Difference between revisions of "1997 AHSME Problems/Problem 20"
(→Solution) |
(→Solution) |
||
| Line 20: | Line 20: | ||
==Solution== | ==Solution== | ||
| − | Notice how the sum of 100 consecutive integers is <math>(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)</math> | + | Notice how the sum of 100 consecutive integers is <math>(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)</math> cancelling out the constants give us <math>100x + 50</math> |
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=19|num-a=21}} | {{AHSME box|year=1997|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 04:29, 19 August 2017
Contents
Problem
Which one of the following integers can be expressed as the sum of 
consecutive positive integers?
Solution
The sum of the first integers is 
.
If you add an integer 
to each of the numbers, you get 
, which is the sum of the numbers from 
to 
.
You're only adding multiples of , so the last two digits will remain unchanged.
Thus, the only possible answer is 
, because the last two digits are 
.
As an aside, if 
, then 
, and the numbers added are the integers from 
to 
.
Solution
Notice how the sum of 100 consecutive integers is 
cancelling out the constants give us 
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
