Difference between revisions of "2004 AMC 10B Problems/Problem 21"
(New page: ==Problem== Let <math>1</math>; <math>4</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the ...) |
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<math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math> | <math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math> | ||
==Solution== | ==Solution== | ||
− | + | ===Solution 1=== | |
The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. | The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. | ||
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</math> multiples of <math>286</math>. There are <math>4008-286=\boxed{\textbf{(A)}\ 3722}</math> distinct numbers in <math>S</math>. | ||
== See also == | == See also == | ||
{{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}} |
Revision as of 15:53, 28 December 2012
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 |