What is the smallest of three consecutive positive integers if the product of the smaller integers is one less than 4 times the largest integer?

The given problem has No Solution.

If the integers are consecutive odd integers, the smallest is 5.

Since we want the smallest of the integers, it is convenient to let the variable x represent that value. Then the other two integers are (x+1) and (x+2).

The problem statement tells us ...

... x(x+1)= 4(x+2) -1 . . . . product of the smallest two is 1 less than 4 times largest

... x² +x = 4x +7 . . . . . . eliminate parentheses, collect terms

... x² -3x = 7 . . . . . . . . . subtract 4x

... x² -3x +2.25 = 9.25 . . . add (3/2)²

... (x -1.5)² = 9.25

... x - 1.5 = √9.25 . . . . . take the square root

... x = 1.5 +√9.25 . . . . . not an integer ⇒ no solution

Alternate Problem

Assuming the consecutive integers are odd, the problem becomes ...

... x(x +2) = 4(x +4) -1

... x² +2x = 4x +15

... x² -2x = 15

... x² -2x +1 = 16 . . . . complete the square

... (x -1)² = 4²

... x -1 = 4 . . . . . . . . . take the square root

... x = 5 . . . . . . the integers are 5, 7, 9

Check

5×7 = 35 = 4×9 -1 . . . . answer checks for consecutive odd integers

Another comment on the original problem

The product of two consecutive integers will be even. The result of multiplying any integer by 4, then subtracting 1 will be odd. There is no way an even result will equal an odd result. In order for the product of two integers to be odd, both must be odd.

3; 9 is 8 to -4

step-by-step explanation: