Difference between revisions of "1967 AHSME Problems/Problem 25"

(Created page with "== Problem == For every odd number <math>p>1</math> we have: <math>\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{...")
 
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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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Given that <math>p</math> is odd, <math>p-1</math> must be even, therefore <math>{{\frac{1}{2}}(p-1)}</math> must be an integer, which will be denoted as n.
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<cmath>(p-1)^n-1</cmath>
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By sum and difference of powers
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<cmath>=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})</cmath>
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<cmath>=(p-2)((p-1)^{n-1}+\cdots+1^{n-1})</cmath>
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<math>p-2</math> divide <math>(p-1)^{{\frac{1}{2}}(p-1)}</math><math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1967|num-b=24|num-a=26}}   
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{{AHSME 40p box|year=1967|num-b=24|num-a=26}}   
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:28, 19 November 2023

Problem

For every odd number $p>1$ we have:

$\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\ \textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad \textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p+1\\ \textbf{(E)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-1$

Solution

Given that $p$ is odd, $p-1$ must be even, therefore ${{\frac{1}{2}}(p-1)}$ must be an integer, which will be denoted as n. \[(p-1)^n-1\] By sum and difference of powers \[=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})\] \[=(p-2)((p-1)^{n-1}+\cdots+1^{n-1})\] $p-2$ divide $(p-1)^{{\frac{1}{2}}(p-1)}$$\fbox{A}$.

See also

1967 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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