The first year after 1999 which is the product of three consecutive integers is:
2004
2040
2046
2052
2184
Please show working out and choose either one of these years
Let the first integer is (x-1) then the 2nd is x, and the 3rd is (x+1)
The product is (x-1)x(x+1)=(x^2-1)x=
x^3-x. If x=13 then x^3-x=13^3-13=
2197-13=2184. (12^3=1728)
2184
To find the first year after 1999 that is the product of three consecutive integers, we can use a systematic approach by trying different consecutive integer sets.
Let's start with the year 2004 and check if it can be expressed as the product of three consecutive integers:
2004 = x(x+1)(x+2)
Expanding this equation, we get:
2004 = x^3 + 3x^2 + 2x
However, there is no integer value of x that satisfies this equation. Therefore, 2004 is not the correct answer.
Let's move on to the next option, 2040:
2040 = x(x+1)(x+2)
Expanding this equation, we get:
2040 = x^3 + 3x^2 + 2x
Again, there is no integer value of x that satisfies this equation. Therefore, 2040 is not the correct answer.
Next, we have 2046:
2046 = x(x+1)(x+2)
Expanding this equation, we get:
2046 = x^3 + 3x^2 + 2x
Similarly, there is no integer value of x that satisfies this equation. Therefore, 2046 is not the correct answer.
Moving on to the next option, 2052:
2052 = x(x+1)(x+2)
Expanding this equation, we get:
2052 = x^3 + 3x^2 + 2x
Again, there is no integer value of x that satisfies this equation. Therefore, 2052 is not the correct answer.
Lastly, we have 2184:
2184 = x(x+1)(x+2)
Expanding this equation, we get:
2184 = x^3 + 3x^2 + 2x
Surprisingly, this equation does have an integer solution. By trial and error, we find that x = 12 satisfies the equation. Therefore, the year 2184 is the first year after 1999 that is the product of three consecutive integers.
So, the correct answer is 2184.
To find the first year after 1999 which is the product of three consecutive integers, we can start by listing out the consecutive integers and calculating their products.
Let's assume the three consecutive integers are:
n, n+1, n+2
We can form an equation by multiplying these three integers:
n * (n+1) * (n+2) = ?
Expanding the equation gives us:
n^3 + 3n^2 + 2n = ?
Now, we need to find the first year after 1999 that satisfies this equation.
Let's substitute different values of n and check if the resulting product is greater than 1999.
For n = 7:
7^3 + 3(7^2) + 2(7) = 658
Since 658 is less than 1999, n = 7 is not the solution.
For n = 8:
8^3 + 3(8^2) + 2(8) = 832
Again, 832 is less than 1999, so n = 8 is not the solution.
Continue this process until we find the correct value. Note that we can skip even values for n as they would result in an even product.
For n = 9:
9^3 + 3(9^2) + 2(9) = 1260
1260 is greater than 1999, which means we found our solution. The year after 1999, which is the product of three consecutive integers, is 2000.
Therefore, the correct answer is not among the options provided.