SOLUTION: Three times the least of three consecutive odd integers exceeds twice the greatest by 19. Find the three integers.

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Question 1193708: Three times the least of three consecutive odd integers exceeds twice the greatest by 19. Find the three integers.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
Answer by ikleyn(52778) About Me  (Show Source):
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Three times the least of three consecutive odd integers exceeds twice the greatest by 19. Find the three integers.
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Let the three consecutive integer numbers be n, (n+2) and (n+4).


Then your equation to find "n" is

    3*n = 2*(n+4) + 19.


Now simplify it and find "n"

    3n = 2n + 8 + 19

    3n - 2n = 8 + 19

       n    =   27.


ANSWER.  The numbers are  27, 29 and 31.

Solved.