Suppose the middle integer is 2a. Then the smallest is 2a-2 and the biggest is 2a+2.
4 times the smallest is 8a-8
So largest subtracted from the smallest is (8a-8) - (2a+2) = 6a-10
So, 6a-10 = 2*2a = 4a so that 2a = 10
So the integers are 8, 10 and 12.
The smallest is 121.
consecutive integers
The smallest six consecutive composite integers are:90, 91, 92, 93, 94, 95.(And 96 is also composite, for a run of seven consecutive.)Is that what you were asking ?
The integers are 47, 49, 51, and 53. 47 * 53 = 2491
1 is the smallest positive integer. But if you include negative integers, there is no smallest.
The smallest is 121.
59,61,63,65 It is 59
consecutive integers
If the largest integer is subtracted from four times the smallest, the result is 4 more than twice the middle integer. Let the smallest integer be x, then the others are x + 2 and x+ 4. Therefore 4x - (x + 4) = 2 (x + 2 ) + 4 Expanding, we get 4x -x -4 = 2x + 4 + 4 Gathering terms: x = 12 Thus the three integers are 12, 14 and 16.
For x, which is the largest integer of nconsecutive positive integers of which the smallest is m:x = m + n - 1
"Consecutive" integers are integers that have no other integer between them.
The sum of three consecutive integers is -72
The smallest six consecutive composite integers are:90, 91, 92, 93, 94, 95.(And 96 is also composite, for a run of seven consecutive.)Is that what you were asking ?
The integers are 47, 49, 51, and 53. 47 * 53 = 2491
1 is the smallest positive integer. But if you include negative integers, there is no smallest.
Let x equal the smallest integer. The sum will be 3x + 6
The first integer is 17.