A Very Nice Math Olympiad Problem | Solve for a, b and c | Algebra

[@kamalmakhtoomnejad]

To solve this equation, it is enough to move the number d148 from base 10 to base 2, which becomes (10010100). Now, if we write this number again in base 10, it becomes:(0×2^0+0×2^1+1×2^2 +0×2^3+1×2^4+0×2^5+0×2^6+1×2^7) by simplifying we have: (2^2+2^4+2^7 ) that the values ​​of a,b,c =(2,4,7) are easily obtained.

[@dan-florinchereches4892]

This will give all integer solutions. We have infinitely many real solutions and I am not sure if question requires them. Consider a simple case Z=2^x+2^y intersected with plane Z=k . The surface is increasing with both X and y and for k >0 there will be a continuous intersection curve hence infinitely many real solutions

[@mchang7901]

This is nice, very clever

[@SpencersAcademy]

Well explained 👏

[@carlosjimenez2848]

Obvio

[@davidseed2939]

yes good reliable method. you can also take 148, then find the highest power of 2 that is smaller than the given number and subtract. ie 148-128=20 and repeat 20-16=4 so we have 148=128+16+4=2^7+2^4+2^2 so x,y,z= 7,4,2

[@mrinaldas9614]

I feel this type of sums could be done using careful observation. Basically we have to distribute 148 in 3 parts, each of which is a power of 2 (2,4,8,16,32 , 64,128 etc) , so that the sum would be 148. Simple iteration would giveth result: 128+16+4. (a,b c are interchangeable.)

[@nabeelahmed6358]

its just binary

[@simonlevett4776]

@@nabeelahmed6358 Is it ?

[@markgideon3997]

Let me show an easier guess-and-check solution. For starters, you need to realise that a, b or c cannot physically be more than 7. If you got 2^8, it’s already 256. So let’s assume a is equal to 7. Then 2^b + 2^c = 148-128=20. Now what is the maximum b or c? It is 4 because 2^5 is already 32. Let’s try b=4. Then 2^c = 20-2^4 = 4. If 2^c = 4, then c=2. So we got the following numbers: 2,4,7. I confess, it’s not really a mathematical “elegant” solution, but it still works

[@NishantChaudhury]

Elegant You care about eelegant This method is amazing someone average like me Thank you

[@davidseed2939]

it is elegant and reliable. And if you know your powers of 2, it’s the quickest because subtraction is the only process.

[@marekzalinski390]

No need to do any trick. The binary representation of 148 is 10010100. There are three 1 in the number, at positions 2^7, 2^4 and 2^2, so the only combination is 7, 4, 2 in any order. If the problem contained more than three powers of 2, more solutions would be possible. E.g. with 2^d added, the solutions will be 6,6,4,2; 7,3,3,2; and 7,4,1,1. As can be seen, the solutions are based on splitting one of the powers of two into two smaller powers of two. That 12 minutes could be directed to explaining binary numbers - a very useful knowledge in this days - and the last 30 seconds, to solving that trivial question.

[@Dr_piFrog]

(7,4,2) 2^7 = 128 2^4 = 16 2^2 = 4 sum = 148

[@starthakog]

Exactly. No need to fully solve like needs.

[@andryvokubadra2644]

2^a + 2^b + 2^c = 148 2^a + 2^b + 2^c = 2²(37) 2^a + 2^b + 2^c = 2²(36 + 1) 2^a + 2^b + 2^c = 2²(32 + 4 + 1) 2^a + 2^b + 2^c = 2²(2^5 + 2² + 2^0) 2^a + 2^b + 2^c = 2^7 + 2⁴ + 2² a = 2,4,7 b = 2,4,7 c = 2,4,7 {a,b,c} = 3*2*1 = 6 combinations of 2,4,7 😊😊😊

[@SpencersAcademy]

Excellent delivery! 👏

[@andryvokubadra2644]

@@SpencersAcademy a,b,c = 2,4,7 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 {a,b,c} = 6 combinations 😊😊😊

[@boguslawszostak1784]

assuming a

[@SpencersAcademy]

Excellent! 👍

[@MARTINWERDER]

6 solutions: (a, b, c) = (7, 4, 2), (7, 2, 4), (4, 2, 7), (4, 7, 2), (2, 7, 4), (2 , 4, 7)

[@xvoidx_yt1723]

Yes

[@9허공]

WLOG, we assume a ≥ b ≥ c => 2^a + 2^b + 2^c = 2^c * ( 2^(a-c) + 2^(b-c) + 1 ) = 2^2*37 => c = 2 => 2^(a-2) + 2^(b-2) + 1 = 37 => 2^(b-2)*( 2^(a-b) + 1 ) = 2^2*9 => b-2=2 => b = 4 => 2^(a-4) + 1 = 9 => 2^(a-4) = 2^3 => a-4 = 3 => a = 3 Answer set (a,b,c) = permutations of (2,4,7) = { (2,4,7),(2,7,4),(4,2,7),(4,7,2),(7,2,4),(7,4,2) }

[@SpencersAcademy]

Amazing 👏

[@nabeelahmed6358]

you know, my first thought was to see how to write the number in binary, because the representation is in powers of 2

[@SpencersAcademy]

Well said 👏

[@simonlevett4776]

So if the powers were all 4, you would use base 4 to find the answer ?

