Difference between revisions of "2002 AMC 10P Problems/Problem 2"
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== Solution 1 == | == Solution 1 == | ||
| + | We can use the sum of an arithmetic series to solve this problem. | ||
| + | |||
| + | Let the first integer equal a. The last integer in this string will be <math>a+10.</math> Plugging in <math>n=11, a_1=a,</math> and <math>a_n=a+10</math> into <math>\frac{n(a_1 + a_n)}{2}=2002,</math> we get: | ||
| + | |||
| + | \begin{align*} | ||
| + | \frac{11(a + a+10)}{2}&=2002 \\ | ||
| + | 11(2a+10)&=4004 \\ | ||
| + | 2a+10&=364 \\ | ||
| + | 2a&=354 \\ | ||
| + | a&=177\\ | ||
| + | \end{align*} | ||
| + | |||
| + | Thus, our answer is <math>\boxed{\textbf{(B) }177}</math> | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2002|ab=P|num-b=1|num-a=3}} | {{AMC10 box|year=2002|ab=P|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 06:58, 15 July 2024
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 a. The last integer in this string will be 
Plugging in 
and 
into 
we get:
\begin{align*} \frac{11(a + a+10)}{2}&=2002 \\ 11(2a+10)&=4004 \\ 2a+10&=364 \\ 2a&=354 \\ a&=177\\ \end{align*}
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. 
