Question

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?

Answer 1

Answer:

The given problem has No Solution.

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

Step-by-step explanation:

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

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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

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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.