Difference between revisions of "2002 AMC 10P Problems/Problem 2"

(Solution 2)
m (Solution 2)
 
Line 37: Line 37:
 
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 + 55 &= 2002 \
11a &= 1936 \
+
11a &= 1947 \
a &= \frac{1936}{11} \
+
a &= \frac{1947}{11} \
a &= 176 \
+
a &= 177 \
 
\end{align*}
 
\end{align*}
  

Latest revision as of 18:59, 31 July 2024

Problem 2

The sum of eleven consecutive integers is $2002.$ What is the smallest of these integers?

$\text{(A) }175 \qquad \text{(B) }177 \qquad \text{(C) }179 \qquad \text{(D) }180 \qquad \text{(E) }181$

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 $a+10.$ Plugging in $n=11, a_1=a,$ and $a_n=a+10$ into $\frac{n(a_1 + a_n)}{2}=2002,$ we get:

11(a+a+10)2=200211(2a+10)=40042a+10=3642a=354a=177

Thus, our answer is $\boxed{\textbf{(B) }177}$

Solution 2

We can directly add everything up since $1 + 2 + \; \dots \; + 10$ is so little.

Similar to the first solution, let the first integer equal $a.$ The last integer in this string will be $a+10.$

a+(a+1)+(a+2)++(a+10)=200211a+(1+2++10)=200211a+55=200211a=1947a=194711a=177

Thus, our answer is $\boxed{\textbf{(B) }177}$

See Also

2002 AMC 10P (ProblemsAnswer KeyResources)
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. AMC logo.png