a, b, c, and d are positive consecutive integers and a < b < c < d.
If the product of the highest and lowest numbers is three more than the square of the second number, what is the value of the third number? Choose one
A) 2
B) 3
C) 4
D) 5
E) 6
ad = bb + 3   We know many squares: 2 squared is 4, 3 squared is 9, 4 squared is 16, 5 squared is 25, 6 squared is 36.   We need to find a sequence where the square is a little smaller than the ad product.   In the sequence 1, 2, 3, 4, the square of b is only 2x2 = 4. The product of the first and last numbers is 1x4 = 4, which is the same. So this can't be our sequence.   In the sequence 3, 4, 5, 6, the square of b is 4x4 = 16. The product of the first and last numbers is 3x6 = 18, which is close, but a little too low.   In the sequence 5, 6, 7, 8, the square of b is 6x6 = 36. The product of the first and last numbers is 5x8 = 40, which is close, but a little too high.   The correct sequence is 4, 5, 6, 7. The square of b is 5x5 = 25, and the product of the first and last numbers is 4x7 = 28, exactly three more than the square.   The answer to the problem is the value of the third number in the sequence, which is 6 (option 6).
<result>E</result>
a, b, c, and d are positive consecutive integers and a < b < c < d.
If the product of the highest and lowest numbers is three more than the square of the second number, what is the value of the third number?