Difference between revisions of "2004 AMC 12A Problems/Problem 10"
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<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>. | ||
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| + | ===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>. | ||
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| + | Solution by franzliszt | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:28, 9 July 2020
Problem
The sum of 
consecutive integers is 
. What is their median?
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 
.
Solution 2
Notice that 
. So, our middle number (median) must be 
since all the other terms can be grouped to form an additional 
copies 
. Adding them would give .
Solution by franzliszt
See also
| 2004 AMC 12A (Problems • Answer Key • Resources) | |
| 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. 
