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
But try the problem first…
Answer: \(3722\)
AMC-10B (2004) Problem 21
Pre College Mathematics
Try with Hints
First hint
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 ……….
Second Hint
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……..
Final Step
\(S=A \cup B\) and \(|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|=4008-286=3722\)
Other useful links
- https://www.cheenta.com/problem-on-semicircle-amc-8-2013-problem-20/
- https://www.youtube.com/watch?v=PfRqs9W8nPQ