Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 iTest Problems/Problem 44"

(Solution to Problem 44 (credit to official solution))
 
(→‎=Problem)
 
(One intermediate revision by the same user not shown)
(No difference)

Latest revision as of 22:27, 6 May 2019

Problem

Now Wendy wanders over and joins Dr. Lisi and her younger siblings. Thinking she knows everything there is about how to work with arithmetic series, she nearly turns right around to walk back home when Dr. Lisi poses a more challenging problem. "Suppose I select two distinct terms at random from the $2008$ term sequence. What's the probability that their product is positive?" If $a$ and $b$ are relatively prime positive integers such that $a/b$ is the probability that the product of the two terms is positive, find the value of $a+b$.

Note: Dr. Lisi’s sequence is $-1776, -1765, -1754 \cdots$

Solution

The common difference of the arithmetic sequence is $11$. To find the number of terms that are negative (and the number of terms that are positive), we can write (and solve) an inequality.

\begin{align*} -1776 + 11(n-1) &< 0 \\ 11(n-1) &< 1776 \\ n-1 &< 161\frac{5}{11} \\ n &< 162\frac{5}{11} \end{align*}

Thus, there are $162$ negative terms and $1846$ positive terms. We can use complementary counting to calculate the probability of getting a negative product because it is easier than adding the probabilities where both numbers are either positive or negative.


There are $\binom{2008}{2}$ ways to choose two numbers from the sequence, and there are $162 \cdot 1846$ ways to choose one positive and one negative number. The probability of having a negative product is $\frac{162 \cdot 1846}{1004 \cdot 2007} = \frac{8307}{55973}$, so the probability of having a positive product is $1 - \frac{8307}{55973} = \frac{47666}{55973}$. Therefore, $a+b = \boxed{103639}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 43 		Followed by:
Problem 45
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_44&oldid=105705"