The sum of three numbers is 127. the second number is 4 times the third. the first number is 7 more than the third. what are the numbers?

  27, 80, 20

Step-by-step explanation:

Let t represent the third number. The second number is 4 times that, 4t. The first number is 7 more than the third, so is (t+7). The sum of the three numbers is ...

  (t+7) + 4t + t = 127

  6t +7 = 127 . . . . . .  collect terms

  6t = 120 . . . . . . . . .subtract 7

  t = 20 . . . . . . . . . . . divide by 6; the 3rd number is 20

  4t = 80 . . . . the second number is 80

  t+7 = 27 . . . the first number is 27

The three numbers are 27, 80, and 20.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18