1974 AHSME Problems/Problem 4

Problem

What is the remainder when $x^{51}+51$ is divided by $x+1$ ?

$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 49 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ }51$

Solution 1

From the Remainder Theorem, the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \boxed{\text{D}}$ .

Solution 2

Notice that $x^{51}+51$ is equal to $(x^{51}+1)+50$ . Since $x^51+1=(x+1)(x^{50}-x^{49}+ \ldots +x^2 - x + 1)$ , this part of the polynomial is divisible by $x+1$ . Thus, the remainder is $\boxed{\textbf{(D)}~50}$ .

~ cxsmi

Video Solution by OmegaLearn

https://youtu.be/Dp-pw6NNKRo?t=256

~ pi_is_3.14

1974 AHSME (Problems • Answer Key • Resources)
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1974 AHSME Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

Video Solution by OmegaLearn

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