Find an integer n such that n has a prime factor that occurs twice AND n plus 1 has a prime factor that occurs three times AND n plus 2 has a prime factor that occurs four times?

Answer 1

Three different prime numbers have to be involved in this question. We should choose them to make life as easy as possible for ourselves: we don't want to be working with big numbers. The three smallest prime numbers are 2, 3 and 5. 5 is the largest, so we will give it the lowest exponent. 52 = 25 33 = 27 24 = 16 This means that we are looking for a multiple of 25, n, such that n+1 divides exactly by 27 and n+2 by 16. In other words, n/27 leaves remainder 26, and n/16 leaves remainder 14. When we divide multiples of 25 by 27, the remainders follow a cycle like this: [0], 25, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 26, 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0... Notice that the cycle is itself of length 27. That is because 27 and 25 do not share any prime factors: you have to multiply 25 by 27 before you get a number that is a multiple of both. When we divide multiples of 25 by 16, the remainders follow a cycle like this: [0], 9, 2, 11, 4, 13, 6, 15, 8, 1, 10, 3, 12, 5, 14, 7, 0... It is our good fortune that the desired remainders occur in 14th position in both series simultaneously, which means we don't need to look any farther. Our value of n is 14 x 25. n = 350 = 5 x 5 x 2 x 7 n + 1 = 351 = 3 x 3 x 3 x 13 n + 2 = 352 = 2 x 2 x 2 x 2 x 2 x 11 Our remainder series aligned very nicely to give us a small answer, at n = 14 x 25. The next valid value of n is when they next align at: n = [(27 x 16) + 14] x 25 = 11150. However, that is using criteria that we specified as we went along. There are likely to be many other smallish solutions to the original question.