The simplest way is to solve 4n + 6 = 138 for the smallest of them.
The integers are 44, 46 and 48.
Do-it-in-your-head method: The middle number MUST be a third of the total so 138 divided by 3 is 46, making your three numbers 44, 46 and 48. Shazam!
The numbers are 45, 46 and 47.
The sets are 277, 279 and 136, 138, 140, 142.
138. What is required is the largest number n such that: 285 = jn + 9 1249 = kn + 7 So subtract the required remainders and then find the hcf of the results: 285 - 9 = 276 1249 - 7 = 1242 Find hcf of 276 and 1242: 1242 / 276 = 4 r 138 276 / 138 = 2 r 0 hcf of 276 and 1242 is 138. Thus 138 is the largest number to divide 285 with a remainder of 9 and divides 1249 with a remainder of 7.
33, 34, 35, 36
Divide the sum of the three consecutive integers by 3: 138/3 = 46. The smallest of these integers will be one less than 46 and the largest will be one more than 46, so the three consecutive integers will be 45, 46, and 47.
138 is, itself, an integer. It is impossible for any integer to lie between two consecutive integers.
45, 46 and 47.
The integers are 44, 46 and 48.
136 + 138+ 140
Let the first lowest number be x, then x + x + 1 + x + 2 + x + 3 = 138 4x + 6 = 138 4x = 138 - 6 4x = 132 x = 33 and the number are 33, 34, 35, and 36, add them and you will get 138
Do-it-in-your-head method: The middle number MUST be a third of the total so 138 divided by 3 is 46, making your three numbers 44, 46 and 48. Shazam!
I assume you mean what are the two consecutive integers. Algebraically; X = integers.X + (X + 1) = 2752X + 1 = 2752X = 274X = 137============solution setorSince the consecutive integers differ by 1, then the first number is 274/2 = 137, so the second one is 1 more, 138.
The numbers are 45, 46 and 47.
The three consecutive even numbers with a sum of 138 are 44, 46 and 48.
The equation to solve this is... x + (x + 2) + (x + 4) = 138 x + x + x + 2 + 4 = 138 3x + 6 = 138 3x = 132 x = 44 x = 44 x + 2 = 46 x + 4 = 48 So, the three consecutive even numbers that add up to 138 are... 44, 46, & 48