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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

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6 votes

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 !


User Spbfox
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