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

(Solution)
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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 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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Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{(A) 3722}</math>.
 
Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{(A) 3722}</math>.
  
 +
==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
 +
 +
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
 +
 +
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
 +
 +
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
 +
 +
total. <math>4008-286=\boxed{(A) 3722}</math>
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 +
~Hithere22702
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:12, 30 April 2020

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$ (Error compiling LaTeX. ! Missing $ inserted.). 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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