Difference between revisions of "1970 AHSME Problems/Problem 22"
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\text{(E) } 600</math> | \text{(E) } 600</math> | ||
− | == Solution == | + | == Solution 1== |
We can setup our first equation as | We can setup our first equation as | ||
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<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
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+ | ==Solution 2== | ||
+ | Expressing as an equation: | ||
+ | |||
+ | |||
+ | The sum of the first 4n positive integers = | ||
+ | |||
+ | |||
+ | We will try to rearrange Equation (1) to give equation (2) | ||
+ | |||
+ | <math>\frac{3n(3n+1)}{2} - \frac{n(n+1)}{2} = 150</math> | ||
+ | |||
+ | <math>=\frac{n(3(3n+1)-(n+1))}{2} = 150 =\frac{n(9n+3-n-1)}{2}</math> | ||
+ | |||
+ | <math>\frac{n(8n+2)}{2}= \frac{2n(4n+1)}{2} = 150</math> | ||
+ | |||
+ | <math>\frac{4n(4n+1)}{2} = 2*150 = 300</math> | ||
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+ | 300 is the answer | ||
+ | |||
+ | <math>\fbox{A}</math> | ||
+ | |||
+ | |||
+ | 〜Melkor | ||
== See also == | == See also == |
Latest revision as of 02:25, 27 April 2024
Contents
[hide]Problem
If the sum of the first positive integers is more than the sum of the first positive integers, then the sum of the first positive integers is
Solution 1
We can setup our first equation as
Simplifying we get
So our roots using the quadratic formula are
Since the question said positive integers, , so
Solution 2
Expressing as an equation:
The sum of the first 4n positive integers =
We will try to rearrange Equation (1) to give equation (2)
300 is the answer
〜Melkor
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.