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Q: What are the three consecutive integers whose sum is 123?

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They are 40, 41 and 42.

The numbers are 39, 41 and 43.

They are: -39 -41 -43 = -123

123/3 = 41 so integers are 39, 41 and 43

There are no consecutive even integers totaling 123 but there are two combinations of odd/even integers. They are 39, 41 and 43 as well as 40, 41 and 42.

123/3 - 2, 123/3 and 123/3 + 2 ie 39, 41 and 43.

There is a set of two consecutive integers that have a sum of 123; one odd and one even. They are 61 and 62.

between which two consecutive integers does the square root lie 123

x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43

39,41,43! Ya

The numbers are 121, 122 and 123.

61 & 62

You are told that:a + b + c = 123You are also told that:b = a + 1c = a + 2Therefore:a + (a + 1) + (a + 2) = 1233a + 3 = 1233a = 120a = 40Therefore, the numbers are 40, 41 and 42.

62 & 61

For three consecutive odd integers to have the sum 123, the equation N + N+2 + N+4 = 123 must be solvable with N being an odd integer.N + N+2 + N+4 = 1233N + 6 = 1233N = 117N = 39Since 39 is an odd integer, the solution is 39, 41, and 43.If N were even, or if N resolved to a non-integer, then the problem would have been incorrectly stated and unsolvable.

The numbers are 40, 41 and 42.

The first ten consecutive composite positive integers are: 114 115 116 117 118 119 120 121 122 123

The numbers are 122, 123, 124 as well as 121, 123, 125.

39, 41 and 43. or 123/3 -2, 123/3 and 123/3 +2

The numbers are 39, 41 and 43.

"Average" number must be 369/3 ie 123 so numbers are 122, 123 and 124

The sum of three consecutive odd integers, starting with N, is expressed as N + (N+2) + (N+4). If that sum is 363, then you have 3N + 6 = 363. Solve for N and you have 119.Since N (119) is odd, the answer and question are valid. (If N had been even, the question would have been invalid, and the answer would have been meaningless. This test is simply a sanity check.) The three numbers are 119, 121, and 123.

7272 * * * * * WRONG! Sum from 23 to 123 = (sum from 1 to 123) minus (sum from 1 to 22) = 123*124/2 - 22*23/2 = 7626 - 253 = 7373. NOT 7272!

39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43

Algebraically. X = integer. X + (X + 1) + (X + 2) = - 123 3X + 3 = - 123 3X = - 126 X = - 42 X + 1 = - 41 X + 2 = - 40 - 42 + - 42 + - 40 = -123 =======