Difference between revisions of "1973 AHSME Problems/Problem 21"
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==See Also== | ==See Also== | ||
| − | {{AHSME | + | {{AHSME 30p box|year=1973|num-b=20|num-a=22}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| − | [[Category: | + | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 13:02, 20 February 2020
Problem
The number of sets of two or more consecutive positive integers whose sum is 100 is
Solution
If the first number of a group of 
consecutive numbers is 
, the 
number is 
. We know that the sum of the group of numbers is 
, so
![\[\frac{n(2a+n-1)}{2} = 100\]](http://latex.artofproblemsolving.com/8/8/8/888e4e5bedfe5956ed2b875653f2ca53257710a5.png)
![\[2a+n-1=\frac{200}{n}\]](http://latex.artofproblemsolving.com/3/a/6/3a645021b6390df0191b0afb383682688c3efab3.png)
![\[2a = 1-n + \frac{200}{n}\]](http://latex.artofproblemsolving.com/1/9/7/1976836b347a3efafab3f84aaaf719c47e2bee52.png)
We know that and are positive integers, so we check values of that are a factor of 
. Of these values, the only ones that result in a positive integer is when 
or when 
, so there are 
sets of two or more consecutive positive integers whose sum is .
See Also
| 1973 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
