Let the two integers be m-1 & m, so the lesser integer is m-1, and the greater integer is m:
(m-1) + 3*m = 43 --> 4*m -1 = 43 --> 4*m = 44, m = 11 & m-1 = 10, so 10 & 11
Check: 11 times 3 is 33, add 10 is 43
They are 5 and 6 because 5+(2*6) = 17
Let the smaller be n, then the larger is n+1; and: n + 4(n+1) = 59 → n + 4n + 4 = 59 → 5n = 55 → n = 11 → the two consecutive integers are 11 and 12.
A quotient of integers is the result of dividing one integer by another. When dividing two integers, the result may be a whole number if the division is exact, or a decimal/fraction if there is a remainder. For example, when dividing 10 by 2, the quotient is 5, which is also an integer.
The question refers to two factors of 132 that are consecutive integers. The answer is 11 & 12, as 11 x 12 = 132 This can be solved as follows. Let n be the smaller of the two numbers then (n + 1) is the other number. n(n + 1) = 132 n2 + n = 132 n2 + n - 132 = 0............which can be factored as (n + 12 )(n - 11) = 0 As we are only concerned with the positive integer result then this occurs when n - 11 = 0, that is when n = 11, thus (n + 1) = 12. NOTE : Integers can also be negative and this applies to the other solution when n + 12 = 0, so n = -12 and consequently (n + 1) = -11 giving the result that -12 x -11 = 132.
Let x represent the smaller of the two integers.Since integers are also members of the set of whole numbers, then the next largest consecutive integer must be (x+1)Translating the question into a mathematical equation:x + 2(x + 1) = 26x + 2x + 2 = 26add the "ex's"3x + 2 = 26subtract 2 from both sides of the equation3x = 24divide both sides of the equation by 3, to solve for xx = 8, and x + 1 = 9the solution is 8 and 9 [8 + (2*9)] = 26;; [8 + 18] = 26
The integers are the sqare root of 121 and 81 to the 2/4 power.
There is not such set of integers.
3(n + 2) = n - 10 3n + 6 = n - 10 2n + 6 = -10 2n = -16 n = -8 -8 and -6
There is no set of two consecutive integers having a product of 14. Product means the result of multiplication.
12 and 13.
The integers are 10 and 11.
The sum of any three consecutive odd integers is going to give an odd result. It is impossible for the sum of an odd number of odd integers to equal an even number.
This is quickly solved by trial-and-error. Try multiplying two consecutive integers. If the result it too high, try again with lower numbers. If the result is too low, try again with higher numbers.
This would be impossible as, seeing as one of the integers would be odd, the result would be an odd number.
That's not possible. Adding two odd integers, the result will always be even.
55 and 57
Let's represent the two consecutive integers as x and x + 1. The sum of these two integers is x + (x + 1) = 2x + 1. We are given that this sum equals -30.03, so 2x + 1 = -30.03. Solving for x, we get x = (-30.03 - 1) / 2 = -31.03 / 2 = -15.515. Therefore, the two consecutive integers are -15 and -14.