Numbers are n and n + 2, so product = n2 + 2n and sum = 2n + 2.
ie n2 + 2n + 1 = 6n + 6
or n2 - 4n - 5 = 0
Factors are (n + 1)(n - 5) making n = 5 and the integers 5 & 7.
Check: 5 x 7 = 35; 5 + 7 = 36. OK
There is also an answer of n = -1, making the integers -1 and 1, product being -1 and sum zero.
The let statement is: let the smallest of the three integers be x.
The product of three negative integers is negative. This is because multiplying two negative integers results in a positive integer, and then multiplying that positive integer by another negative integer yields a negative result. For example, if the integers are -2, -3, and -4, their product is -24.
the product of two consicutive multiple of three integers
The sum of three consecutive integers can be expressed using ( l ) as the middle integer. The three integers can be represented as ( (l-1) ), ( l ), and ( (l+1) ). Therefore, the sum is ( (l-1) + l + (l+1) = 3l ).
There is no set of three consecutive odd or even integers whose sum is negative 31.
The sum of three consecutive integers is -72
The integers are 5 and 7.
9240 is the product of the three consecutive integers 20, 21, and 22.
There must be three consecutive integers to guarantee that the product will be divisible by 6. For the "Product of three consecutive integers..." see the Related Question below.
10-11-12
51
This would be impossible - since the mean of the three integers would have to be an integer, and if you divide -56 by 3, you do not get an integer.
The let statement is: let the smallest of the three integers be x.
The product of three negative integers is negative. This is because multiplying two negative integers results in a positive integer, and then multiplying that positive integer by another negative integer yields a negative result. For example, if the integers are -2, -3, and -4, their product is -24.
the product of two consicutive multiple of three integers
270
Not sure what thress is. If three, then there is no answer since the sum (or product) of any three consecutive integers must be divisible by 3.