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MathGroup Archive 2003

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Re: A difficult problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40307] Re: A difficult problem
  • From: "Bill Bertram" <wkb at ansto.gov.au>
  • Date: Sun, 30 Mar 2003 20:15:08 -0500 (EST)
  • Organization: Australian Nuclear Science and Technology Organisation
  • References: <b66cuo$ftd$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Billy Yeung" <eg_ysyaa at stu.ust.hk> wrote in message
news:b66cuo$ftd$1 at smc.vnet.net...
> I have a difficult problem:
> Given a constant integer X (which may be very large) , you no need to find
the value of X
> and
> X mod 8 = n1
> X mod 16 = n2
> the problem is that find n2 if n1 is know?

Not so difficult!
 Suppose you write X = 8 i + (Xmod 8) and  X = 16 j + (Xmod 16) then,

  j = i/2 + (n1 - n2)/16.

Since  0 <=  n1 < 8, 0 <= n2 < 16 and j must be an integer, it is easily
shown that
for i even, n2 = n1. Whilst for i odd, n2 = n1+8.

Cheers,
  Bill




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