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 <u>x (x + 1) = 5 (x+2) - 5</u> 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: <u>x² - 4x - 5 = 0</u>
You could solve that by factoring it, or use the quadratic equation.
Factored, it says that (x + 1) (x - 5) = 0
From which <em>x = -1</em> and <em>x = +5</em>
We only want the positive results, so our three consecutive integers are
5, 6, and 7 .
To answer the question, the smallest one is <em><u>5 </u></em>.
