Difference between revisions of "2002 AMC 10P Problems/Problem 2"
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a + (a + 1) + (a + 2) + \; \dots \; + (a + 10) &= 2002 \ | a + (a + 1) + (a + 2) + \; \dots \; + (a + 10) &= 2002 \ | ||
11a + (1 + 2 + \; \dots \; + 10) &= 2002 \ | 11a + (1 + 2 + \; \dots \; + 10) &= 2002 \ | ||
− | 11a + | + | 11a + 55 &= 2002 \ |
− | 11a &= | + | 11a &= 1947 \ |
− | a &= \frac{ | + | a &= \frac{1947}{11} \ |
− | a &= | + | a &= 177 \ |
\end{align*} | \end{align*} | ||
Latest revision as of 18:59, 31 July 2024
Contents
[hide]Problem 2
The sum of eleven consecutive integers is What is the smallest of these integers?
Solution 1
We can use the sum of an arithmetic series to solve this problem.
Let the first integer equal The last integer in this string will be Plugging in and into we get:
Thus, our answer is
Solution 2
We can directly add everything up since is so little.
Similar to the first solution, let the first integer equal The last integer in this string will be
Thus, our answer is
See Also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.