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Math Olympiad | A Nice Exponential Problem | VIJAY Maths

VIJAY Maths
2024 ж. 3 Қаң.
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Пікірлер

  • 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).

    @shaankundu7249@shaankundu72493 ай бұрын

    • คิดเหมือนกัน

      @piggypiggy9743@piggypiggy97433 ай бұрын

    • For this I needed 1 minute.

      @manfredbottchergo@manfredbottchergo3 ай бұрын

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

    @myverypersonalstuff@myverypersonalstuff4 ай бұрын

    • 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.

      @dorienjames5276@dorienjames52764 ай бұрын

    • @@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: kzhead.info/sun/naehmcpuomWVi6s/bejne.html Do I guess right if I think you watch 3blue1brown, Numberphile and Mathologer?

      @myverypersonalstuff@myverypersonalstuff4 ай бұрын

    • 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.

      @thomasw.eggers4303@thomasw.eggers43033 ай бұрын

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

      @breadfor_bobross4055@breadfor_bobross40553 ай бұрын

    • 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?

      @MrEliseoD@MrEliseoD3 ай бұрын

  • Excellent, nice and wonderful illustration, thanks sir.

    @ezzatabdo5027@ezzatabdo50272 ай бұрын

  • 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.

    @luiscostacarlos@luiscostacarlos3 ай бұрын

    • 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!

      @Hans_Magnusson@Hans_Magnusson3 ай бұрын

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

      @joseemanuelmonteiro@joseemanuelmonteiro3 ай бұрын

    • Rápido e rasteiro. Parabéns.

      @joseemanuelmonteiro@joseemanuelmonteiro3 ай бұрын

  • 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

    @lechaiku@lechaikuАй бұрын

  • 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.

    @mikezilberbrand1663@mikezilberbrand16633 ай бұрын

    • I‘m at the same point👍

      @RodgersHolmes-nf8vb@RodgersHolmes-nf8vb3 ай бұрын

  • 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.

    @andrewlayton9760@andrewlayton97603 ай бұрын

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

      @rathinaveluthiruvenkatam6203@rathinaveluthiruvenkatam62033 ай бұрын

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

    @slacklineapuntacorvo@slacklineapuntacorvo3 ай бұрын

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

      @sauronbadeye@sauronbadeye3 ай бұрын

    • ​@@sauronbadeyeЯ ошибся. 10010100 - правильный ответ.

      @user-it6fh7hy6t@user-it6fh7hy6t3 ай бұрын

  • 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

    @davidbrisbane7206@davidbrisbane72063 ай бұрын

    • 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🙏🏻🤗

      @Sharonjrainey@Sharonjrainey3 ай бұрын

    • @@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).

      @davidbrisbane7206@davidbrisbane72063 ай бұрын

  • Parabéns pela excelente solução

    @robisonluiz5826@robisonluiz58263 ай бұрын

  • 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.

    @kiranjoshi2705@kiranjoshi27052 ай бұрын

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

    @manojkantsamal4945@manojkantsamal49454 ай бұрын

  • Wow. Fantastic solution.

    @davisnganga6266@davisnganga62663 ай бұрын

    • Thank you

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 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

    @erapesmobile4216@erapesmobile42163 ай бұрын

  • Excellent

    @rongdhonu677@rongdhonu677Ай бұрын

    • Thank you

      @vijaymaths5483@vijaymaths5483Ай бұрын

  • a=2, b=4, c=7

    @manojkantsamal4945@manojkantsamal49454 ай бұрын

    • Nice👍☺

      @vijaymaths5483@vijaymaths54834 ай бұрын

  • Sure is excellent and nice, thanks sir.

