(무지성 대입)

[@NgsM111]

If this was asked to solve it by showing the way too in exam lr olympiad💀

[@NIKHIL-mm1tc]

It took me less than a minute to solve with simply breaking 148 into 3 numbers which are 128, 16 and 4 (necessarily power of 2s)

[@vijaymaths5483]

I think it work's only in compitative exams , any way thanks for visiting ⚘️

[@ryanchiang9587]

128, 16, 4

[@kimobrien.]

c must be less than 7 because 2^7 is 128. 148 minus 128 is 20 twenty is 16 plus 4 so the answer is c=7 B=4 and a=2.

[@random_IIITK]

I simply did it in 5 seconds by hit n trial

[@vijaymaths5483]

Good, very good 👍

[@LearningAcademy-jp9gg]

Observation:- a = 2 , b = 4 and c = 7

[@user-it6fh7hy6t]

Блогер! Вы знакомы с двоичной системой исчисления? Решение занимает ровно минуту.А если решать устно,то 15 секунд.

[@bilalhussain2498]

a=2 , b=4 and c=7 are the correct values

[@vijaymaths5483]

Exactly 👏

[@ashutoshbajpay7639]

128,16,4 ऐसे ही आजाएगा यानी 2,4,7

[@stephenbryant7189]

The only words I understand are the English words. But 4 * 37 in English is not 4 "into" 37 . It is 4 times 37. The British Raj would be mortified.

[@vijaymaths5483]

Thank you sir for correction 😀

[@Frisbieinstein]

I got this in a few seconds.

[@vijaymaths5483]

Good, very good 👏

[@KipIngram]

7, 4, 2

[@vijaymaths5483]

Due to the condition a

[@chrisbarrington108]

Did it in 30 seconds by converting to binary, then saw your solution...

[@vijaymaths5483]

You are genius :) Thanks for watching !!

[@radmehrhakhamanesh6816]

guessing it would’ve been easier

[@vijaymaths5483]

😃

[@ajitnarainsingh6490]

Able to read only by micriscope !

[@vijaymaths5483]

Rotate your phone and watch video horizontally.

[@ramudukuruva3145]

148= 128+ 16+4 a=2,b=4, c=7

[@vijaymaths5483]

👍👍👍

[@1fessor]

a,b,c=7,4,2

[@vijaymaths5483]

Yes, u r right ✅️ thanks for watching

[@user-cl1km6mj7i]

2 4 7

[@vijaymaths5483]

Thanks for watching!!

[@user-bz1rj7uo5t]

Как же он это расписывает и вроде не было условия, что a, b, c - натуральные)

[@dhirenpatel3780]

1 4 8

[@benjaminchang1382]

why explained into complicate

[@ramudukuruva3145]

2 pow1+ 2 pow 4 + 2 pow +2 pow 7

[@vijaymaths5483]

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