You could solve it at 1:35. Just rewrite it as k^3 - k^2 = 125 - 25. Here you could just take the square root of 25 or the cube root of 125 to get k = 5
Assume a positive integral solution, since there's no evidence of imaginary or fractional nos. k^3 must be significantly greater than 100. 4 doesn't work. 5 does. It's rigorous, because k has a specific value, and is not part of a class of values. I cannot believe this was an Oxford entrance exam question. More clickbait. I'm on a mission to crucify these stupid problems.
You could solve it at 1:35. Just rewrite it as k^3 - k^2 = 125 - 25. Here you could just take the square root of 25 or the cube root of 125 to get k = 5
10^10 2^5^2^5 1^1^2^1 2^1 (k ➖ 2k+1).
Assume a positive integral solution, since there's no evidence of imaginary or fractional nos. k^3 must be significantly greater than 100. 4 doesn't work. 5 does. It's rigorous, because k has a specific value, and is not part of a class of values. I cannot believe this was an Oxford entrance exam question. More clickbait.
I'm on a mission to crucify these stupid problems.
This guy makes it look like its so complicated but its actually ez. Oxford Entrance Exam? I'm 13... (Edit: No offense)