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

(Solution: solution 2 isn't really correct as the positions of the terms which appear in both sequences change after shifting.)
(Solution)
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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>.
  
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{(A) 3722}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:34, 26 January 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

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 $|S|=4008-|A\cap B|=\boxed{(A) 3722}$.

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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