For each positive integer \(n\), \(\text{LCM}(1,2,\ldots,n)\) is the least common multiple of \(1, 2, \ldots, n.\) That is, the smallest positive integer divisible by each of \(1, 2, \ldots, n.\)
Determine all positive integers \(n\), with \(1\leq n \leq 100\) such that \[\text{LCM}(1, 2, \ldots, n) = \text{LCM}(1, 2, \ldots, n+4)\]
Note: In solving this problem, it might be helpful to know that we can calculate the LCM of a set of positive integers by
determining the prime factorization of each integer in the set,
determining the list of prime numbers that occur in these prime factorizations,
determining the highest power of each prime number from this list that occurs in the prime factorizations, and
multiplying these highest powers together.
For example, \(\text{LCM}(1, 2, 3, 4, 5, 6, 7, 8) = 2^3\cdot 3^1 \cdot 5^1 \cdot 7^1 = 840\), since the prime factorizations of \(2\), \(3\), \(4\), \(5\), \(6\), \(7\), and \(8\) are \(2\), \(3\), \(2^2\), \(5\), \(2\cdot 3\), \(7\), and \(2^3\), respectively.
Since \(\text{LCM}(1, 2, \ldots, n)\) is the least common multiple of \(1, 2, \ldots, n\) and \(\text{LCM}(1, 2, \ldots, n+4)\) is the least common multiple of \(1, 2, \ldots, n, n+1, n+2, n+3, n+4\), then \(\text{LCM}(1, 2, \ldots, n) \neq \text{LCM}(1, 2, \ldots, n+4)\) if either
there are prime factors that occur in \(n+1\), \(n+2\), \(n+3\), \(n+4\) that don’t occur in any of \(1, 2, \ldots, n\), or
there is a higher power of a prime that occurs in the factorizations of one of \(n+1\), \(n+2\), \(n+3\), \(n+4\) that doesn’t occur in any of \(1, 2, \ldots, n.\)
For (i) to occur, consider a prime \(p\) that is a divisor of one of \(n+1\), \(n+2\), \(n+3\), \(n+4\), and none of \(1, 2, \ldots, n.\). This means that the smallest positive integer that has \(p\) as a divisor is one of the integers \(n+1\), \(n+2\), \(n+3\), \(n+4\), which in fact means that this integer equals \(p.\) (The smallest multiple of a prime \(p\) is \(1\cdot p = p\) itself.)
Thus, for (i) to occur, one of \(n+1\), \(n+2\), \(n+3\), \(n+4\) is a prime number.
For (ii) to occur, consider a prime power \(p^k\), where \(k\) is an integer \(>1\), that is a divisor of one of \(n+1\), \(n+2\), \(n+3\), \(n+4\) and none of \(1, 2, \ldots, n.\) This means that the smallest positive integer that has \(p^k\) as a divisor is one of the integers \(n+1\), \(n+2\), \(n+3\), \(n+4\), which in fact means that this integer equals \(p^k.\) (The smallest multiple of \(p^k\) is \(p^k\) itself.)
Therefore, \(\text{LCM}(1, 2, \ldots, n) \neq \text{LCM}(1, 2, \ldots, n+4)\) whenever one of \(n+1\), \(n+2\), \(n+3\), \(n+4\) is a prime number or a prime power.
In other words, \(\text{LCM}(1, 2, \ldots, n) = \text{LCM}(1, 2, \ldots, n+4)\) whenever none of \(n+1\), \(n+2\), \(n+3\), \(n+4\) is a prime number or a prime power.
Therefore, we want to determine the positive integers \(n\) with \(1\leq n\leq 100\) for which none of \(n+1\), \(n+2\), \(n+3\), \(n+4\) is a prime number or a prime power.
The prime numbers less than or equal to \(104\) are
\(2\), \(3\), \(5\), \(7\), \(11\), \(13\), \(17\), \(19\), \(23\), \(29\), \(31\), \(37\), \(41\), \(43\), \(47\), \(53\), \(59\), \(61\), \(67\), \(71\), \(73\), \(79\), \(83\), \(89\), \(97\), \(101\), \(103\)
The prime powers, with exponent \(>1\), less than or equal to \(104\) are
\(2^2=4\), \(2^3=8\), \(2^4=16\), \(2^5=32\), \(2^6=64\), \(3^2=9\), \(3^3=27\), \(3^4=81\), \(5^2=25\), \(7^2=49\)
Therefore, we want to count the positive integers \(n\) with \(1\leq n\leq 100\) for which none of \(n+1\), \(n+2\), \(n+3\), \(n+4\) appear in the list
\(2\), \(3\), \(4\), \(5\), \(7\), \(8\), \(9\), \(11\), \(13\), \(16\), \(17\), \(19\), \(23\), \(25\), \(27\), \(29\), \(31\), \(32\), \(37\), \(41\), \(43\), \(47\), \(49\), \(53\), \(59\), \(61\), \(64\), \(67\), \(71\), \(73\), \(79\), \(81\), \(83\), \(89\), \(97\), \(101\), \(103\)
For four consecutive integers to not occur in this list, we need a difference between adjacent numbers to be at least \(5.\)
The values of \(n\) that satisfy this condition are \(n=32\), \(53\), \(54\), \(73\), \(74\), \(83\), \(84\), \(89\), \(90\), \(91\), \(92.\) (For example, \(54\) is a value of \(n\) that works since none of \(55\), \(56\), \(57\), \(58\) appears in the list.)
Therefore, there are \(11\) values of \(n\) with \(1\leq n \leq 100\) for which \(\text{LCM}(1, 2, \ldots, n) = \text{LCM}(1, 2, \ldots, n+4).\)