Math Olympiad | A Nice Algebra Problem | 95% Failed to solve!

[@wes9627]

c=floor(ln2320/ln2)=11; b=floor[ln(2320-2^11)/ln2]=8; a=ln(2320-2^11-2^8)/ln2=4.

[@abbeydudeuk]

1^4 = 1^6 base is same but power is not then how u equated a=4?

[@adribber]

2^a + 2^b + 2^c = 2320 (a < b < c)
2^a + 2^b + 2^c = 2^4*5*29
Let, a = k , b = k + m , c = k + n (k, m , n ∈ N)
2^k + 2^(k + m) + 2^(k + n) = 2^4*5*29
2^k + 2^k*2^m + 2^k*2^n = 2^4*5*29
2^k(2^m + 2^n + 1) = 2^4*5*29
2^k : even number,
2^m , 2^n : even number,
2^m + 2^n + 1 : odd number
∴ 0 ≤ k ≤ 4
if, k = 0
2^0*(2^m + 2^n + 1) = 2^4*5*29
1*(2^m + 2^n + 1) = 2^4*5*29
odd number * odd number ≠ even number Failed
∴ k ≠ 0
if, k = 1
2^1*(2^m + 2^n + 1) = 2^4*5*29
2*(2^m + 2^n + 1) = 2^4*5*29
(2^m + 2^n + 1) = 2^3*5*29
odd number ≠ even number Failed
∴ k ≠ 1
if, k = 2
2^2*(2^m + 2^n + 1) = 2^4*5*29
(2^m + 2^n + 1) = 2^2*5*29
odd number ≠ even number Failed
∴ k ≠ 2
if, k = 3
2^3*(2^m + 2^n + 1) = 2^4*5*29
(2^m + 2^n + 1) = 2*5*29
odd number ≠ even number Failed
∴ k ≠ 3
if, k = 4
2^4*(2^m + 2^n + 1) = 2^4*5*29
(2^m + 2^n + 1) = 5*29
odd number = odd number Pass
∴ k = 4
Recall, a = k
∴ a = 4
∴ 2^m + 2^n + 1 = 5*29 = 145
2^m + 2^n = 144
2^m + 2^n = 2^4*3^2
0 ≤ m ≤ 4
if, m = 0
2^0 + 2^n = 2^4*3^2
1 + 2^n = 2^4*3^2
2^n = 2^4*3^2 - 1
even number ≠ odd number Failed
∴ m ≠ 0
if, m = 1
2^1 + 2^n = 2^4*3^2
2 + 2^n = 2^4*3^2
2^n = 2^4*3^2 - 2
2^n = 142 = 2*71 (n ∈ N) Failed
∴ m ≠ 1
if, m = 2
2^2 + 2^n = 2^4*3^2
4 + 2^n = 2^4*3^2
2^n = 2^4*3^2 - 4
2^n = 140 = 2^2*5*7 (n ∈ N) Failed
∴ m ≠ 2
if, m = 3
2^3 + 2^n = 2^4*3^2
8 + 2^n = 2^4*3^2
2^n = 2^4*3^2 - 8
2^n = 136 = 2^3*17 (n ∈ N) Failed
∴ m ≠ 3
if, m = 4
2^4 + 2^n = 2^4*3^2
2^n = 2^4*3^2 - 2^4
= 2^4(9 - 1)
= 2^4*8
= 2^4*2^3
= 2^7
∴ n = 7 Pass
∴ m = 4
∴ k = 4 , m = 4 , n = 7
Recall, a = k , b = k + m , c = k + n
a = 4 , b = 8 , c = 11
Verifying
2^a + 2^b + 2^c =? 2320
2^4 + 2^8 + 2^11 =? 2320
16 + 256 + 2048 =? 2320
2320 = 2320 Pass
∴ a = 4 , b = 8 , c = 11