Integers - ACT Math

Recall that when you multiply by an even number, you get an even product.

Therefore, we know the following from the first statement:

is even or is even or both and are even.

For the second, we know this:

Since is even, therefore, can be either even or odd. (Regardless of what it is, we can get an even value for .)

Based on all this data, we can tell nothing necessarily about . If is even, then is even, even if is odd. However, if is odd while is even, then will be even.

the student has four choices for the first question, but only three choices for each of the remaining questions since he does not choose answers with the same letter twice in a row. From the multiplicative counting principle, there are 4 × 3 × 3 × 3 × 3 x 3 = 972 ways Steve can answer the six questions.

It is said that the three numbers are different. So the number of lock combinations is 30_P_3 = 24 360.

The key here is for any integers and , that means that no matter what you set and equal to you will get an odd number. An odd number is not divisible by 2, also it is an even number plus and odd number. The only expression that satisfies this is . will always be even, so will , but is always odd so the combination gives us an odd number, always.

The hand could have 2, 3, 4, or 5 black cards. There are 12 black cards and 12 red cards, so the numbers of combinations for the four cases are as follows.

2 black cards: C(12, 2) × C(12, 3) = 14 520

3 black cards: C(12, 3) × C(12, 2) = 14 520

4 black cards: C(12, 4) × C(12, 1) = 5940

5 black cards: C(12, 5) × C(12, 0) = 792

The total number of euchre hands that have at least two black cards is the total of these four cases, 35 772.

For this problem, align and solve by adding the ones digit , tens digit , and hundreds digit . This also means that you have to add to the in the thousands place to get .

Rewrite the question into a mathematical expression.

Add the ones digit.

Add the tens digit.

Combine the tens digit and the ones digit. The answer is .

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.

Rewrite and use common factors to simplify.

Rewrite the question in a mathematical expression.

Add the ones digit.

Since this number is larger than , carry over the in tens digit when adding the next term.

Add the tens digit with the carry over.

Combine the tens digit and the ones digit. The answer is .

In order to get an odd result from an addition, we must have one odd and one even number, thus you know from the first point about that only one of the two values is odd. Now, to get an even result, you can have two evens or two odds. So, let's presume that has two odd values, this means that must be even. Thus, you have:

Now, if we presume that has two even values, we must then know that is odd. Thus, we have:

First of all, you can eliminate the two answers that say that a given value is positive or negative. This cannot be told from our data. Next, it cannot be the case that is even. It will always be odd (hence, the correct answer is this). Finally, it cannot be that is even always. In the second case above, you will have two even numbers added together, given you an even. Then, you will add in an odd, giving you an odd.

This problem can be solved using common factors.

Rewrite using factors.

Rewrite by using factors. Simplify until the answer cannot be reduced any further.

Rewrite by using common factors. Reduce to the simplest form.

There are two ways to solve this problem. First you can do so algebraically by dividing both sides by 13:

But, there is another way, which if you understand odd numbers, is even faster. Of all the answers above, only one is odd. You know, given the equation, that must be odd--any odd number multiplied by an odd number will yeild an odd number. If you multiply an odd number by an even number, you will get an even number.

There are two ways to approach this problem:

1. Use the rule that states that any two odd numbers multiplied together will yield another odd number.

Using this rule, only one answer is an odd number (29) which will yield another odd number (493) when multiplied by the given odd number (17).

2. Solve algebraically:

There are two ways to approach this problem:

1. Use the rule that states that any two odd numbers multiplied together will yield another odd number.

Using this rule, only one answer is an odd number (85) which will yield another odd number (7,735) when multiplied by the given odd number (91).

2. Solve algebraically:

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.

15_P_15 = 15!

= 1.307 674 368 × 1012