112/4 = 28 so two odd numbers before 28 and two after.
25 + 27 + 29 + 31 = 112
so 31 is the greatest.
x + (x+ 2) + (x+4) + (x+6) = 112; x + 12=112; x = 25; x+6 = largest = 31
It is 31.
The numbers are 55, and 57. Two consecutive integers have an odd sum.
112 and 113
As there are two consecutive integers then one must be an even number and the other an odd number. If the numbers are y and y + 1 then if y is even, y + 1 is odd and if y is odd then y + 1 is even.The product of an even integer and an odd integer is always even.The question therefore has no answer.121 = 112 but this is not what the question has asked.
Adding all integers from 33 to 112 inclusive gives you 5800.
111 + 112 + 113 = 336
The numbers are 55, and 57. Two consecutive integers have an odd sum.
The numbers are 11, 13, 15 and 17.
112 and 113
112+92 is 202.
110 + 111 + 112 = 333
The larger is 114, and the other is 112.
The greatest common multiple of any set of integers is infinite.
There has to be 2 or more integers to find the gcf
9 and 11 92 = 81 112 = 121 81+121 = 202
We began by assuming that the 4 odd integers are consecutive, and everythingworked just fine. So, although there may well be more than one correct answer tothe question as it's stated, we have a hunch they're supposed to be consecutive,and that detail was accidentally left out of the question.Call the 4 consecutive odd integers (2x - 3), (2x - 1), (2x + 1), and (2x + 3).Product of the two larger ones = (4x2 + 8x + 3)Product of the two smaller ones = (4x2 - 8x + 3)Smaller product + 112 = larger product . . . . . (4x2 - 8x + 3) + 112 = (4x2 + 8x + 3)Eliminate parentheses . . . . . 4x2 - 8x + 115 = 4x2 + 8x + 3Subtract (4x2 + 3) from each side . . . . . -8x + 112 = 8xAdd 8x to each side . . . . . 112 = 16xDivide each side by 16 . . . . . x = 7The consecutive odd integers are: 11, 13, 15, and 17.Check:11 x 13 = 14315 x 17 = 255255 - 143 = 112 Yippee!
11,13,15,17solve this: n(n-2)=(n-4)(n-6)+112n is the highest of the four.solving, you get:n^2 -2n=n^2-10n+1368n=136n=136/8n=17so, the four are: 11, 13, 15, 17.
91, 98, 105, 112