Difference between revisions of "2002 AMC 10P Problems/Problem 2"
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\begin{align*} | \begin{align*} | ||
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 + 66 &= 2002 \\ | 11a + 66 &= 2002 \\ | ||
11a &= 1936 \\ | 11a &= 1936 \\ | ||
| Line 42: | Line 42: | ||
a &= 176 \\ | a &= 176 \\ | ||
\end{align*} | \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}} | ||
Latest revision as of 17:25, 15 July 2024
Contents
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:
\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 
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
\begin{align*} a + (a + 1) + (a + 2) + \; \dots \; + (a + 10) &= 2002 \\ 11a + (1 + 2 + \; \dots \; + 10) &= 2002 \\ 11a + 66 &= 2002 \\ 11a &= 1936 \\ a &= \frac{1936}{11} \\ a &= 176 \\ \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. 
