Difference between revisions of "2004 AMC 10B Problems/Problem 21"
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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=== | + | ===<h>Solution 2<h>=== |
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 20:04, 7 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 
.
<h>Solution 2<h>
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. 
