Explaining the Finger Multiplication Table in Base n

mariouma
Messages
3
Reaction score
0
please i need your help, this question is from my midterm exam ineed solution very quick, it is due next week!
"your finger can provide an (n-1) multiplication table in base n, 2<=n<=10 as follow:
hold n fingers in front of you. to multiply (n-1) by k in base n, lower the kth finger from the left. your answer is (ab)in base n or (b) in base n if a=0, where ab is a string, a is the number of fingers to the left of the finger you lowered, and b is the number of fingers to the right."
explain why this works without listing all possible cases.
 
Physics news on Phys.org
We have (n-1)k = nk -k = (k-1)n + (n-k). "a" will be (k-1), and "b" will be (n-k). The number of finger in the left of the kth finger is (k-1), right? So it's "a". And similarly, the number of finger in the right is (n-k), so it's "b".
"nk-k = (k-1)n + (n-k)" comes from the idea that change "-" to "+", because (ab) = a*(n)+b.
Is this help you?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top