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

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick.

Please feel free to share your ideas in the comment section.

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[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 by simplifying we have: (2^2+2^4+2^7 ) that the values ​​of a, b, c =(2, 4, 7) are easily obtained.

[markgideon]

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

[marekzalinski]

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.

[mrinaldas]

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.)

[허공]

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) }

[kavitapatidar]

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

[nabeelahmed]

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

[MARTINWERDER]

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

[Dr_piFrog]

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

[boguslawszostak]

assuming a<b<c

so a=2 b=a+2=4 b+3=7

[xvoidx_yt]

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

[davidbrisbane]

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.

[ilafya]

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

[simonlevett]

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

[andryvokubadra]

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 😊😊😊

[pascallemesle]

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

[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

[RealQinnMalloryu]

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

[omm-oi]

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