Integers - SAT Math

Pick any even integer (2, 4, 6, etc.) to represent x. The only value that is odd is 3_x_ + 1. Any number multiplied by an even integer will be even. When an even number is added and subtracted to that product, the result will be even as well. 3_x_ + 1 is the only choice that adds an odd number to the product.

Add the ones digit.

Since the number is 10 or greater, the tens digit is the carryover.

Add the tens digit with the carryover.

Add the hundreds digit with the carryover.

Combine all the ones digits in our calculations.

The answer is .

Consecutive odd integers differ by 2. If the smallest integer is x, then

x + (x + 2) + (x + 4) = 93

3x + 6 = 93

3x = 87

x = 29

The three numbers are 29, 31, and 33, the largest of which is 33.

If the largest of the three consecutive odd integers is d, then the three numbers are (in descending order):

d, d – 2, d – 4

This is true because consecutive odd integers always differ by two. Adding the three expressions together, we see that the sum is 3_d_ – 6.

Since and must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where . This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.

Add the ones digits:

Since there is no tens digit to carry over, proceed to add the tens digits:

The answer is .

If there are students taking Greek, then there are or students taking Latin. However, the question asks how many total students there are in the school, so you must add these two values together to get:

or total students.

Add the ones digit.

Since the there is a tens digit, use that as the carryover to the next term.

Add the tens digit including the carryover.

The hundreds digit is 7.

Combine the ones digit of each calculation in order.

The answer is:

Add the ones digit.

Carry over the one from the tens digit to the next number.

Add the tens digit with the carry over.

Carry over the one from the tens digit to the hundreds digit.

Add the hundreds digit with the carry over.

The thousands digit has no carry over. The second number has no thousands digit. This means that the thousands is one. Combine all the ones digits from each of the previous calculations.

The correct answer is:

While I & II can be true, examples can be found that show they are not always true (for example, 22/22 is odd and 42/22 is even).

III is always true – a square even number is always divisible by four, and the distributive property tell us that adding two numbers with a common factor gives a sum that also has that factor.

S consists of all even integers. If we were to increase each of these even numbers by 2, then we would get another set of even numbers, because adding 2 to an even number yields an even number. In other words, T also consists entirely of even numbers.

In order to find the mean of T, we would need to add up all of the elements in T and then divide by however many numbers are in T. If we were to add up all of the elements of T, we would get an even number, because adding even numbers always gives another even number. However, even though the sum of the elements in T must be even, if the number of elements in T was an even number, it's possible that dividing the sum by the number of elements of T would be an odd number.

For example, let's assume T consists of the numbers 2, 4, 6, and 8. If we were to add up all of the elements of T, we would get 20. We would then divide this by the number of elements in T, which in this case is 4. The mean of T would thus be 20/4 = 5, which is an odd number. Therefore, the mean of T doesn't have to be an even number.

Next, let's analyze the median of T. Again, let's pretend that T consists of an even number of integers. In this case, we would need to find the average of the middle two numbers, which means we would add the two numbers, which gives us an even number, and then we would divide by two, which is another even number. The average of two even numbers doesn't have to be an even number, because dividing an even number by an even number can produce an odd number.

For example, let's pretend T consists of the numbers 2, 4, 6, and 8. The median of T would thus be the average of 4 and 6. The average of 4 and 6 is (4+6)/2 = 5, which is an odd number. Therefore, the median of T doesn't have to be an even number.

Finally, let's examine the range of T. The range is the difference between the smallest and the largest numbers in T, which both must be even. If we subtract an even number from another even number, we will always get an even number. Thus, the range of T must be an even number.

Of choices I, II, and III, only III must be true.

The answer is III only.

Take a known common factor of two and rewrite the fraction.

Dividing the number 143 into 16, the coefficient is 8 since:

There is a remainder of fifteen, which is .

Combining the coefficient and the remainder as a mixed fraction, this can be rewritten as:

The answer is:

I is not always true because a negative number multiplied by 7 will give a number that is more negative than the original. II is true because adding 5 to any number will increase the value. III is true because squaring any number will increase the magnitude of the value, and squaring a negative number will make it positive.

