

Click here to see ALL problems on Numbers Word Problems Question 478538: Prove that the product of three consecutive integers is divisible by 6; of four consecutive integers by 24.

Answer by Edwin McCravy(18169) (Show Source):

Theorem: If n is a positive integer, then n(n+1)(n+2) is divisible by 3 Proof by induction: 1*2*3 = 6, which is divisible by 6. Assume that for n = k, k(k+1)(k+2) is divisible by 6 We need to show that, based on that assumption, (k+1)(k+2)(k+3) is also divisible by 6. (k+1)(k+2)(k+3) = (k+1)(k+2)k + (k+1)(k+2)3 = k(k+1)(k+2) + 3(k+1)(k+2). By induction hypothesis, the first term is divisible by 6, and the second term 3(k+1)(k+2) is divisible by 6 because it contains a factor 3 and one of the two consecutive integers k+1 or k+2 is even and thus is divisible by 2. Thus it is divisible by both 3 and 2, which means it is divisible by 6. The theorem is proved since the sum of two multiples of 6 is also a multiple of 6. --------------------- Theorem: If n is a positive integer, then n(n+1)(n+2)(n+3) is divisible by 24. Proof by induction: 1*2*3*4 = 24, which is divisible by 24. Assume that for n = k, k(k+1)(k+2)(k+3) is divisible by 24 We need to show that (k+1)(k+2)(k+3)(k+4), based on that ssumption, is also divisible by 24. (k+1)(k+2)(k+3)(k+4) = (k+1)(k+2)(k+3)k + (k+1)(k+2)(k+3)4 = k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3). By induction hypothesis, the first term is divisible by 24, and the second term 4(k+1)(k+2)(k+3) is divisible by 24 because it contains a factor 4 and by the preceding theorem (k+1)(k+2)(k+3) is divisible by 6, so 4(k+2)(k+3)(k+4) is divisible by 24. Therefore (k+1)(k+2)(k+3)(k+4) is divisible by 24 since it is the sum of two multiples of 24. Edwin You can put this solution on YOUR website!

