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The sum of 4 consecutive odd integers if 112. What is the greatest of the four integers?
Let's denote the smallest odd integer as x. The next three consecutive odd integers would then be x+2, x+4, and x+6. The sum of these four integers is x + (x+2) + (x+4) + (x+6) = 4x + 12. Given that the sum is 112, we have 4x + 12 = 112. Solving for x, we get x = 25. Therefore, the greatest of the four integers is x + 6 = 25 + 6 = 31.
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What are 2 consecutive integers that have the sum of 112?
The numbers are 55, and 57. Two consecutive integers have an odd sum.
What are 2 consecutive integers whose sum is 225?
112 and 113
The product of two consecutive integers is 121 Find the numbers?
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.
What is the sum of the sequence 33 34 35 36 ... 112?
Adding all integers from 33 to 112 inclusive gives you 5800.
Three consecutive numbers that have a product of 336?
111 + 112 + 113 = 336
What are 2 consecutive integers that have the sum of 112?
The numbers are 55, and 57. Two consecutive integers have an odd sum.
Find four consecutive odd integers if the product of the two smaller integers is 112 less than the product of the two large integers?
The numbers are 11, 13, 15 and 17.
What are 2 consecutive integers whose sum is 225?
112 and 113
If the sum of the squares of two consecutive odd positive integers is 202 what are the integers?
112+92 is 202.
3 consecutive integers whose sum is 333?
110 + 111 + 112 = 333
What is the larger of two consecutive even integers if their sum is 226?
The larger is 114, and the other is 112.
What is greatest common factor of 112?
There has to be 2 or more integers to find the gcf
The sum of the squares of two consecutive odd integers is 202 find the integers?
9 and 11 92 = 81 112 = 121 81+121 = 202
What is four odd integers if the product of the two smaller integers is 112 less than the product of the two larger integers?
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!
Find four consecutive odd integers if the product of the two smaller integers is 112 less than the product of the two larger integers?
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.
What are the four consecutive multiples of 7 whose sum is 406?
91, 98, 105, 112
What are 4 consecutive odd integers where the product of the two smaller integers is 112 less than the product of the two larger integers?
11, 13, 15 and 17(143 is 112 less than 255)The equation is (A)(A+2) = (A+4)(A+6) - 112A2 + 2A = A2 +10A +24 -1122A = 10A - 888A = 88A = 112,5,6
