Hmmm.... For a^{11} I could use ssxsx : s means "square the number", x means "multiply with the starting number" It would be 5 multiplications... I assume that this could be faster and less error prone than your way...
Yeah that's insane. Why go 5 to the 5th equals 3125 but stop there? , square 3125 and substract 9 for gods sake you can go to 3125 but don't know how to setup a multiplication on paper?
This method is incredibly convoluted. I think it would have just been faster to compute 5^11.
I would just write 100,000,000,000 and divide it by 2 eleven times. And subtract 9 of course
Jesus. Just multiply times five twelve times and save yourself all that junk.
It's comforting to know that even math instructors forget what they wanted to carry over :)
Hmmm....
For a^{11} I could use ssxsx :
s means "square the number", x means "multiply with the starting number"
It would be 5 multiplications... I assume that this could be faster and less error prone than your way...
It's literally easier to do it the normal way
5^11^1 ➖ 3^2 5^1^1 ➖ 3^1 1^1 ➖ 3^1 3^1 (x ➖ 3x+1).
49
Yeah that's insane. Why go 5 to the 5th equals 3125 but stop there? , square 3125 and substract 9 for gods sake you can go to 3125 but don't know how to setup a multiplication on paper?