Let the two consecutive integers be x and x+1. The product of these two integers is x(x+1). Setting this equal to 132 gives us the equation x(x+1) = 132. By expanding the left side of the equation, we get x^2 + x = 132. Rearranging terms, we have x^2 + x - 132 = 0. This is a quadratic equation that can be factored as (x+12)(x-11) = 0. Therefore, the two consecutive integers are 11 and 12, and their product is 132.
42, 44, 46.
Divide the sum of the three consecutive odd integers by 3: 402/3 = 134. The smallest of these integers will be two less than 134 and the largest will be two more than 134, so the three consecutive odd integers will be 132, 134, and 136.
The numbers are -134, -132 and -130.
4n - 132 = 88 + 132 +132 4n = 220 / / 4 4 n = 55
22 x 6 = 132
The numbers are 65 and 67.
43, 44, 45
132/3 = 44 43, 44, 45
3 consecutive even integers whose sum is 396 are 130, 132 and 134.
11 & 12 are the integers whose product is 132.
42, 44, 46.
Divide the sum of the three consecutive odd integers by 3: 402/3 = 134. The smallest of these integers will be two less than 134 and the largest will be two more than 134, so the three consecutive odd integers will be 132, 134, and 136.
64 65 66 67
The numbers are -134, -132 and -130.
11 and 12
X is the 1st integerX+1 would be the 2ndX+2 would be the 3rdX+(X+1)+(X+2) = 1323X+3=1323X+3-3=132-33X=129X=129/3 = 43So the consecutive integers are 43, 44, 45
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.