    @ezzatabdo5027@ezzatabdo50272 ай бұрын

    • Thanks a lot 🎈

      @vijaymaths5483@vijaymaths54832 ай бұрын

  • ತುಂಬಾ ಸುಂದರವಾಗಿ ಬಿಡಿಸಿ ತಿಳಿಸಲಾಗಿದೆ 🎉ಎಷ್ಟು ವರ್ಣೀಸಿದರೂ ಸಾಲದು ❤

    @prabhushettysangame6601@prabhushettysangame66012 ай бұрын

  • Great problem thank for solution😊

    @urmilas5356@urmilas53564 ай бұрын

    • Most welcome 😊

      @vijaymaths5483@vijaymaths54834 ай бұрын

  • 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. 😉

    @Zane_Zaminsky@Zane_Zaminsky4 ай бұрын

    • Phản logic!!!!!

      @doanhnguyenquoc4707@doanhnguyenquoc47074 ай бұрын

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

      @helioxghost@helioxghost3 ай бұрын

  • Excellent teacher!

    @joseagra7725@joseagra77253 ай бұрын

    • Thank you ⚘️

      @vijaymaths5483@vijaymaths54833 ай бұрын

    • Bad teacher!

      @sauronbadeye@sauronbadeye3 ай бұрын

  • Very nice question sir.

    @prajnaparamitapatra9722@prajnaparamitapatra97224 ай бұрын

    • Thank you for feedback :) :)

      @vijaymaths5483@vijaymaths54834 ай бұрын

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

    @ItsAnkush108@ItsAnkush1083 ай бұрын

    • Very good 👍👍 👍👍

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 🙏 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.

    @A.V.N.SRIRAM@A.V.N.SRIRAM3 ай бұрын

  • 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

    @dkajjuaxz@dkajjuaxz3 ай бұрын

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

      @zorg6020@zorg60202 ай бұрын

  • 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

    @igorsidorin3585@igorsidorin3585Ай бұрын

  • Very good solution.

    @HanhTran-tl8dm@HanhTran-tl8dm3 ай бұрын

    • Lot of thanks

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 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.

    @deema1974@deema19743 ай бұрын

    • О русскоязычный, решил тоже так

      @erapesmobile4216@erapesmobile42163 ай бұрын

  • Excellent professor!!

    @Rocio62154@Rocio621542 ай бұрын

    • Thank you

      @vijaymaths5483@vijaymaths54832 ай бұрын

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

    @manuelrivero124@manuelrivero1243 ай бұрын

  • 👏👏👏

    @veerakitchen1946@veerakitchen194615 күн бұрын

  • muy bonito, aprendí buenas cosas!!!!

    @sirjuliusdeviscensus114@sirjuliusdeviscensus1144 ай бұрын

    • Thanks for watching!!

      @vijaymaths5483@vijaymaths54834 ай бұрын

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

    @manojkantsamal4945@manojkantsamal49454 ай бұрын

    • Thanks a ton :)

      @vijaymaths5483@vijaymaths54834 ай бұрын

  • Оригинальный процесс решения!! Спасибо!!

    @user-ux5uy3gs7j@user-ux5uy3gs7j3 ай бұрын

  • Nice & community comment interesting also🙏🏻🤔

    @Sharonjrainey@Sharonjrainey3 ай бұрын

    • Thanks for visiting ⚘️

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 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.

    @quakers200@quakers2004 ай бұрын

  • 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

    @sergey_r@sergey_r3 ай бұрын

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

    @harrydonnison5343@harrydonnison53433 ай бұрын

  • Such an easy problem cannot be a math olympaid problem.

    @ramanma9915@ramanma99153 ай бұрын

  • 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

    @grzegorzkondracki4630@grzegorzkondracki46303 ай бұрын

    • 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

      @jorgepinonesjauch8023@jorgepinonesjauch80233 ай бұрын

    • @@jorgepinonesjauch8023 Thx 🙂

      @grzegorzkondracki4630@grzegorzkondracki46303 ай бұрын

    • Así es,por la "cuenta la vieja"

      @l.muriel598@l.muriel5983 ай бұрын

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

    @rathinaveluthiruvenkatam6203@rathinaveluthiruvenkatam62033 ай бұрын

  • 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

    @williammarshal4043@williammarshal40432 ай бұрын

  • 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)

    @user-lc3er8bb8d@user-lc3er8bb8d3 ай бұрын

  • 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

    @krishnannambiar5988@krishnannambiar59882 ай бұрын

  • Excelente, mil gracias

    @jamalopez33@jamalopez332 ай бұрын

    • Thanks you !

