Question

How many ways can David pick four of the first twelve positive integers such that no two of the numbers he picks are consecutive?

Answer 1

Answer:

70 times

Step-by-step explanation:

5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 The smallest integer in our sequence is 2n = 2*17 = 34 Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70