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

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==Problem==
 
==Problem==
  
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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 union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>?
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Let <math>1</math>; <math>4</math>; <math>7</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>23</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>?
  
 
<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==
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===Solution 1===
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==Solution 1==
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 +
The terms in the first sequence are defined by <math>n \equiv 1 \pmod3</math>, while the terms in the second sequence are defined by <math>n \equiv 2 \pmod 7.</math> We seek to find the solutions to this system of modular congruences.
 +
 
 +
Letting <math>n = 3a + 1 = 7b + 2</math> for nonnegative integers <math>a,b</math> we solve the congruence \begin{align*} 3a+1 &\equiv 7b+2 \pmod 3, \\ 1 &\equiv b + 2 \pmod 3, \\ b &\equiv 2 \pmod 3.\end{align*}
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 +
For nonnegative integer <math>c,</math> we have that <math>b = 3c+2</math>. Substituting back into our original equation, we have <math>n = 21c + 16,</math> or <math>n \equiv 16 \pmod {21}.</math>
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Now we have to find the largest term in the smaller sequence (first sequence), which is <math>2003 \cdot 3 + 1 = 6010.</math> Dividing <math>\frac{6010}{21}</math> gives that there are <math>286</math> terms in common between the sequences.
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 +
By PIE, the answer is just <math>2004 + 2004 - 286 = \boxed{(A) 3722}.</math>
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 +
==Solution 2==
 
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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Thus the good values of <math>l</math> are <math>\{1,4,7,\dots,856\}</math>, and their count is <math>858/3 = 286</math>.
 
Thus the good values of <math>l</math> are <math>\{1,4,7,\dots,856\}</math>, and their count is <math>858/3 = 286</math>.
  
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Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>.
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Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{(A) 3722}</math>.
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===Solution 2===
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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>.
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==Solution 3==
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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
 +
 
 +
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  
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<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
 +
 
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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
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 +
terms in total. <math>4008-286=\boxed{(A) 3722}</math>
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 +
~kempwood
  
 
== 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}}
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{{MAA Notice}}

Latest revision as of 08:39, 5 July 2024

Problem

Let $1$; $4$; $7$; $\ldots$ and $9$; $16$; $23$; $\ldots$ 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 terms in the first sequence are defined by $n \equiv 1 \pmod3$, while the terms in the second sequence are defined by $n \equiv 2 \pmod 7.$ We seek to find the solutions to this system of modular congruences.

Letting $n = 3a + 1 = 7b + 2$ for nonnegative integers $a,b$ we solve the congruence \begin{align*} 3a+1 &\equiv 7b+2 \pmod 3, \\ 1 &\equiv b + 2 \pmod 3, \\ b &\equiv 2 \pmod 3.\end{align*}

For nonnegative integer $c,$ we have that $b = 3c+2$. Substituting back into our original equation, we have $n = 21c + 16,$ or $n \equiv 16 \pmod {21}.$

Now we have to find the largest term in the smaller sequence (first sequence), which is $2003 \cdot 3 + 1 = 6010.$ Dividing $\frac{6010}{21}$ gives that there are $286$ terms in common between the sequences.

By PIE, the answer is just $2004 + 2004 - 286 = \boxed{(A) 3722}.$

Solution 2

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 3

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

~kempwood

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

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