Properties of Integers - GMAT Quantitative

From statement (1), we know that the possible value of would be 4 and 5. From statement (2), we know that the possible value of would be 4 and 6. Putting the two statements together, we know that only satisfies both conditions. Therefore both statements together are sufficient.

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.

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.

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.

raised to an odd power must have the same sign as , so, if is positive, then is also positive. But either a positive number or a negative number raised to an even power must be positive. Therefore, being positive is inconclusive.

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

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 :

Example:

To find :

In each case, 3 and 5 are factors of , and in each case, .

The answer is that both statements together are insufficient.

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.

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.

(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.

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.

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.

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 :

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 :

Then

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.

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 A*B is an integer. Let's see what happens if B is 2. If B is 2, then 2.5*2 = 5 is an integer, but 2*2 = 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 = x*B. 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.

If the last digit of is 1, then the last digit of is either 1 or 9.

If the last digit of is 1, however, the last digit of must be 1.

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.

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.

The product of two or more numbers is 0 if and only if at least one of the factors is 0.

If you know the first statement, you know that neither nor is equal to 0, and that either , or both are equal to 0. You have narrowed the answer down to one or two.

If you know the second statement, you know that and are nonzero, but you know nothing about . You have narrowed the answer down to none or one.

If you know both statements, though, you know that is the only one of the four numbers equal to zero, so you have answered the question.

The product of two or more numbers is 0 if and only if at least one of the numbers is 0.

Each of the individual statements eliminates only two possible answers to the question; Statement 1 eliminates and , and Statement 2 eliminates and . The two together, however, eliminate all possible answers except for , leaving that to be the number worth zero.

Neither of these statements help you, individually or together. If three of the four numbers are equal to zero, then both products will equal zero. There is no way of knowing which of the numbers is non-zero.