      @vijaymaths5483@vijaymaths54832 ай бұрын

  • 이런 문제는 대입해서 조합 찾는게 가장 빠름

    @user-rm4xj1ot3u@user-rm4xj1ot3u3 ай бұрын

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

    @Mordorian_Orque@Mordorian_Orque3 ай бұрын

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

    @juliovasquezdiaz2432@juliovasquezdiaz24323 ай бұрын

  • 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.

    @user-ph4iv9yd6t@user-ph4iv9yd6t3 ай бұрын

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

    @DonaldKim33@DonaldKim332 ай бұрын

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

    @joseampueroacuna7662@joseampueroacuna76623 ай бұрын

  • Por tanteo: a = 2; b = 4 y c = 7

    @jllaury75@jllaury753 ай бұрын

  • 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

    @Niyakun@Niyakun3 ай бұрын

  • 👏🏾👏🏼👏🏽👏👏🏿👏🏻

    @joseangelocidreiradaconcei2505@joseangelocidreiradaconcei25054 ай бұрын

    • Thank you

      @vijaymaths5483@vijaymaths54834 ай бұрын

  • Sensacional

    @betocoelho4863@betocoelho48633 ай бұрын

    • Thanks for watching !!

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 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

    @severinodearaujoferreira9996@severinodearaujoferreira99963 ай бұрын

  • 😮

    @ToanPham-wr7xe@ToanPham-wr7xe3 ай бұрын

  • Thanks 🙏

    @crazyindianvines1472@crazyindianvines14724 ай бұрын

    • Welcome

      @vijaymaths5483@vijaymaths54834 ай бұрын

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

    @user-it6fh7hy6t@user-it6fh7hy6t3 ай бұрын

  • 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.

    @josejuanvallschafer3006@josejuanvallschafer30063 ай бұрын

  • nice

    @razzp1155@razzp11553 ай бұрын

    • Thanks a lot

      @vijaymaths5483@vijaymaths54833 ай бұрын

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

    @toly1961@toly19613 ай бұрын

  • Thank you

    @amazingmathpv@amazingmathpvАй бұрын

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

    @NgsM111@NgsM1112 ай бұрын

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

    @beyeu123@beyeu1233 ай бұрын

  • 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

    @Ditt_ty@Ditt_ty3 ай бұрын

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

    @mohammadhosseinian5986@mohammadhosseinian59863 ай бұрын

    • 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.

      @thomasw.eggers4303@thomasw.eggers43033 ай бұрын

  • 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.

    @kimobrien.@kimobrien.3 ай бұрын

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

    @user-nd7th3hy4l@user-nd7th3hy4l3 ай бұрын

    • Не можем: а

      @user-it6fh7hy6t@user-it6fh7hy6t3 ай бұрын

  • well explained ✅

    @superiorlyrics8326@superiorlyrics83264 ай бұрын

    • Glad you liked it

      @vijaymaths5483@vijaymaths54834 ай бұрын

  • 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! 😅😅😅😅

    @sauronbadeye@sauronbadeye3 ай бұрын

    • Very well 👏

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 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)

    @NIKHIL-mm1tc@NIKHIL-mm1tc3 ай бұрын

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

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 128, 16, 4

    @ryanchiang9587@ryanchiang95873 ай бұрын

  • 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.

