1967 AHSME Problems/Problem 25: Difference between revisions

Revision as of 23:26, 18 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

$\fbox{A}$ 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 $=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})$ $=(p-2)((p-1)^{n-1}+\cdots+1^{n-1})$ Therefore $p-2$ divide $(p-1)^{{\frac{1}{2}}(p-1)}$

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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

@@ Line 10: / Line 10: @@
 == Solution ==
 <math>\fbox{A}</math>
+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.
+<cmath>(p-1)^n-1</cmath>
+By sum and difference of powers
+<cmath>=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})</cmath>
+<cmath>=(p-2)((p-1)^{n-1}+\cdots+1^{n-1})</cmath>
+Therefore <math>p-2</math> divide <math>(p-1)^{{\frac{1}{2}}(p-1)}</math>
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1967 AHSME Problems/Problem 25: Difference between revisions

Revision as of 23:26, 18 November 2023

Problem

Solution

See also