Difference between revisions of "1965 AHSME Problems/Problem 27"

Revision as of 20:14, 18 July 2024

Problem

When $y^2 + my + 2$ is divided by $y - 1$ the quotient is $f(y)$ and the remainder is $R_1$ . When $y^2 + my + 2$ is divided by $y + 1$ the quotient is $g(y)$ and the remainder is $R_2$ . If $R_1 = R_2$ then $m$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ - 1 \qquad \textbf{(E) }\ \text{an undetermined constant}$

Solution

Let $h(y)=y^2+my+2$

$h(y)=y^2+my+2=f(y)(y-1)+R_1$

$h(y)=y^2+my+2=g(y)(y+1)+R_2$

$h(1)=3+m=R_1$

$h(-1)=3-m=R_2$

$m=0$

Thus, the answer is $\boxed{\textbf{(A) }0}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 Let <math>h(y)=y^2+my+2</math>
-<math>h(y)=y^2+my+2=f(y)(y-1)R_1</math>
+<math>h(y)=y^2+my+2=f(y)(y-1)+R_1</math>
-h(y)=<math>y^2</math>+my+2=g(y)(y+1)<math>R_2</math>
+<math>h(y)=y^2+my+2=g(y)(y+1)+R_2</math>
-h(1)=3+m=<math>R_1</math>
+<math>h(1)=3+m=R_1</math>
-h(-1)=3-m=<math>R_2</math>
+<math>h(-1)=3-m=R_2</math>
-m=0
+<math>m=0</math>
-The answer is <math>\boxed{A}</math>
+Thus, the answer is <math>\boxed{\textbf{(A) }0}</math>.
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Difference between revisions of "1965 AHSME Problems/Problem 27"

Revision as of 20:14, 18 July 2024

Problem

Solution

See Also