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Smallest of 3 consecutive positive intergers if the product of the smaller two intergers is 5 less than 5 times the largest integer

Sagot :

AL2006

What a delightful little problem ! (Partly because I could see
right away how to do it, and had the answer in a few minutes,
after a lot of impressive-looking algebra on my scratch-paper.)

Three consecutive integers are . . . x,  x+1,  and  x+2

The smallest two are . . . x  and  x+1
    Their product is . . . . . x(x+1)

5 times the largest one is . . . 5(x+2)
      5 less than that is . . . . . . 5(x+2)-5

Now, the conditions of the problem say that    x (x + 1) = 5 (x+2) - 5
THAT's the equation we have to solve, to find 'x' .

       Eliminate parentheses:    x² + x = 5x + 10 - 5
            Combine like terms:    x² + x = 5x + 5
Subtract 5x from each side:    x² - 4x = 5
Subtract  5  from each side:   x² - 4x - 5 = 0

You could solve that by factoring it, or use the quadratic equation.

Factored, it says that    (x + 1) (x - 5) = 0

From which       x = -1
             and      x = +5

We only want the positive results, so our three consecutive integers are

             5,  6,  and  7 .   

To answer the question, the smallest one is    5 .

Check:

   5 x 6  ? = ?  (7 x 5) - 5

       30  ? = ?  (35) - 5

         30   =   30

             yay !   


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