Difference between revisions of "2006 AIME A Problems/Problem 7"

(solution, box)
(Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Find the number of [[ordered pair]]s of [[positive]] [[integer]]s <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a [[zero]] [[digit]].
+
An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> \mathcal{C} </math> to the area of shaded region <math> \mathcal{B} </math> is 11/5. Find the ratio of shaded region <math> \mathcal{D} </math> to the area of shaded region <math> \mathcal{A}.  </math>
 +
 
 +
[[Image:2006AimeA7.PNG]]
  
 
== Solution ==
 
== Solution ==

Revision as of 14:30, 25 September 2007

Problem

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is 11/5. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}.$

2006AimeA7.PNG

Solution

There are $\frac{1000}{10} = 100$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities for $a$ are when $a$ or $b$ have a 0 in the tens digit, or if the tens digit of $a$ is 0 or 9. Excluding the hundreds, which were counted above already, there are $9 \cdot 2 = 18$ numbers in every hundred numbers that have a tens digit of 0 or 9, totaling $10 \cdot 18 = 180$ such numbers. However, the numbers from 1 to 9 and 991 to 999 do not have 0s, so we must subtract $18$ from that to get $162$. Therefore, there are $1000 - (100 + 162) = 738$ such ordered pairs.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions