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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

Check the Answer


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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