Find three consecutive odd integers such that 5 times the sum of all three is 42 more than the product of the first and second integers

Question

Martin Steele

Mathematics

22 February, 20:24

Find three consecutive odd integers such that 5 times the sum of all three is 42 more than the product of the first and second integers

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Ashlee · Answer 1

The three integers are x, x+2, x+4

5 (3x+6) = 42+x (x+2)

15x+30=x^2+2x+42

x^2 - 13x + 12=0

(x-12) (x-1) = 0

x = 12 or 1

since the integers have to be odd we exclude 12

integers are 1,3,5