Difference between revisions of "1997 AHSME Problems/Problem 20"
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==Solution== | ==Solution== | ||
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The sum of the first <math>100</math> integers is <math>\frac{100\cdot 101}{2} = 5050</math>. | The sum of the first <math>100</math> integers is <math>\frac{100\cdot 101}{2} = 5050</math>. | ||
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Thus, the only possible answer is <math>\boxed{A}</math>, because the last two digits are <math>50</math>. | Thus, the only possible answer is <math>\boxed{A}</math>, because the last two digits are <math>50</math>. | ||
| − | As an aside, if <math>5050 + 100k = 1627384950</math>, then <math>k = 16273799</math>, and the numbers added are the integers from <math>16273800</math> to <math>16273899</math>. | + | As an aside, if <math>5050 + 100k = 1627384950</math>, then <math>k = 16273799</math>, and the numbers added are the integers from <math>16273800</math> to <math>16273899</math>. |
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==Solution== | ==Solution== | ||
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Cancelling out the constants give us <math>100x + 50</math>. | Cancelling out the constants give us <math>100x + 50</math>. | ||
| − | Looking over at the possible | + | 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}} | ||
Latest revision as of 08:59, 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. 
