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find three consecutive odd integers such that the sum of the least integer and the middle integer 19 more than the great integer
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Let A = 2n+1 be the first (the least) of the three consecutive odd integers;
B = 2n+3 be the second of the three consecutive odd integers; and
C = 2n+5 be the third.
The condition requires
A + B = C + 19, or
(2n+1) + (2n+3) = (2n+5) + 19.
Simplify:
2n + 1 + 2n + 3 = 2n + 5 + 19,
2n + 4 = 24,
2n = 24 - 4 = 20,
n = = 10.
So, your numbers are A = 2n+1 = 2*10+1 = 21; B = 21+2 = 23; and C = 23+2 = 25.
Check. A + B = 21 + 23 = 44.
44 - 19 = 25. Correct !
Answer. The numbers are 21, 23 and 25.