Properties of Integers - GMAT Quantitative

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Question

A whole number has four digits. True or false: The integer is divisible by 5.

Statement 1: The last digit is 5.

Statement 2: The sum of the digits is 25.

Answer

Assume Statement 1 alone. A necessary and sufficient condition for a whole number to be divisible by 5 is that its last digit be either 0 or 5. By Statement 1, the number meets this requirement, so it is divisible by 5. Statement 2, which gives the digit sum, is therefore, irrelevant.

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Question

A whole number has four digits. True or false: The integer is divisible by 9.

Statement 1: The sum of the digits is 26.

Statement 2: The last two digits are 98.

Answer

A necessary and sufficient condition for a whole number to be divisible by 9 is that the sum of its digits must be divisible by 9. Statement 1 provides proof that this condition is not met, since 26 is not a multiple of 9. Statement 2 provides irrelevant information.

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Question

, , and are integers. Is odd?

(1) is a prime number and

(2) is odd

Answer

Statement (1) tells us that is an odd number. Since we don’t know if and are odd numbers, we cannot decide the sign of .

Statement (2) tells us that is an even number, since . When we do the multiplication, the product will be even if one or more of the integers are even. So with statement (2), we know for sure that is even.

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Question

Is $\dpi{100} \small x+y$ odd?

(1) $\dpi{100} \small x$ is odd

(2) $\dpi{100} \small x-y$ is even

Answer

For statement (1), we only know that $\dpi{100} \small x$ is odd but we have no idea about $\dpi{100} \small y$ . If $\dpi{100} \small y$ is odd, then $\dpi{100} \small x+y$ is even. If $\dpi{100} \small y$ is even, then $\dpi{100} \small x+y$ is odd. Therefore we have no clear answer to the question using this condition. For statement (2), since $\dpi{100} \small x-y$ is even, we know that $\dpi{100} \small x$ and $\dpi{100} \small y$ are either both odd or both even, therefore we know for sure that $\dpi{100} \small x+y$ is even and the answer to this question is “no”.

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Question

If is an integer and $\dpi{100} \small 3<k<7$ , what is the value of ?

(1) $\dpi{100} \small k$ is a factor of 20.

(2) $\dpi{100} \small k$ is a factor of 24.

Answer

From statement (1), we know that the possible value of $\dpi{100} \small k$ would be 4 and 5. From statement (2), we know that the possible value of $\dpi{100} \small k$ would be 4 and 6. Putting the two statements together, we know that only $\dpi{100} \small k=4$ satisfies both conditions. Therefore both statements together are sufficient.

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Question

What is the remainder when the two digit, positive integer is divided by 3?

(1) The sum of the digits of is 3.

(2) The difference of the digits is 3.

Answer

For statement (1), there are 3 possible two digit positive integers: 30, 12, 21. The remainder when these numbers are divided by 3 is 0. Therefore, statement (1) is sufficient. For statement (2), there are lots of different combinations of integers that have different remainders when divided by 3. For example, the remainder of 30 is 0, but the remainder of 14 is 2. Therefore, statement (2) is insufficient.

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Question

If is a positive integer, is divisible by 6?

1. The sum of the digits of is divisible by 6

2. is even

Answer

Statement 1: Numbers whose digits sum to a number divisible by 3 are divisible by 3, but the same does not apply to sums of 6. This indicates that is divisble by 3 but is not sufficient at proving is divisible by 6.

Statement 2: Though all multiples of 6 are even, not all even numbers are multiples of 6.

Together: The fact that is a multiple of 3 and even is sufficient evidence for the conclusion that x is divisible by 6.

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Question

Is positive, negative, or zero?

$\small x^{3}$ is positive.

$\small x^{4}$ is positive.

Answer

raised to an odd power must have the same sign as , so, if $\small x^{3}$ is positive, then is also positive. But either a positive number or a negative number raised to an even power must be positive. Therefore, $\small \small x^{4}$ being positive is inconclusive.

Therefore, the correct choice is that Statement 1, but not Statement 2, is sufficient.

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Question

The greatest common factor of 32 and a number is 16. What is ?

3 is also a factor of .

5 is also a factor of .

Answer

That cannot be determined, even if both statements are known to be true, can be proved by demonstrating two examples of that fit these conditions. We can do this by comparing the prime factorizations of 32 and .

Example:

To find :

$32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

$240 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5$

$GCF (32, 240) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$

Example:

To find :

$32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

$720 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5$

$GCF (32, 720) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$

In each case, 3 and 5 are factors of , and in each case, $\small GCF (32, B) = 16$ .

The answer is that both statements together are insufficient.

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Question

What is the area of a rhombus in square inches?

One of its angles measures $30^\circ$

One of its sides measures 10 inches

Answer

As is true of any other parallelogram, the area of the rhombus is the base multiplied by the height. The common sidelength alone can be used to determine the base, but without the angles, the height cannot be determined. Using trigonometry, the angle can be used to determine the height relative to the base, but without the base, the height is unknown.

If we know both of the given statements, then part of one base, an altitude from an endpoint of the opposite base, and one adjacent side form a 30-60-90 triangle. The hypotenuse of that triangle is 10 inches, and the altitude is half that, or 5 inches. This makes the area 50 square inches.

The answer is that both statements together are sufficient to answer the question, but neither statement alone.

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Question

Data Sufficiency Question

Out of 100 students, 60 took French and 25 took German. How many students took neither?

1. 15 students took Spanish

2. 7 students took both French and German

Answer

Statement 1 does not tell us anything about the number of students taking French or German. The information from statement 2 is sufficient, if 60 took French, 25 took German, and 7 took both, we can calculate the number that took neither.

