59,61,63,65
It is 59
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.
Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.
The simplest way is to solve 4n + 6 = 138 for the smallest of them.
14, 15, 16 and 17.
There are four consecutive odd integers: 81, 83, 85 and 87.
The smallest is -1
The integers are 47, 49, 51, and 53. 47 * 53 = 2491
No.
The let statement is: let the smallest of the three integers be x.
If n is the smallest of the four integers, their sum is 4n+6.
The sum is four.
The integers would begin with 10.
9,11,13
17 , 19 , 21 and 23 are the odd integers whose sum is 80 and the least integer is 17
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.
Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.
The simplest way is to solve 4n + 6 = 138 for the smallest of them.