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

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.

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.

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.)

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) }

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

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

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

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