How to multiply even numbers

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ACT Math › How to multiply even numbers

Questions 1 - 5

The product of three consecutive nonzero integers is taken. Which statement must be true?

$\textup{The product must be even.}$

$\textup{The third consecutive integer must be even.}$

$\textup{The product must be odd.}$

$\textup{Two of the three integers must be even.}$

$\textup{Two of the three integers must be odd.}$

Explanation

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.

$E \cdot O \cdot E = E$

$O \cdot E \cdot O = E$

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?

Explanation

To start, let's calculate the total philosophers:

Ockham: * , or

Aquinas: * , or

Scotus: divided by , or

Therefore, the total number is:

Which of the following integers has an even integer value for all positive integers and ?

$b^{2}+d^{2}$

$d^{2}+bd$

Explanation

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.

$\textup{If } abc\textup{ is even and }ac\textup{ is odd, which of the following must be true? Presume that }a\textup{,}b\textup{,}c\textup{ are integers.}$

$\textup{is even}$

$\textup{is odd}$

$\textup{is even}$

Explanation

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.

is even

Therefore, which of the following must be true about ?

It could be either even or odd.

It must be even.

It must be odd.

Explanation

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.