Difference between revisions of "1970 AHSME Problems/Problem 22"
(→Solution 2) |
(→Solution 2) |
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==Solution 2== | ==Solution 2== | ||
Expressing as an eqaution: | Expressing as an eqaution: | ||
− | \begin{equation}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150. | + | |
− | The sum of the first 4n positive integers = | + | The sum of the first 4n positive integers = |
+ | \begin{equation}\frac{4n(4n+1)}{2}\end{equation} | ||
We will try to rearrange Equation (1) to give equation (2) | We will try to rearrange Equation (1) to give equation (2) |
Revision as of 02:19, 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 eqaution:
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 |
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