Difference between revisions of "1980 AHSME Problems/Problem 2"
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<math>\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72</math> | <math>\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72</math> | ||
| − | == Solution == | + | == Solution 1== |
It becomes <math> (x^{8}+...)(x^{9}+...) </math> with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or <math>\boxed{(D)}</math> | It becomes <math> (x^{8}+...)(x^{9}+...) </math> with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or <math>\boxed{(D)}</math> | ||
| + | ==Solution 2== | ||
| + | |||
| + | First note that given a polynomial <math>P(x)</math> and a polynomial <math>Q(x)</math>: | ||
| + | |||
| + | |||
| + | <math>deg(P(x))^n = ndeg(P(x))</math> and <math>deg(P(x)Q(x)) = deg(P(x))+deg(Q(x))</math>. | ||
| + | |||
| + | |||
| + | |||
| + | We let <math>x^2+1=P(x)</math> and <math>x^3+1 = Q(x)</math>. | ||
| + | |||
| + | Hence <math>deg(P(x)) = 2</math> and <math>deg(Q(x)) = 3</math> | ||
| + | |||
| + | So <math>deg(P(x))^4) = 4\cdot2 = 8</math> and <math>deg(Q(x)^3) = 3\cdot3 = 9</math> | ||
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| + | |||
| + | Now we let <math>P(x)^4 = R(x)</math> and <math>Q(x)^3 = S(x)</math> | ||
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| + | We want to find <math>deg(R(x)S(x)) = deg(R(x))+deg(S(x)) = 9+8 = 17</math>. | ||
| + | |||
| + | So the answer is '''(D)''' 17. | ||
| + | |||
| + | *Solution 2 by mihirb | ||
== See also == | == See also == | ||
{{AHSME box|year=1980|num-b=1|num-a=3}} | {{AHSME box|year=1980|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:39, 20 March 2018
Contents
Problem
The degree of 
as a polynomial in 
is
Solution 1
It becomes 
with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or 
Solution 2
First note that given a polynomial 
and a polynomial 
:
and 
.
We let 
and 
.
Hence 
and 
So 
and 
Now we let 
and 
We want to find 
.
So the answer is (D) 17.
- Solution 2 by mihirb
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
