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> | + | |
+ | 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> | ||
+ | <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}} | + | {{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 23:28, 18 November 2023
Problem
For every odd number we have:
Solution
Given that is odd, must be even, therefore must be an integer, which will be denoted as n. By sum and difference of powers divide .
See also
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 |
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