When an integer, \(N\), is divided by \(10\), \(11\), or \(12\), the remainder is \(5\). If \(N>5\), what is the smallest possible value of \(N\)?
Solution 1
When \(N\) is divided by \(10\), \(11\), or \(12\), the remainder is \(5.\) This means that \(M=N-5\) is divisible by each of \(10\), \(11\), and \(12.\) Since \(M\) is divisible by each of \(10\), \(11\), and \(12\), then \(M\) is divisible by the least common multiple of \(10\), \(11\), and \(12.\)
Since \(10=2\times 5\), \(12 = 2^2 \times 3\), and \(11\) is prime, then to find the least common multiple, we calculate the product of the highest powers of each of the prime factors that occur in the given numbers. It follows that the least common multiple of \(10\), \(11\), and \(12\) is \(2^2 \times 3 \times 5 \times 11 = 660.\)
Since \(N>5\) and \(M=N-5\), we can conclude that \(M>0\). Since \(M\) is divisible by \(660\), then the smallest possible value for \(M\) is \(660\). Then \(N=660+5=665.\)
Solution 2
When \(N\) is divided by \(10\), \(11\), or \(12\), the remainder is \(5.\) This means that \(M=N-5\) is divisible by each of \(10\), \(11\), and \(12.\) Since \(M\) is divisible by \(10\) and \(11\), then \(M\) must be divisible by the least common multiple of \(10\) and \(11\), which is \(110.\) We test the first few multiples of \(110\) until we obtain one that is divisible by \(12.\)
The integers \(110\), \(220\), \(330\), \(440\), and \(550\) are not divisible by \(12\), but \(660\) is. Therefore, \(M\) is divisible by \(660.\) Since \(N>5\) and \(M=N-5\), we can conclude that \(M>0.\) Therefore, the smallest possible value for \(M\) is \(660.\) Then \(N=660+5=665.\)