Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Let's determine the values of \( A \) and \( L \) such that the number \( 13AL016 \) is divisible by 11.
### Divisibility Rule for 11:
A number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is either 0 or a multiple of 11.
Given number: 13AL016
### Step-by-Step Solution:
1. Identify the digits in odd and even positions:
- Odd positions (1st, 3rd, 5th, 7th): 1, A, 0, 6
- Even positions (2nd, 4th, 6th): 3, L, 1
2. Sum of the digits in the odd positions:
[tex]\[ \text{Sum of odd positions} = 1 + A + 0 + 6 = 7 + A \][/tex]
3. Sum of the digits in the even positions:
[tex]\[ \text{Sum of even positions} = 3 + L + 1 = 4 + L \][/tex]
4. Set up the equation for divisibility by 11:
According to the rule, we need:
[tex]\[ (7 + A) - (4 + L) = 0 \quad \text{or} \quad \text{a multiple of 11} \][/tex]
Let's first solve for the equation being equal to zero:
[tex]\[ (7 + A) - (4 + L) = 0 \][/tex]
[tex]\[ 7 + A - 4 - L = 0 \][/tex]
[tex]\[ 3 + A - L = 0 \][/tex]
[tex]\[ A - L = -3 \][/tex]
5. Solve for the variables:
[tex]\[ A - L = -3 \][/tex]
One possible solution set could be:
\( A = 2 \) and \( L = 5 \)
Let's check:
[tex]\[ 2 - 5 = -3 \][/tex]
This satisfies the equation.
Another potential solution set (considering multiples of 11):
[tex]\[ A - L = 8 \quad \text{(since \( A - L = -3 + 11 \))} \][/tex]
[tex]\[ A = 8, \quad L = 0 \][/tex]
Also,
[tex]\[ A - L = -3 + 22 = 19 \][/tex]
[tex]\[ (A = 1) & (L = -2), but L and A can only be single digit numbers from 0-9. Having iterated through potential candidates, Only the valid solutions within the constraints \(0-9\) range are: - \( A = 2 \) and \( L = 5 \) - \( A = 8 \) and \( L = 0 \) Therefore, the set of values for \(A \) and \( L \) that make \(13AL016\) divisible by 11 are: \[ A = 2 \quad \text{and} \quad L = 5 \][/tex]
or
\ [tex]\[ A = 8 \quad \text{and} \quad L = 0 \][/tex]
### Divisibility Rule for 11:
A number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is either 0 or a multiple of 11.
Given number: 13AL016
### Step-by-Step Solution:
1. Identify the digits in odd and even positions:
- Odd positions (1st, 3rd, 5th, 7th): 1, A, 0, 6
- Even positions (2nd, 4th, 6th): 3, L, 1
2. Sum of the digits in the odd positions:
[tex]\[ \text{Sum of odd positions} = 1 + A + 0 + 6 = 7 + A \][/tex]
3. Sum of the digits in the even positions:
[tex]\[ \text{Sum of even positions} = 3 + L + 1 = 4 + L \][/tex]
4. Set up the equation for divisibility by 11:
According to the rule, we need:
[tex]\[ (7 + A) - (4 + L) = 0 \quad \text{or} \quad \text{a multiple of 11} \][/tex]
Let's first solve for the equation being equal to zero:
[tex]\[ (7 + A) - (4 + L) = 0 \][/tex]
[tex]\[ 7 + A - 4 - L = 0 \][/tex]
[tex]\[ 3 + A - L = 0 \][/tex]
[tex]\[ A - L = -3 \][/tex]
5. Solve for the variables:
[tex]\[ A - L = -3 \][/tex]
One possible solution set could be:
\( A = 2 \) and \( L = 5 \)
Let's check:
[tex]\[ 2 - 5 = -3 \][/tex]
This satisfies the equation.
Another potential solution set (considering multiples of 11):
[tex]\[ A - L = 8 \quad \text{(since \( A - L = -3 + 11 \))} \][/tex]
[tex]\[ A = 8, \quad L = 0 \][/tex]
Also,
[tex]\[ A - L = -3 + 22 = 19 \][/tex]
[tex]\[ (A = 1) & (L = -2), but L and A can only be single digit numbers from 0-9. Having iterated through potential candidates, Only the valid solutions within the constraints \(0-9\) range are: - \( A = 2 \) and \( L = 5 \) - \( A = 8 \) and \( L = 0 \) Therefore, the set of values for \(A \) and \( L \) that make \(13AL016\) divisible by 11 are: \[ A = 2 \quad \text{and} \quad L = 5 \][/tex]
or
\ [tex]\[ A = 8 \quad \text{and} \quad L = 0 \][/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.