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The smallest consecutive three numbers that total 120 are... 39, 40 & 41.

Q: The sum of three consecutive number is 120 what is the smallest of the three numbers?

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Divide the sum of the three consecutive numbers by 3: 75/3=25. The smallest of these numbers will be one less than 25 and the largest will be one more than 25, so the three consecutive numbers will be 24, 25, and 26.

consecutive means each number in the sequence is 1 more than the previous # Therefore (Algebraically) (x) + (x+1) + (x+2) =72 3x + 3 = 72 x = 23, x+1 =24 x+2 = 25 The three numbers are: 23 24 and 25. The smallest is 23.

That isn't possible. The three consecutive number are assumed to be integers; the sum of three consecutive integers is always a multiple of 3 (try it out).

Let be x the smallest number. So, the second number will be x + 1, and the third one will be x + 2. So, x + (x + 1) + (x + 2) = 18 3x + 3 = 18 subtract 3 to both sides; 3x = 15 divide by 3 to both sides; x = 5 Thus, the smallest number is 5. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Couldn't you also just use the factors of 18: 1, 2, 3, 6, 9, 18 Prime Numbers: 3 x 3 x 2 Smallest Number is then 2. You are right, 2 is the smallest prime number in the prime factorization of 18. But, the question is that three consecutive numbers are added up to 18, and 3 x 3 x 2 is a multiplication. However, 2 and 3 are consecutive so the third number will be 4 not 3. I didn't catch the consecutive part. Good call and thanks for the input.

The answer to that question is 49,50,51

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55

91 are.

55

The sum is always 3 more than (triple the smallest number).

25 (25+27+29=81)

55