Difference between revisions of "1974 AHSME Problems/Problem 4"
(Created page with "==Problem== What is the remainder when <math> x^{51}+51 </math> is divided by <math> x+1 </math>? <math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 49 \qqu...") |
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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 == | ||
| + | https://youtu.be/Dp-pw6NNKRo?t=256 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1974|num-b=3|num-a=5}} | {{AHSME box|year=1974|num-b=3|num-a=5}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
