Difference between revisions of "1974 AHSME Problems/Problem 4"
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<math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 49 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ }51 </math> | <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>. | 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 == | == Video Solution by OmegaLearn == |
Latest revision as of 00:17, 9 June 2024
Problem
What is the remainder when is divided by ?
Solution 1
From the Remainder Theorem, the remainder when is divided by is .
Solution 2
Notice that is equal to . Since , this part of the polynomial is divisible by . Thus, the remainder is .
~ cxsmi
Video Solution by OmegaLearn
https://youtu.be/Dp-pw6NNKRo?t=256
~ pi_is_3.14
See Also
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 |
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