Difference between revisions of "2004 AMC 10B Problems/Problem 21"

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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>
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==Solution 1==
 
==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>.  
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==Solution 2==
 
==Solution 2==
We can start by finding the first non-distinct term from both sequences. We find that that number is <math>16</math>. Now, to find every other
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We can start by finding the first non-distinct term from both sequences. We find that that number is <math>16</math>. Now, to find every  
  
non-distinct terms, we can just keep adding 21. We know that the last terms of both sequences are <math>1+3\cdot 2003</math> and <math>9+7\cdot
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other non-distinct terms, we can just keep adding <math>21</math>. We know that the last terms of both sequences are <math>1+3\cdot 2003</math> and  
  
2003</math>. Clearly, <math>1+3\cdot 2003</math> is smaller and that is the last possible common term of both sequences. Now, we can create the
+
<math>9+7\cdot 2003</math>. Clearly, <math>1+3\cdot 2003</math> is smaller and that is the last possible common term of both sequences. Now, we can  
  
inequality <math>16+21k \leq 1+3\cdot 2003</math>. Using the inequality, we find that there are <math>286</math> common terms. There are 4008 terms in
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create the inequality <math>16+21k \leq 1+3\cdot 2003</math>. Using the inequality, we find that there are <math>286</math> common terms. There are 4008  
  
total. <math>4008-286=\boxed{(A) 3722}</math>
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terms in total. <math>4008-286=\boxed{(A) 3722}</math>
  
 
~Hithere22702
 
~Hithere22702

Latest revision as of 23:37, 24 May 2021

Problem

Let $1$; $4$; $\ldots$ and $9$; $16$; $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$?

$\mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007$

Solution 1

The two sets of terms are $A=\{ 3k+1 : 0\leq k < 2004 \}$ and $B=\{ 7l+9 : 0\leq l<2004\}$.

Now $S=A\cup B$. We can compute $|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|$. We will now find $|A\cap B|$.

Consider the numbers in $B$. We want to find out how many of them lie in $A$. In other words, we need to find out the number of valid values of $l$ for which $7l+9\in A$.

The fact "$7l+9\in A$" can be rewritten as "$1\leq 7l+9 \leq 3\cdot 2003 + 1$, and $7l+9\equiv 1\pmod 3$".

The first condition gives $0\leq l\leq 857$, the second one gives $l\equiv 1\pmod 3$.

Thus the good values of $l$ are $\{1,4,7,\dots,856\}$, and their count is $858/3 = 286$.

Therefore $|A\cap B|=286$, and thus $|S|=4008-|A\cap B|=\boxed{(A) 3722}$.

Solution 2

We can start by finding the first non-distinct term from both sequences. We find that that number is $16$. Now, to find every

other non-distinct terms, we can just keep adding $21$. We know that the last terms of both sequences are $1+3\cdot 2003$ and

$9+7\cdot 2003$. Clearly, $1+3\cdot 2003$ is smaller and that is the last possible common term of both sequences. Now, we can

create the inequality $16+21k \leq 1+3\cdot 2003$. Using the inequality, we find that there are $286$ common terms. There are 4008

terms in total. $4008-286=\boxed{(A) 3722}$

~Hithere22702

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
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

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