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

Latest revision as of 16:28, 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

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.

@@ 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>
-== 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>.
+===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
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