I)xy is Even*Odd is Even. II) x-y is Even+/-Odd is Odd. III) 3x is Odd*Even =Even, 2y is Even*Odd=Even, Even + Even = Even. Therefore only II is Odd.

I. xyz = even * odd * even = even

II. 2x + 3y = even*even + odd*odd = even + odd = odd

III. z2 – y = even * even – odd = even – odd = odd

Therefore only I will give an even number.

Since we are only told that "at least" one of the numbers is even, we could have one even and one odd integer OR we could have two even integers.

Even plus odd is odd, but even plus even is even, so x + y could be either even or odd.

Even times odd is even, and even times even is even, so xy must be even.

There are certain patterns that can be used to predict whether the product or sum of numbers will be odd or even. The sum of two odd numbers is always even, as is the sum of two even numbers. The sum of an odd number and an even number is always odd. In multiplication the product of two odd numbers is always odd. While the product of even numbers, as well as the product of odd numbers multiplied by even numbers is always even. So for this problem we need to find scenarios where the only possibile answers are even. can only result in even numbers no matter what positive integers are used for and , because must can only result in even products; the same can be said for . The rules provide that the sum of two even numbers is even, so is the answer.

In order to solve this problem, it will help us to find all of the possible scenarios of a, b, _a_2, and _b_2. We need to make use of the following rules:

1. The product of two even numbers is an even number.

2. The product of two odd numbers is an odd number.

3. The product of an even and an odd number is an even number.

The information that we are given is that _ab_2 is an even number. Let's think of _ab_2 as the product of two integers: a and _b_2.

In order for the product of a and _b_2 to be even, at least one of them must be even, according to the rules that we discussed above. Thus, the following scenarios are possible:

Scenario 1: a is even and _b_2 is even

Scenario 2: a is even and _b_2 is odd

Scenario 3: a is odd and _b_2 is even

Next, let's consider what possible values are possible for b. If _b_2 is even, then this means b must be even, because the product of two even numbers is even. If b were odd, then we would have the product of two odd numbers, which would mean that _b_2 would be odd. Thus, if _b_2 is even, then b must be even, and if _b_2 is odd, then b must be odd. Let's add this information to the possible scenarios:

Scenario 1: a is even, _b_2 is even, and b is even

Scenario 2: a is even, _b_2 is odd, and b is odd

Scenario 3: a is odd, _b_2 is even, and b is even

Lastly, let's see what is possible for _a_2. If a is even, then _a_2 must be even, and if a is odd, then _a_2 must also be odd. We can add this information to the three possible scenarios:

Scenario 1: a is even, _b_2 is even, and b is even, and _a_2 is even

Scenario 2: a is even, _b_2 is odd, and b is odd, and _a_2 is even

Scenario 3: a is odd, _b_2 is even, and b is even, and _a_2 is odd

Now, we can use this information to examine choices I, II, and III.

Choice I asks us to determine if _a_2 must be even. If we look at the third scenario, in which a is odd, we see that _a_2 would also have to be odd. Thus it is possible for _a_2 to be odd.

Next, we can analyze a_2_b. In the first scenario, we see that _a_2 is even and b is even. This means that a_2_b would be even. In the second scenario, we see that _a_2 is even, and b is odd, which would still mean that a_2_b is even. And in the third scenario, _a_2 is odd and b is even, which also means that a_2_b would be even. In short, a_2_b is even in each of the possible scenarios, so it must always be even. Thus, choice II must be true.

We can now look at ab. In scenario 1, a is even and b is even, which means that ab would also be even. In scenario 2, a is even and b is odd, which means that ab is even again. And in scenario 3, a is odd and b is even, which again means that ab is even. Therefore, ab must be even, and choice III must be true.

The answer is II and III only.

The word "product" refers to the answer of a multiplication problem. Since 2 times 8 equals 16, it is a valid pair of numbers.

If is an odd integer then we can plug 1 into and solve for yielding 13. 13 is prime, meaning it is only divisible by 1 and itself.