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

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 \(|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|=4008-286=3722\)