Prove that there are exactly 10 sets of three consecutive positive integers with the least less than 1000 such that the least is divisible by 3, the middle is divisible by 5, and the greatest is divisible by 7.

38+105= 143, 39+104=143, and you will find then adding the numbers on opposite sides, moving inwards yields the answer 138 every time. because of this, you’re going to multiply 138 times the number of pairs, which is 34. therefore, 34*138= 4692(i’m sorry if it’s the wrong numbers, but it’s the right concept)