Difference between revisions of "1974 AHSME Problems/Problem 4"

Latest revision as of 01:17, 9 June 2024

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)
Preceded by Problem 3	Followed by Problem 5
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.

@@ Line 4: / Line 4: @@
 <math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 49 \qquad \mathrm{(D) \  } 50 \qquad \mathrm{(E) \  }51  </math>
-==Solution==
+==Solution 1==
 From the [[Remainder Theorem]], the remainder when <math> x^{51}+51 </math> is divided by <math> x+1 </math> is <math> (-1)^{51}+51=-1+51=50, \boxed{\text{D}} </math>.
+==Solution 2==
+Notice that <math>x^{51}+51</math> is equal to <math>(x^{51}+1)+50</math>. Since <math>x^51+1=(x+1)(x^{50}-x^{49}+ \ldots +x^2 - x + 1)</math>, this part of the polynomial is divisible by <math>x+1</math>. Thus, the remainder is <math>\boxed{\textbf{(D)}~50}</math>.
+~ cxsmi
+== Video Solution by OmegaLearn ==
+https://youtu.be/Dp-pw6NNKRo?t=256
+~ pi_is_3.14
 ==See Also==
 {{AHSME box|year=1974|num-b=3|num-a=5}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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