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Number system

  1. Apr 2, 2007 #1
    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.
  2. jcsd
  3. Apr 2, 2007 #2
    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?
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