Let x equal the lowest number of the three.
Therefore:
x + x + 1 + x + 2 = 228
3x + 3 = 228
3x = 228 - 3
3x = 225
x = 75
Thus the 3 numbers are 75, 76 & 77.
They are: 75+76+77 = 228
This has no answer. The sum of three odd integers is always odd.
Let "a" be the first of the four numbers. Then we are told:a + (a + 1) + (a + 2) + (a + 3) = 234∴ 4a + 6 = 234∴ 4a = 228∴ a = 57So the numbers are 57, 58, 59 and 60
Let the first one be x, the second be x + 1, and the third one is x + 2. So we have,x + x+1 + x +2 = 2283x + 3 = 228 subtract 3 to both sides3x = 225 divide by 3 to both sidesx = 75x + 1 = 75 + 1 = 76x + 2 = 75 + 2 = 77Thus, the three consecutive integers are 75, 76, and 77.
228 divides by these numbers: 1, 2, 3, 4, 6, 12, 19, 38, 57, 76, 114, 228.
The numbers are 226, 227 and 228.
The three numbers added together will equal 228. The average of the three numbers will be 228 / 3 = 76. If you take one even integer from either side of 76, you get 74 and 78.Therefore, the three consecutive even integers which equal 228 are 74, 76 and 78.
They are: 75+76+77 = 228
The middle number must be the average, so 681/3 = 227 and the other two numbers are 226 and 228.
The sum of three consecutive multiples of 6 is 666, the multiples are 216, 222 and 228.
This has no answer. The sum of three odd integers is always odd.
The sum of any two consecutive integers must be an odd number. So this question is incorrect.
This would be impossible as, seeing as one of the integers would be odd, the result would be an odd number.
Let "a" be the first of the four numbers. Then we are told:a + (a + 1) + (a + 2) + (a + 3) = 234∴ 4a + 6 = 234∴ 4a = 228∴ a = 57So the numbers are 57, 58, 59 and 60
If two numbers have a sum of 31 and a product of 228, then the two numbers are 12 and 19. Adding them will give you 31, and multiplying them gives you 228.
Let the second of the three consecutive multiples of 6 be 6n Then the first is 6n - 6 and the last is 6n + 6; and: (6n - 6) + 6n + (6n + 6) = 666 → 18n = 666 → n = 37 → the consecutive multiples of 6 which sum to 666 are 216, 222, 228
Let the first one be x, the second be x + 1, and the third one is x + 2. So we have,x + x+1 + x +2 = 2283x + 3 = 228 subtract 3 to both sides3x = 225 divide by 3 to both sidesx = 75x + 1 = 75 + 1 = 76x + 2 = 75 + 2 = 77Thus, the three consecutive integers are 75, 76, and 77.