Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.