Difference between revisions of "1976 AHSME Problems/Problem 19"
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==Solution== | ==Solution== | ||
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| + | We know that <math>p(1)=3</math> and <math>p(3)=5</math>. We write <math>p(x)</math> as <math>p(x)=q(x)(x-1)(x-3)+r(x)</math>, where <math>r(x)=ax+b</math>. Plugging in <math>x=1</math>, we get <math>a+b=3</math>. Plugging in <math>x=3</math>, we know <math>3a+b=5</math>. We have a systems of equations, where we can solve that <math>a=1</math> and <math>b=2</math>. So, our answer is <math>1(x)+(2)=x+2\Rightarrow \textbf{(B)}</math>. ~MathJams | ||
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| + | ==1976 AHSME Problems== | ||
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| + | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Latest revision as of 18:50, 12 July 2020
A polynomial 
has remainder three when divided by 
and remainder five when divided by 
.
The remainder when is divided by 
is
Solution
We know that 
and 
. We write as 
, where 
. Plugging in 
, we get 
. Plugging in 
, we know 
. We have a systems of equations, where we can solve that 
and 
. So, our answer is 
. ~MathJams
1976 AHSME Problems
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