Roots as integers
We want to find out if
is an integer.
Statement 1 tells us that
is an integer. We can use the rule
, like this:
So we know that 6
is an integer. So since 6 is an integer, we know that
is also an integer.
Thus, statement 1 is sufficient.
Now let's look at statement 2. It tells us that
is an integer. This means that 3x + 4 must be a perfect square, like 4, 9,16, or 25.
So let's test some values. Let's say that 3x + 4 = 16.
3x + 4 = 16
3x = 12
x = 4
In this case,
= 2, so x is an integer.
Now let's test another possibility:
3x + 4 = 25
3x = 21
x = 7
, which is not an integer.
Thus, with statement 2,
may or may not be an integer. So statement 2 by itself is insufficient, and the answer is A.
Frequently Asked Questions
Q: In the first statement: if x is
, then the √x would be
by 6 results in an integer. So the √x (which is in this case
) does not have to be an integer. Am I wrong?
A: Notice the setup here: "If x is a positive integer..." so we're only concerned about the times when x is a positive integer. In your example, you use x =
but that's not allowed by the constraints of the question. If you're confused about the wording, we can rephrase it so it's a bit clearer:
x is a positive integer. Is √(x) an integer?
The wording in these DS questions can be really tricky, so make sure you're always paying close attention to your constraints!
Q: I still cannot understand why statement 1 is sufficient. Why MUST √x be an integer? Just to demonstrate, what I mean: 6 × 1.5 = 9. In this case √x can be equal to 1.5. Why can 1.5, or any other such decimal, not be equal to the square root of an integer is what I want to know?
A: Okay, here is the proof! :)
Remember only the square roots of perfect squares (i.e. 9, 36, 81, etc.) are integers. Square roots of any other number are irrational numbers, meaning they are a number followed by decimals that continue until infinity. Let's say we have a non-irrational number that is not an integer.
So it can be expressed as a fraction in lowest terms:
a and b are integers, and since the fraction is in lowest terms, a and b share no common factors except 1.
Note that any non-irrational decimal can be expressed as a fraction in lowest terms this way.
So for 1.5,
, and we get
be an integer?
No, it can't.
We already said that a and b have no common factors besides 1. So if they have no common factors, then,
must also be a fraction because b² cannot divide evenly into a².
So it is impossible to square a fraction in lowest terms
to get an integer.
which is not an integer.
The only ways to square a number to get an integer is:
1) if the original number itself is an integer
2) if the original number is the square root of an integer such as √2 or √5 But a square root of a non-perfect square is an irrational number. We cannot multiply an irrational number by an integer such as 6 to get another integer. We will get another irrational number.
Therefore, "√x" can only be EITHER an integer — if x is a perfect square — or an irrational number if x is NOT a perfect square.
From statement (1), we are told √(36x) is an integer, so , √36 × √x is an integer, which means that 6√x is also an integer. Thus it is only possible for √x to be an integer too. √x will never end in a clean "0.5," or any other decimal that will turn 6 into an integer.
So, remember this rule: All square-roots of integers that don't come out evenly are irrationals. If x is an integer, there are only two possibilities:
(1) x is a perfect square, so √x is an integer or
(2) √x is an irrational number.
If the later is the case, then 6 times an irrational number will be another irrational number, so 6√x couldn't possibly be an integer. It is absolutely impossible to take the square root of an integer and have that result in a rational decimal, such as something with a 2 or 3 or 6 in the denominator.
Fun fact: it was Mr. Pythagoras who first proved that √2 is an irrational number. This is a very deep idea in mathematics.
Watch the lessons below for more detailed explanations of the concepts tested in this question. And don't worry, you'll be able to return to this answer from the lesson page.