If x is a positive integer such that 3x + 1 is a perfect square and 4x + 1 is a perfect cube, what is the value of x?
To find the value of x, we need to use a bit of algebraic manipulation.
Let’s assume that 3x + 1 is equal to a perfect square, say y^2. This means that:
3x + 1 = y^2
Now we can rearrange this equation to get:
3x = y^2 – 1
3x = (y + 1)(y – 1)
Since x is a positive integer, y must also be an odd positive integer (because y^2 is odd and 3 is odd). This means that y + 1 and y – 1 are consecutive even integers.
The only way for 3x to be the product of two consecutive even integers is if x is itself an even integer. So let’s assume that x = 2n, where n is some positive integer.
If x = 2n, then we can rewrite the second equation as:
4(2n) + 1 = 8n + 1 = z^3
Where z is some positive integer. Now we can solve for n by using trial and error or by observing that 8n+1 is one less than a multiple of 9, so it must be a perfect cube only if n=1. Therefore:
n = 1, x = 2n = 2, y^2 = 3x + 1 = 7, and z^3 = 4x + 1 = 9.
So the value of x is 2.
"Ready for a challenging SAT Writing question? Can you correctly identify the error in this sentence: 'The team of volunteers was divided in their opinions about the best way to tackle the project.'
The error in the sentence is the use of “in” after “divided.” The correct preposition to use would be “on,” so the sentence should read: “The team of volunteers was divided on their opinions about the best way to tackle the project.”