True or fals: a number is divisible by 8 if it is divisble by 4 and divisible by 2?

And how many variations of 3_2_ are divisible by six?

"True or fals: a number is divisible by 8 if it is divisble by 4 and divisible by 2?"

False.
This logic works if the two numbers are prime to each other, i.e. there are no common factors between the two numbers.
For example, it is true to say that a number is divisible by 6 if it is divisible by 3 and divisible by 2.

If the number 3A2B needs to be divisible by 6 (where A and B stand for undertermined digits), then it has to be divisible by 3 and divisible by 2.
The rule of divisibility for 2 is that the last digit (B) must be even.
The rule of divisibility for 3 is that the sum of all digits must be divisible by 3.
So enumerate the possibilities for A for each possible value of B, as follows:
A=0
B=1 (3120), B=4 (3420) B=7 (3720)
A=2
B=2 (3222), B=5 (3522) B=8 (3822)
A=4
...
A=6
...
A=8
...
How many variations are possible?

True! A number is indeed divisible by 8 if it is divisible by 4 and divisible by 2. That's like saying, "Hey, if you can divide by 4 and still have something left to divide by 2, then you're divisible by 8 too!"

As for the variations of 3_2_ that are divisible by six, let's have some fun with it. If we want our number to be divisible by six, it needs to be divisible by both 2 and 3. So, let's see what we can do:

If we put a 3 in the first slot, we need 2_ _ in the other slots. The only possibilities are 2 0 and 2 6.

If we put a 3 in the second slot, we need _ 2 _ as the other numbers. Here, we have 2 2 and 2 8 as options.

Finally, if we put a 3 in the last slot, then we need _ _ 2 as the other numbers. The options are 0 2 and 6 2.

So, in total, we have 6 variations of 3_2_ that are divisible by six. Just remember, math and humor can go hand in hand, but not always for division problems!

True, a number is divisible by 8 if it is divisible by 4 and divisible by 2.

To determine the number of variations of 3_2_ that are divisible by six, we need to consider the divisibility rules of 6. A number is divisible by 6 if it is divisible by both 2 and 3.

Let's break it down step-by-step:

Step 1: The number needs to end with an even digit, since it needs to be divisible by 2. So we have the following options: 3_2_, where "_" represents a digit.

Step 2: To be divisible by 3, the sum of its digits must be divisible by 3. The digits can be chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

Step 3: Let's consider all the possibilities for the number of variations. Each digit can take any value from 0 to 9, resulting in 10 options for the first digit and 10 options for the third digit.

Step 4: For the second digit, it must be an even number to maintain divisibility by 2. So we have 5 options: {0, 2, 4, 6, 8}.

Step 5: For the fourth digit, it can be any digit from 0 to 9, resulting in 10 options.

Step 6: Multiply the number of options for each digit together: 10 * 5 * 10 = 500.

Therefore, there are 500 variations of 3_2_ that are divisible by six.

The statement "a number is divisible by 8 if it is divisible by 4 and divisible by 2" is true.

To understand why this is true, we need to consider the properties of divisibility. A number is divisible by another number if it can be divided evenly without any remainder.

Let's break this down:

1. Divisibility by 2: A number is divisible by 2 if its last digit is even (ends in 0, 2, 4, 6, or 8). For example, 16, 24, and 108 are all divisible by 2 because their last digits are 6, 4, and 8, respectively.

2. Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. For example, 132, 448, and 932 are all divisible by 4 because the numbers formed by their last two digits (32, 48, and 32) are all divisible by 4.

Now, if a number is divisible by both 2 and 4, it means that its last digit must be even, and the number formed by its last two digits must be divisible by 4. This also means that the number formed by its last three digits is divisible by 8.

Regarding the second question, "How many variations of 3_2_ are divisible by six?" let's break it down step-by-step:

1. The number must be divisible by 2. This requires that the last digit is even. There are five options for the last digit: 0, 2, 4, 6, and 8.

2. The number must be divisible by 3. This requires that the sum of all its digits is divisible by 3. The digit sum of 3_2_ is 3 + 2 = 5. In order to make it divisible by 3, the remaining two digits need to add up to a multiple of 3.

Let's analyze each combination of the remaining two digits:

- 30: The next digit can be either 0, 3, or 6 since 0 + 3 + 2 + 0 = 5 + 3 + 0 + 2 = 10, which is a multiple of 3. So, 3020, 3030, 3060 are divisible by 6.
- 33: The next digit can be either 0, 3, 6, or 9 since 0 + 3 + 2 + 3 = 5 + 3 + 3 + 2 = 13, which is not a multiple of 3. So, 3320, 3330, 3360, and 3390 are not divisible by 6.
- 36: The next digit can be either 0, 3, 6, or 9 since 0 + 3 + 2 + 6 = 5 + 3 + 6 + 2 = 16, which is not a multiple of 3. So, 3620, 3630, 3660, and 3690 are not divisible by 6.
- 39: The next digit can be either 0, 3, 6, or 9 since 0 + 3 + 2 + 9 = 5 + 3 + 9 + 2 = 19, which is not a multiple of 3. So, 3920, 3930, 3960, and 3990 are not divisible by 6.
- 60: The next digit can be either 0, 3, 6, or 9 since 0 + 3 + 2 + 9 = 5 + 3 + 9 + 2 = 19, which is not a multiple of 3. So, 6020, 6030, 6060, and 6090 are not divisible by 6.
- 63: The next digit can be either 0, 3, 6, or 9 since 0 + 3 + 2 + 9 = 5 + 3 + 9 + 2 = 19, which is not a multiple of 3. So, 6320, 6330, 6360, and 6390 are not divisible by 6.

In summary, there are three variations of 3_2_ that are divisible by six: 3020, 3030, and 3060.