Three sequences each have first term \(3\). In the first sequence, each term after the first term is \(15\) more than the previous term. In the second sequence, each term after the first term is \(16\) more than the previous term. In the third sequence, each term after the first term is \(18\) more than the previous term. The three sequences continue indefinitely.
Determine all numbers between \(3\) and \(2025\), inclusive, that are common to all three sequences.
Solution 1
One could write out each sequence to a term less than or equal to \(2025\). You would write \(135\) terms of the first sequence, \(127\) terms of the second sequence and \(113\) terms of the third sequence. At this point you would compare the three sequences to find three numbers, \(3\), \(723\), and \(1443\), that are common to all three sequences. This is not a practical solution if you are solving the problem using pencil and paper. However, a solver could write a computer program that would easily handle this problem.
Solution 2
Notice that the number \(147\) occurs in both the second and third sequences. This number is \(147-3=144\) greater than the first number common to both sequences. What is the significance of \(144\)? It is the Least Common Multiple (LCM) of \(16\) and \(18\). The number \(16\) written in terms of prime factors is \(2\times 2\times 2\times 2 = 2^4\) and the number \(18\) written in terms of prime factors is \(2\times 3\times 3 = 2\times 3^2\). To find the LCM of \(16\) and \(18\), we determine the highest power that appears on each prime number in the two prime factorizations, and multiply all primes to the highest power together. Since the highest power of \(2\) in the two factorizations is \(2^4\) and the highest power of \(3\) in the two factorizations is \(3^2\), we have that the LCM of \(16\) and \(18\) is \(2^4\times 3^2 = 144.\)
Now, we can create a fourth sequence that starts with \(3\), and each term after the first term is \(144\) greater than the previous term. The first few terms of this sequence are \(3\), \(147\), \(291\), \(435\), \(579\), \(723\), \(867.\)
What numbers are common to this fourth sequence and the first sequence? We need to find the LCM of \(15\), the amount that each term in the first sequence increases by, and \(144\), the amount that each term in the fourth sequence increases by. The number \(15\) written in terms of prime factors is \(3\times 5\) and the number \(144\) written in terms of prime factors is \(2^4\times 3^2\). Therefore, the LCM of \(15\) and \(144\) is \(2^4\times 3^2 \times 5=720.\)
Now, we create a fifth sequence that starts with \(3\), and each term after the first term is \(720\) greater than the previous term. This sequence contains all numbers that would be common to each of the three given sequences. The first few terms would be \(3\), \(723\), \(1443\), \(2163.\)
Therefore, there are three numbers between \(3\) and \(2025\), inclusive, common to all three sequences. These numbers are \(3\), \(723\), and \(1443.\)
Solution 3
This solution builds on the ideas in Solution 2. To solve the problem, we need to find the Least Common Multiple, or LCM, of \(15\), \(16\), and \(18\). First, we write each of the three numbers as a product of their prime factors. \[\begin{aligned} 15&=3 \times 5\\ 16&=2 \times 2 \times 2 \times 2 = 2^4\\ 18&=2 \times 3 \times 3 = 2\times 3^2 \end{aligned}\] The LCM of \(15\), \(16\) and \(18\) is equal to the product of the highest power that appears on each prime number in the three prime factorizations. Since the highest power of \(2\) in the three factorizations is \(2^4\), the highest power of \(3\) is \(3^2\), and the highest power of \(5\) is \(5^1\), we have that the LCM of \(15\), \(16\), and \(18\) is \(2^4\times 3^2 \times 5 = 720.\)
We can now determine the numbers between \(3\) and \(2025\) that would be common to all three sequences. The numbers are \(3\), \(3+720=723\) and \(723+720=1443\). If we were to continue, the next common number would be \(1443+720=2163\), which is greater than \(2025.\)
Therefore, there are three numbers between \(3\) and \(2025\), inclusive, common to all three sequences. These numbers are \(3\), \(723\), and \(1443.\)