When subtracting even numbers, we will know that the ones digit will also be an even number.

Borrow a one from the tens digit of 64, since we cannot subtract the ones digits.

After borrowing, we can then subtract.

Subtract the tens place.  The tens place of 64 becomes a 5. 

Combine the two numbers.

The answer is .

Pick any even integer (2, 4, 6, etc.) to represent x. The only value that is odd is 3x + 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. 3x + 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 .

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)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.

In order to solve this problem, it will help us to find all of the possible scenarios of a, b, a², and b². 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 ab² is an even number. Let's think of ab² as the product of two integers: a and b².

In order for the product of a and b² 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 b² is even

Scenario 2: a is even and b² is odd

Scenario 3: a is odd and b² is even

Next, let's consider what possible values are possible for b. If b² 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 b² would be odd. Thus, if b² is even, then b must be even, and if b² is odd, then b must be odd. Let's add this information to the possible scenarios:

Scenario 1: a is even, b² is even, and b is even

Scenario 2: a is even, b² is odd, and b is odd

Scenario 3: a is odd, b² is even, and b is even

Lastly, let's see what is possible for a². If a is even, then a² must be even, and if a is odd, then a² must also be odd. We can add this information to the three possible scenarios:

Scenario 1: a is even, b² is even, and b is even, and a² is even

Scenario 2: a is even, b² is odd, and b is odd, and a² is even

Scenario 3: a is odd, b² is even, and b is even, and a² is odd

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

Choice I asks us to determine if a² must be even. If we look at the third scenario, in which a is odd, we see that a² would also have to be odd. Thus it is possible for a² to be odd.

Next, we can analyze a²b. In the first scenario, we see that a² is even and b is even. This means that a²b would be even. In the second scenario, we see that a² is even, and b is odd, which would still mean that a²b is even. And in the third scenario, a² is odd and b is even, which also means that a²b would be even. In short, a²b 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.