Divisibility Rules for 2, 4 and 10: Arranging the digits

To solve this problem, let's explore how to form a number divisible by 2 using the digits 2, 1, and 3. A number is divisible by 2 if the last digit is an even number, which, from our given digits, is only the digit 2.

Let's examine all possible permutations of these digits and identify which numbers are divisible by 2:

• Permutation 1: 213, where the last digit is 3 (not divisible by 2).
• Permutation 2: 231, where the last digit is 1 (not divisible by 2).
• Permutation 3: 123, where the last digit is 3 (not divisible by 2).
• Permutation 4: 132, where the last digit is 2 (divisible by 2).
• Permutation 5: 312, where the last digit is 2 (divisible by 2).
• Permutation 6: 321, where the last digit is 1 (not divisible by 2).

From our examination, the numbers 132 and 312 both end in 2, making them divisible by 2. Therefore, the correct choice according to the problem's options is to select both permutations that satisfy the condition.

Therefore, the solution to the problem is Answer b and c.

To solve this problem, we need to apply the rule of divisibility by 2:

• A number is divisible by 2 if its last digit is even.

We are given the digits: 5, 3, 7, 4.

Among these digits, the only even digit is 4.

Therefore, to form a number that is divisible by 2, 4 must be the last digit.

From the provided multiple-choice options, we will select the number that ends with 4:

• Reading each choice, we find that choice 4, "7534," has 4 as the last digit and thus is divisible by 2.

Therefore, the correct number is 7534.

To solve this problem, we'll follow these steps:

• Step 1: Identify and understand the requirement for divisibility by 2.
• Step 2: Determine the potential numbers by rearranging the digits to satisfy the divisibility rule.
• Step 3: Confirm by checking the last digit of the arranged number.

Now, let's work through each step:
Step 1: A number is divisible by 2 if its last digit is even. Among the given digits (8, 1, 3, 7), only 8 is even.
Step 2: Rearrange the digits so that the last digit is 8. Potentially, using all digits, the number 1378 can be formed.
Step 3: Check that 1378 ends with 8, confirming it is divisible by 2.

Therefore, the solution to the problem is 1378.

To solve this problem, we should organize the given digits to make a number that is divisible by 10. According to the rule of divisibility by 10, a number must end in 0.

Let's follow these steps:

• Step 1: Ensure the number ends in 0 by placing 0 at the unit's place.
• Step 2: Arrange the remaining digits (5, 1, and 2) to form the largest possible number.
• Step 3: Arrange 5, 1, and 2 in descending order, resulting in 521.
• Step 4: Combine the arranged digits with 0 at the end: This gives 5210.

The number 5210 ends with 0, making it divisible by 10.

Therefore, the solution to the problem is $5120$.

To solve this problem, we need to understand the divisibility rule for $10$. A number is divisible by $10$ if and only if it ends in $0$.

Given the digits $2$, $3$, and $5$, we need to form a number ending with $0$. However, none of these digits is $0$. Therefore, using only the digits $2$, $3$, and $5$, it is impossible to create a number that ends in $0$.

This means it's impossible to rearrange these digits to form a number divisible by $10$.

Therefore, the correct answer is It is impossible.

To solve this problem, we will apply the divisibility rule for 4:

A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Therefore, we need to consider pairs from the digits $1, 2, 3,$ and $4$ that can form numbers divisible by 4. Let's list them:

• 12: Not divisible by 4.
• 21: Not divisible by 4.
• 32: Divisible by 4 ($32 \div 4 = 8$).
• 42: Divisible by 4 ($42 \div 4 = 10.5$ - Incorrect!).

From the list above, the only pair that forms a number divisible by 4 is $32$. By setting the last two digits as $32$, we use the remaining digits $1$ and $4$ as the preceding two digits to form the complete number.

Possible numbers using this order include $x132$, where $x$ is any combination of the remaining $1, 4$.

Let's attempt to form a number:
Arrange $3, 1, 2, 4$: $31$ and $24$, yielding 3124.
Verify if it matches our condition as laid in the answer choices.

Looking through our options, only $3124$ fits a number divisible by 4.

Thus, the number 3124 fulfills the condition of divisibility by 4.

Therefore, the solution to the problem is $3124$.