[@nabeelahmed6358]

​@@simonlevett4776 yeah ig

[@ilafya]

Il suffit écrire le nombr148 en base 2 Ie en somme se puissance de 2 ou148=10010100

[@davidbrisbane7206]

What are the positive integer values of a, b and c such that 333 = 3^a + 3^b + 3^c? Notice 333 (base 10) = 110100 (base 3) So a = 5, b = 4, c = 2 (corrected) and all combinations of these values.

[@SpencersAcademy]

You did a great job. Could you please check your result?

[@davidbrisbane7206]

@@SpencersAcademy Thanks. c = 2

[@RealQinnMalloryu4]

{74+74+74}=222 1^1^2 1^2 (abc ➖ 2abc+1).

[@xvoidx_yt1723]

Why did we raise it to 2^a. Can you please explain?

[@pascallemesle]

there are 6 solutions (2,4,7), (2,7,4), (4,2,7), (4,7,2), (7,2,4) and (7,4,2)

[@SpencersAcademy]

Excellent! 👍

[@simonlevett4776]

I thought there was only one solution.

[@AlphaAnirban]

​@@simonlevett4776 Technically there is only one solution to this. But you dont know the value of a,b,c individually. You just know that ANY ONE of them has to be 2, while the other two have to be 4 and 7. But we dont know WHICH of the three will have these values. For all we know, anyone could have the value of 2 or 4 or 7. So we list down all possibilities rather than just giving a single answer.

[@kavitapatidar2811]

How do you say that the even and odd numbers are same in both sides

[@kavitapatidar2811]

Right question

[@key_board_x]

2^(a) + 2^(b) + 2^(c) = 148 2^(a + c - c) + 2^(b + c - c) + 2^(c) = 148 2^(c + a - c) + 2^(c + b - c) + 2^(c) = 148 [2^(c) * 2^(a - c)] + [2^(c) * 2^(b - c)] + 2^(c) = 148 2^(c) * [2^(a - c) + 2^(b - c) + 1] = 148 ← there is an odd number in the second bracket 2^(c) * [2^(a - c) + 2^(b - c) + 1] = 4 * 37 2^(c) * [2^(a - c) + 2^(b - c) + 1] = 2^(2) * 37 → you can deduce that: 2^(c) = 2^(2) → c = 2 [2^(a - c) + 2^(b - c) + 1] = 37 2^(a - c) + 2^(b - c) = 36 [2^(a) * 2^(- c)] + [2^(b) * 2^(- c)] = 36 [2^(a) * 1/2^(c)] + [2^(b) * 1/2^(c)] = 36 → recall: c = 2 [2^(a) * 1/4] + [2^(b) * 1/4] = 36 (1/4) * [2^(a) + 2^(b)] = 36 2^(a) + 2^(b) = 144 2^(a + b - b) + 2^(b) = 144 2^(b + a - b) + 2^(b) = 144 [2^(b) * 2^(a - b)] + 2^(b) = 144 2^(b) * [2^(a - b) + 1] = 144 ← there is an odd number in the second bracket 2^(b) * [2^(a - b) + 1] = 16 * 9 2^(b) * [2^(a - b) + 1] = 2^(4) * 9 → you can deduce that: 2^(b) = 2^(4) → b = 4 [2^(a - b) + 1] = 9 2^(a - b) = 8 2^(a - b) = 2^(3) a - b = 3 a = b + 3 → recall: b = 4 → a = 7

[@SpencersAcademy]

Fantabulous! 👍

[@riteshgulati7110]

L am

[@simonlevett4776]

How do you calculate these when the numbers are different, ie 2, 3 & 5 for example.

[@andryvokubadra2644]

@@simonlevett4776 umumnya soal olimpiade terlihat rumit namun sebenarnya mudah, karena menggunakan trik. Ya jika soal olimpiade muncul seperti yang anda sebutkan, maka mau tak mau berjuanglah secara 'brute force' 😁😁😁

[@omm-o6i]

2:37 Sir if in a case b=a (let say) then 2 ᷨ/2 ͣwill become 1 then (1+1+2 ͨ ̄ ͣ) will become even

[@SpencersAcademy]

Yes, you're right. But in this case, a is not equal to b