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Difference between revisions of "1961 AHSME Problems/Problem 5"
(Solution to Problem 5)
 
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Let <math>S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1</math>. Then S equals:
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== Problem ==
  
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(A) <math>(x-2)^4</math>
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Let <math>S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1</math>. Then <math>S</math> equals:
  
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(B) <math>(x-1)^4</math>
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<math>\textbf{(A)}\ (x-2)^4 \qquad
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\textbf{(B)}\ (x-1)^4 \qquad
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\textbf{(C)}\ x^4 \qquad
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\textbf{(D)}\ (x+1)^4 \qquad
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\textbf{(E)}\ x^4+1 </math>
  
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(C) <math>x^4</math>
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==Solution==
  
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(D) <math>(x+1)^4</math>
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Let <math>y = x-1</math>.  Substitution results in
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<cmath>S = y^4 + 4y^3 + 6y^2 + 4y + 1</cmath>
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<cmath>S = (y+1)^4</cmath>
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Substituting back results in
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<cmath>S = x^4</cmath>
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The answer is <math>\boxed{\textbf{(C)}}</math>.  This problem can also be solved by traditionally expanding and combining like terms (though it would take much longer).
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==See Also==
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{{AHSME 40p box|year=1961|num-b=4|num-a=6}}
  
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(E) <math>x^4+1</math>
 
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:54, 17 May 2018

Problem

Let $S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$. Then $S$ equals:

$\textbf{(A)}\ (x-2)^4 \qquad \textbf{(B)}\ (x-1)^4 \qquad \textbf{(C)}\ x^4 \qquad \textbf{(D)}\ (x+1)^4 \qquad \textbf{(E)}\ x^4+1$

Solution

Let $y = x-1$. Substitution results in \[S = y^4 + 4y^3 + 6y^2 + 4y + 1\] \[S = (y+1)^4\] Substituting back results in \[S = x^4\] The answer is $\boxed{\textbf{(C)}}$. This problem can also be solved by traditionally expanding and combining like terms (though it would take much longer).

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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


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