Difference between revisions of "2004 AMC 12A Problems/Problem 10"

(create)
 
(Solution)
 
(One intermediate revision by one other user not shown)
Line 4: Line 4:
 
<math>\text {(A)}\ 7 \qquad \text {(B)}\ 7^2\qquad \text {(C)}\ 7^3\qquad \text {(D)}\ 7^4\qquad \text {(E)}\ 7^5</math>
 
<math>\text {(A)}\ 7 \qquad \text {(B)}\ 7^2\qquad \text {(C)}\ 7^3\qquad \text {(D)}\ 7^4\qquad \text {(E)}\ 7^5</math>
  
== Solution ==  
+
== Solutions ==
 +
 
 +
===Solution 1===  
 
The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the [[mean]], so the median is <math>\frac{7^5}{49} = 7^3\ \mathrm{(C)}</math>.
 
The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the [[mean]], so the median is <math>\frac{7^5}{49} = 7^3\ \mathrm{(C)}</math>.
 +
 +
===Solution 2===
 +
Notice that <math>49\cdot7^3=7^5</math>. So, our middle number (median) must be <math> 7^3\ \mathrm{(C)}</math> since all the other terms can be grouped to form an additional <math>48</math> copies <math>7^3</math>. Adding them would give <math>7^5</math>. 
 +
 +
Solution by franzliszt
  
 
== See also ==
 
== See also ==
Line 11: Line 18:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:28, 9 July 2020

Problem

The sum of $49$ consecutive integers is $7^5$. What is their median?

$\text {(A)}\ 7 \qquad \text {(B)}\ 7^2\qquad \text {(C)}\ 7^3\qquad \text {(D)}\ 7^4\qquad \text {(E)}\ 7^5$

Solutions

Solution 1

The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the mean, so the median is $\frac{7^5}{49} = 7^3\ \mathrm{(C)}$.

Solution 2

Notice that $49\cdot7^3=7^5$. So, our middle number (median) must be $7^3\ \mathrm{(C)}$ since all the other terms can be grouped to form an additional $48$ copies $7^3$. Adding them would give $7^5$.

Solution by franzliszt

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png