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Given that the number \( 13AL016 \) is divisible by 11, find the values of \( A \) and \( L \).

[tex]\[ A = \quad \][/tex]
[tex]\[ L = \quad \][/tex]


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]