a, b, c, d, and e are positive consecutive odd integers and a < b < c < d < e.
If the sum of a+c+e is four less than the square of b, what is the product of de?
Choose the correct choice from the following answers.
A) 25  B) 35  C) 53  D) 99  E) 143
a+c+e = bb - 4   Because we know the numbers are consecutive odd integers, we can rewrite the list as versions of the first number:   b = a+2   c = a+4   d = a+6   e = a+8   Therefore we can rewrite the original problem: a+(a+4)+(a+8) = (a+2)(a+2) - 4   Simplify: 3a + 4 + 8 = (aa + 4a + 4) - 4   Simplify: 3a + 12 = aa + 4a   Swap (just switch the sides in their entirety): aa + 4a = 3a + 12   Separate the number from the variables: aa + 4a - 3a = 12   Simplify: aa + a = 12   So the square of a, plus another a, equals 12. We can try out a few numbers quickly:   1x1 = 1, +1 = not 12   5x5 = 25, +5 = not 12   3x3 = 9, +3 = there we go!   So a = 3, and the sequence of consecutive odd numbers is 3, 5, 7, 9, 11.   The answer is the product of the last two numbers: 9x11 = 99 (option 99).
<result>D</result>
a, b, c, d, and e are positive consecutive odd integers and a < b < c < d < e.
If the sum of a+c+e is four less than the square of b, what is the product of de?