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# Arithmetic Progression | AMC-10B, 2004 | Problem 21

Try this beautiful problem from Algebra based on Arithmetic Progression.

## Arithmetic Progression - AMC-10B, 2004- Problem 21

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 $S$?

• $3478$
• $3722$
• $3378$

### Key Concepts

algebra

AP

Divisior

Answer: $3722$

AMC-10B (2004) Problem 21

Pre College Mathematics

## Try with Hints

There are two AP series .....

Let A=$\{1,4,7,10,13........\}$ and B=$\{9,16,23,30.....\}$.Now we have to find out a set $S$ which is the union of first $2004$ terms of each sequence.so if we construct a set form i.e $A=\{3K+1,where 0\leq k <2004\}$ and B=$\{7m+9, where 0\leq m< 2004\}$.Now in A and B total elements=4008.

Now $S=A \cup B$ and $|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|$

Now we have to find out $|A \cap B|$

Can you now finish the problem ..........

To find out $|A \cap B|$ :

Given set A=$\{1,4,7,10,13........\}$ and B=$\{9,16,23,30.....\}$.Clearly in the set A $1$ is the first term and common difference $3$.and second set i.e B first term is $9$ and common difference is $7$.

Now $|A \cap B|$ means there are some terms of $B$ which are also in $A$.Therefore $7m+9 \in A$ $\Rightarrow$ $1\leq 7m+9 \leq 3\cdot 2003 + 1$, and $7m+9\equiv 1\pmod 3$".

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

Therefore $m$= $\{1,4,7,\dots,856\}$, and number of digits in $m$= $858/3 = 286$.

can you finish the problem........

$S=A \cup B$ and $|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|=4008-286=3722$

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