    @stephenbryant7189@stephenbryant71893 ай бұрын

    • Thank you sir for correction 😀

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • It takes much time to solve

    @abhirampanda5462@abhirampanda54623 ай бұрын

  • Long time 😮

    @banitasethy2323@banitasethy23233 ай бұрын

  • 7,4,2(무지성 대입)

    @i_am_a_gugugu@i_am_a_gugugu3 ай бұрын

  • If the commentator could speak English (phonetics teacher strongly recommended)it would be more pleasant to watch.

    @arming.9840@arming.98403 ай бұрын

  • Able to read only by micriscope !

    @ajitnarainsingh6490@ajitnarainsingh64903 ай бұрын

    • Rotate your phone and watch video horizontally.

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • I simply did it in 5 seconds by hit n trial

    @random_IIITK@random_IIITK3 ай бұрын

    • Good, very good 👍

      @vijaymaths5483@vijaymaths54833 ай бұрын

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

    @LearningAcademy-jp9gg@LearningAcademy-jp9gg2 ай бұрын

  • Türkiye learning that in highschool

    @oguzkaganatmaca9892@oguzkaganatmaca98923 ай бұрын

  • 148=4x37, 37=1+4+32. Done.

    @bingwang7889@bingwang78893 ай бұрын

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

    @user-bz1rj7uo5t@user-bz1rj7uo5t3 ай бұрын

  • guessing it would’ve been easier

    @radmehrhakhamanesh6816@radmehrhakhamanesh68163 ай бұрын

    • 😃

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • ,a=1. b=2c=4

    @HimuBora-me8nh@HimuBora-me8nh2 ай бұрын

  • 128,16,4 ऐसे ही आजाएगा यानी 2,4,7

    @ashutoshbajpay7639@ashutoshbajpay76392 ай бұрын

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

    @chrisbarrington108@chrisbarrington1083 ай бұрын

    • You are genius :) Thanks for watching !!

      @vijaymaths5483@vijaymaths54833 ай бұрын

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

    @DProgram-xb9pp@DProgram-xb9pp3 ай бұрын

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

    @bilalhussain2498@bilalhussain24983 ай бұрын

    • Exactly 👏

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 148= 128 + 16 + 4 = 2^7 + 2^4 + 2^2 so what? a,b, and c = 7,4,2 in the order you want

    @MrBmarx@MrBmarx3 ай бұрын

  • A2, b4, c8

    @KUNDANKUMAR-vv2qw@KUNDANKUMAR-vv2qw3 ай бұрын

    • No, c=7

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • This is so trivial - it has nothing to do with math Olympiads.

    @rainerausdemspring3584@rainerausdemspring35843 ай бұрын

  • why explained into complicate

    @benjaminchang1382@benjaminchang13823 ай бұрын

  • 7, 4 y 2 (sin ver el video)

    @emmanuel_zunguero@emmanuel_zunguero2 ай бұрын

    • Good ,very good 👍

      @vijaymaths5483@vijaymaths54832 ай бұрын

  • It can be answered without pen

    @rohitnayek7733@rohitnayek77333 ай бұрын

    • Good, very good 👍

      @vijaymaths5483@vijaymaths54833 ай бұрын

  • 1 4 8

    @dhirenpatel3780@dhirenpatel37803 ай бұрын

  • Too complicated. Just choose the exponent for 2 to get the result under 148. It's 7. So 2^7=124. 148-124=20=2^4+2^2. That"s the solution

    @alexanderkochetov2552@alexanderkochetov25523 ай бұрын

    • Slight typo, 2^7=128 not 124

      @breadfor_bobross4055@breadfor_bobross40553 ай бұрын

  • A=2,b=4,c=7

    @asifikh1736@asifikh17363 ай бұрын

  • NEW WORLD

    @sonatya5978@sonatya59783 ай бұрын

    • ?

      @vijaymaths5483@vijaymaths54833 ай бұрын

    • @vijaymaths5483 your video is very interesting experience

      @sonatya5978@sonatya59783 ай бұрын
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