Difference between revisions of "1969 AHSME Problems/Problem 34"
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− | <math>x^{100}=(x-2)(x-1)Q(x)+ax+b</math>, | + | <math>x^{100}=(x-2)(x-1)Q(x)+ax+b</math>, |
− | <math>2^{100}=(2-2)(2-1)Q(x)+2a+b</math>, | + | <math>2^{100}=(2-2)(2-1)Q(x)+2a+b</math>, |
− | <math>2^{100}=2a+b</math>. | + | <math>2^{100}=2a+b</math>. |
+ | |||
+ | Similarly, | ||
+ | <math>1^{100}=a+b</math>. | ||
− | |||
Solving this system of equations gives <math>a=2^{100}-1</math> and <math>b=2-2^{100}</math>. Thus, <math>R=ax+b=(2^{100}-1)x+(2-2^{100})</math>. Expanding and combining <math>x</math> terms, we see that the answer is <math>\fbox{B}</math>. | Solving this system of equations gives <math>a=2^{100}-1</math> and <math>b=2-2^{100}</math>. Thus, <math>R=ax+b=(2^{100}-1)x+(2-2^{100})</math>. Expanding and combining <math>x</math> terms, we see that the answer is <math>\fbox{B}</math>. |
Revision as of 17:00, 10 July 2015
Problem
The remainder obtained by dividing
by
is a polynomial of degree less than
. Then
may be written as:
Solution
Let the polynomial be the quotient when
is divided by
, and let the remainder
, for some real
and
. Then we can write:
. Since it is hard to deal with
(it is of degree 98!), we factor
as
so we can eliminate
by plugging in
values of
and
.
,
,
.
Similarly,
.
Solving this system of equations gives and
. Thus,
. Expanding and combining
terms, we see that the answer is
.
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.