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Question

This five-digit number has two digits missing:

______

If both blanks are two be filled with the same digit, what is that digit?

Statement 1: The number is divisible by 5.

Statement 2: The number is not divisible by 3.

Answer

There are two different choices we can make for the common digit that would make both statements true, 0 and 5. This makes the last digit 0 or 5, making Statement 1 true, Also, this makes the digit sum either

or

Either way, the digit sum is not a multiple of 3, and the number itself is not a multiple of 3.

Therefore, the two statements together do not provide enought information to answer the question definitively.

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Question

Does the integer have at least 4 different positive prime factors?

(1) $\frac{x}{14}$ is an integer.

(2) $\frac{x}{15}$ is an integer.

Answer

(1) This statement tells us that 2 and 7 are prime factors of . This information is not sufficient.

(2) This statement tells us that 3 and 5 are prime factors of . this information is not sufficient.

Considering both (1) and (2), we have four positive prime factors of between them.

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Question

If and are both integers, evaluate .

Statement 1: .

Statement 2: and are both prime integers.

Answer

There are infinitely many primes, and several integers between 13 and 23, so knowing just one of these statements is not enough. But only two integers in the stated range - 17 and 19 - are prime, so knowing both statements tells you that and are 17 and 19, respectively. Subsequently, you can add them to get 36.

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Question

If a positive integer is divided by 2, what is the remainder?

Statement 1: If $N^{2 }$ is divided by 2, the remainder is 1.

Statement 2: If is divided by 4, the remainder is 3.

Answer

The question is the same as asking whether is odd or even.

Statement 1 says that the square of is odd. If we know this, then we know that is odd, since the square of an even number is even.

Statement 2 says that is 3 greater than a multiple of 4; this makes odd.

Therefore, either statement alone tells us that is an odd number.

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Question

You are given two positive integers, and . Is their product divisible by 7?

Statement 1: is divisible by 7.

Statement 2: is divisible by 7.

Answer

If we only know that is divisible by 7, then may or may not be a multiple of 7.

Example 1: If , then , which is not divisible by 7.

Example 2: If , then , which is divisible by 7.

A similar argument shows that if we only know that is divisible by 7, then may or may not be a multiple of 7.

Suppose we know both.

If we know that is divisible by 7, then, when is divided by 7, the remainder is 1, and we can write for some integer $Q_{1}$ :

$M = 7Q_{1} + 1$

If we know that is divisible by 7, then, when is divided by 7, the remainder is 6, and we can write for some integer $Q_{2}$ :

$N = 7Q_{2} + 6$

Then

$M N = \left (7Q_{1} + 1 \right) \left ( 7Q_{2} + 6 \right )$

$= \7Q_{1} \left ( 7Q_{2} + 6 \right ) + 1 \left ( 7Q_{2} + 6 \right )$

$= \7Q_{1} \left ( 7Q_{2} + 6 \right ) + 7Q_{2} + 6$

$= 7( 7Q_{1}Q_{2} + 6Q_{1} \right + Q_{2} )+ 6$

for some integer quotient

So, when is divided by 7, the remainder is 6, and is not a multiple of 7. This makes the two statements together sufficient information.

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Question

Let A-B = C. If B is an integer not equal to 0, is C an integer?

1. A/B is an integer

2. A*B is an integer

Answer

Let's first look at what the question is asking for. They want us to determine if C is an integer. Since we know that B is an integer, C will be an integer only if A is an integer. If A not an integer, C will not be an integer. So from statements 1 and 2, we want to see if we can prove if A is definitely an integer.

First, let's try statement 2, which says that AB is an integer. Let's see what happens if B is 2. If B is 2, then 2.52 = 5 is an integer, but 22 = 4 is also an integer. So A in this case could be an integer or a not. So statement 2 alone is not sufficient to get our answer.

Now, let's go back to statement 1, which says A/B is an integer. Let's name another variable x. Let x = A/B. If x=A/B we can rewrite this as A = xB. But since we know that x is an integer (given in the statement) and B is a non-zero integer (given in the question), A is therefore an integer! (The product of two integers is ALWAYS an integer.) This statement alone is enough to prove that A is an integer.

Since, as discussed before, A is definitely an integer, and B is an integer. An integer minus an integer will always be an integer. Therefore C is an integer, and statement 1 is sufficient to answer the question.

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Question

What is the last digit of a positive integer ?

Statement 1: The last digit of $N^{2}$ is 1.

Statement 2: The last digit of $N^{3}$ is 1.

Answer

If the last digit of $N^{2}$ is 1, then the last digit of is either 1 or 9.

If the last digit of $N^{3}$ is 1, however, the last digit of must be 1.

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Question

This six-digit number has two digits missing:

______

If the blanks are to be filled with the same digit, what is that digit?

Statement 1: The number is divisible by 4.

Statement 2: The number is divisible by 3.

Answer

If the number is divisible by 4, then the last two digits make a number that is divisible by 4; this allows that digit to be 0, 4, or 8.

If the number is divisible by 3, then its digit sum is divisible by 3. This allows the comon digit to be 1, 4, or 7:

Neither statement alone narrows the common digit to one possibility, but if both statements are true, the only possibility becomes 4.

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Question

How many negative numbers are in the set $\left { A,B,C,D,E,F\right }$ ?

Statement 1:

Statement 2:

Answer

Statement 1 tells you that and are of unlike sign, that and are of unlike sign, and that and are of unlike sign; that is, exactly one number from each pair is negative. Therefore, there are three negative numbers.

Statement 2 tells you that of , , and , there can be either exactly zero or two negative numbers; and that of , , and , there can be exactly one or three negative numbers. This means that the number of negative numbers among the six can be as few as one or as many as five.

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