Difference between revisions of "1997 AHSME Problems/Problem 20"
(→Solution) |
(→Solution) |
||
Line 23: | Line 23: | ||
Cancelling out the constants give us <math>100x + 50</math>. | Cancelling out the constants give us <math>100x + 50</math>. | ||
− | Looking over at the | + | Looking over at the list of possible values, we quickly realise that the only possible solution is <math>\boxed{A}</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 04:34, 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 .
Looking over at the list of possible values, we quickly realise that the only possible solution is
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.