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Explanation:
Let A,B,C be the three positive consecutive integers such that A is the smallest, B is in the middle (aka median), and C is the largest.
If A = x, then B = x+1 and C = x+2. The x is some positive whole number. Eg: x = 2 leads to A = 2, B = 3, C = 4
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"The product of the smallest and largest is 17 more than three times the median integer" means
(smallest)*(largest) = 3*(median)+17
A*C = 3*B+17
x(x+2) = 3(x+1)+17
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Let's solve for x.
x(x+2) = 3(x+1)+17
x^2+2x = 3x+3+17
x^2+2x = 3x+20
x^2+2x-3x-20 = 0
x^2-x-20 = 0
(x+4)(x-5) = 0
x+4 = 0 or x-5 = 0
x = -4 or x = 5
We made x be positive, so ignore x = -4
The only practical solution here is x = 5
A = x = 5
B = x+1 = 5+1 = 6
C = x+2 = 5+2 = 7
The three consecutive integers are: 5, 6, 7
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Check:
A*C = 3*B + 17
5*7 = 3*6 + 17
35 = 18 + 17
35 = 35
Answer is confirmed.
Three consecutive positive integers following the given condition will be 5, 6 and 7.
Let three consecutive integers are x, (x + 1), (x + 2).
Given in the question → "product of the smallest and the largest is 17 more than 3 times the median integer".
Median number amongst these integers → x
Therefore, expression for the statement will be,
x(x + 2) = 3(x + 1) + 17
x² + 2x = 3x + 3 + 17
x² + 2x - 3x - 20 = 0
x² - x - 20 = 0
x² - 5x + 4x - 20 = 0
x(x - 5) + 4(x - 5) = 0
(x + 4)(x - 5) = 0
x = -4, 5
Therefore, consecutive integers will be
x = 5
(x + 1) = 6
(x + 2) = 7
Hence, three positive consecutive integers will be 5, 6 and 7.
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