Even / Odd Numbers - ACT Math

If Steve has 10 chores, Jeff has 7 chores (3 less than Steve). Tom has twice as many as Steve, so Tom has 14 chores.

Create an equation for the situation described and solve:

Create an equation to decribe the above situation and solve.

Total presents given out is represented by .

Presents given to Sally is represented by .

Presents given to Billy is represented by .

Thus our equation becomes,

Always start by gathering your facts. You know that if is odd, then neither of those numbers are even. (If either were, then the product would be even.) Now, the rules for subtracting are just like adding. If the difference of two numbers is even, then they must either both be odd or both be even. Thus if is even and we know that is odd, then we know that must be odd. Thus if is odd, it is also true that is even, for both and are even.

If Steve has 10 chores, Jeff has 7 chores (3 less than Steve). Tom has twice as many as Steve, so Tom has 14 chores.

Create an equation for the situation described and solve:

Create an equation to decribe the above situation and solve.

Total presents given out is represented by .

Presents given to Sally is represented by .

Presents given to Billy is represented by .

Thus our equation becomes,

Always start by gathering your facts. You know that if is odd, then neither of those numbers are even. (If either were, then the product would be even.) Now, the rules for subtracting are just like adding. If the difference of two numbers is even, then they must either both be odd or both be even. Thus if is even and we know that is odd, then we know that must be odd. Thus if is odd, it is also true that is even, for both and are even.

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.

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 a group of philosophers, are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

To start, let's calculate the total philosophers:

Ockham: * , or

Aquinas: * , or

Scotus: divided by , or

Therefore, the total number is:

We know that when you multiply any integer by an even number, the result is even. Thus, if is odd but when you multiply this by the number is even, we know that must be even. You cannot say anything about the sign value of any of the numbers. Likewise, it is impossible for either or to be even.

Three consecutive integers will include at least one, and possibly two, even integers. Since the addition of even a single even integer to a chain of integer products will make the final product even, we know the product must be even.

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.

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 a group of philosophers, are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

To start, let's calculate the total philosophers:

Ockham: * , or

Aquinas: * , or

Scotus: divided by , or

Therefore, the total number is:

We know that when you multiply any integer by an even number, the result is even. Thus, if is odd but when you multiply this by the number is even, we know that must be even. You cannot say anything about the sign value of any of the numbers. Likewise, it is impossible for either or to be even.

Three consecutive integers will include at least one, and possibly two, even integers. Since the addition of even a single even integer to a chain of integer products will make the final product even, we know the product must be even.

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.

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 .