| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Here's a problem from last night's _Granite State Challenge_. I liked it because I figured out the answer before I realized what I was doing (it's my old competition reflexes). What is the remainder when x^37 + 2x is divided by x+1 ? -- edp
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 436.1 | TOOLS::STAN | Tue Jan 21 1986 21:06 | 4 | ||
37 x +2x=(x+1)Q(x)+R Let x=-1 and we get -1-2=R, so the remainder, R, is -3. | |||||
| 436.2 | R2ME2::GILBERT | Wed Jan 22 1986 12:53 | 5 | ||
I realized that x^37+1 is a multiple of x+1, so x^37+2x = x^37+1 + 2x-1 = (x+1)P(x) + 2x-1 And the remainder when 2x-1 is divided by x+1 is -3. | |||||
| 436.3 | A faster way? | CIMLAB::HAINSWORTH | Many pages make a thick book. | Thu Jul 30 1987 19:29 | 5 |
The remainder when x^37 + 2x is divided by x+1 is the value of the
original expression when x+1 = 0. (-1)^37 + 2(-1) = -3. That's
how I did it.
John
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