Answer :
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]
