1967 AHSME Problems/Problem 25: Difference between revisions
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== Solution == | == Solution == | ||
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. | 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> | <cmath>(p-1)^n-1</cmath> | ||
| Line 15: | Line 15: | ||
<cmath>=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})</cmath> | <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> | <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 == | ||
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.
![\[(p-1)^n-1\]](http://latex.artofproblemsolving.com/9/f/7/9f75ea5e62d6bb26e7057de5c72e63680dbaf361.png)
By sum and difference of powers
![\[=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})\]](http://latex.artofproblemsolving.com/b/7/6/b76aa6938ff70b03e97dd9909a4432636570bd20.png)
![\[=(p-2)((p-1)^{n-1}+\cdots+1^{n-1})\]](http://latex.artofproblemsolving.com/c/f/a/cfa879f9ab85e8c7d4cf47b28ffbd3ae124c38c8.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
