Let the two consecutive integers be n and n+1.
Then, n + (n + 1) < 55
2n + 1 < 55
2n < 55 - 1 : 2n < 54
n < 27
The Inequality Statement can therefore be modified to show that for two consecutive integers to be less than 55 then the smaller integer must be less than 27.
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Two consecutive integers will be 0.5 more and 0.5 less than the quotient of their sum divided by 2. The given sum of the two consecutive integers divided by 2 is -3471.5, so the two consecutive integers are -3472 and -3471.
The pair of consecutive integers which add up to 55 are 27 and 28. Therefore, any consecutive pair of numbers below 27 and 28 add up to a total less than 55.
The answer is, obviously, 9.
Divide the sum of the three consecutive odd integers by 3: -273 /3 = -91. The smallest of these integers will be two less than -91 and the largest will be two more than -91, so the three consecutive odd integers will be -89, -91, and -93.
The only proper factor of 51 that is less than 1/3 of 51 is 3. Divide the sum of three consecutive odd integers by 3: 51/3 = 17. The smallest of these integers will be two less than 17 and the largest will be two more than 17, so the three consecutive odd integers having a sum of 51 are 15, 17, and 19.