51
The integers are 5 and 7.
9,11,13
The second integer is 64 (128/2), so the first is 62 and the third is 66.OrLet the 1st integer be x and the 3rd integer be y.Since even integers differ by 2, then we have:y = x + 4x + y = 128x + x + 4 = 1282x = 124x = 62So the integers are 62, 64, and 66, and their sum is 192
No. There is no set of three consecutive even integers with a sum of 40.
It cannot be done. The basic rules of math. odd integer plus odd integer = even integer. odd integer plus even integer = odd integer. Always. odd integer plus odd integer plus odd integer = odd integer. Always.
The sum of three consecutive integers is -72
They are 14, 16 and 18.
The numbers are 14, 16 and 18.
This would be impossible - since the mean of the three integers would have to be an integer, and if you divide -56 by 3, you do not get an integer.
10-11-12
If the first of these consecutive integers is x, the second integer would be x + 1, and the third integer would be x + 2.Since the sum of the second and the third integer is 17, we can writex + 1 + x + 2 = 172x + 3 = 172x + 3 - 3 = 17 - 32x = 142x/2 = 14/2x = 7Thus, the consecutive integers are 7, 8, and 9.
The let statement is: let the smallest of the three integers be x.
There is no set of three consecutive odd or even integers whose sum is negative 31.
0.75*(2n+1) where n is an integer.
21, 22 and 23.
The integers are 53, 55 and 57.
9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70