Difference between revisions of "2004 AMC 10B Problems/Problem 21"
Kylewu0715 (talk | contribs) m (→Solution 2) |
m (→Solution 2) |
||
Line 19: | Line 19: | ||
Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>. | Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>. | ||
− | === | + | ===Solution 2=== |
Shift down the first sequence by <math>1</math> and the second by <math>9</math> so that the two sequences become <math>0,3,6,\cdots,6009</math> and <math>0,7,14,\cdots,14028</math>. The first becomes multiples of <math>3</math> and the second becomes multiples of <math>7</math>. Their intersection is the multiples of <math>21</math> up to <math>6009</math>. There are <math>\lfloor \frac{6009}{21} \rfloor=286</math> multiples of <math>21</math>. There are <math>4008-286=\boxed{\textbf{(A)}\ 3722}</math> distinct numbers in <math>S</math>. | Shift down the first sequence by <math>1</math> and the second by <math>9</math> so that the two sequences become <math>0,3,6,\cdots,6009</math> and <math>0,7,14,\cdots,14028</math>. The first becomes multiples of <math>3</math> and the second becomes multiples of <math>7</math>. Their intersection is the multiples of <math>21</math> up to <math>6009</math>. There are <math>\lfloor \frac{6009}{21} \rfloor=286</math> multiples of <math>21</math>. There are <math>4008-286=\boxed{\textbf{(A)}\ 3722}</math> distinct numbers in <math>S</math>. | ||
Revision as of 15:55, 9 October 2019
Problem
Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ?
Solution
Solution 1
The two sets of terms are and .
Now . We can compute . We will now find .
Consider the numbers in . We want to find out how many of them lie in . In other words, we need to find out the number of valid values of for which .
The fact "" can be rewritten as ", and ".
The first condition gives , the second one gives .
Thus the good values of are , and their count is .
Therefore , and thus .
Solution 2
Shift down the first sequence by and the second by so that the two sequences become and . The first becomes multiples of and the second becomes multiples of . Their intersection is the multiples of up to . There are multiples of . There are distinct numbers in .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.