Question

Show that an integer is divisible by 11 if and only if the alternating sum (add first digit, subtract the second, add the third, subtract the fourth etc) of its digits is divisible by 11
Need an answer by tomorrow if possible please

Answer 1

Answer and Step-by-step explanation:

Suppose that we have a number y which is a positive integer and that:

y = $x_n...x_5x_4x_3x_2x_1x_0$

Where;

$x_{0}$ = digit at 10⁰ => one's place (units place)

$x_1$ = digit at 10¹ => 10's place (tens place)

$x_{2}$ = digit at 10² => 100's place (hundreds place)

$x_{3}$ = digit at 10³ => 1000's place (thousands place)

.

.

.

$x_{n}$ = digit at 10ⁿ place

Then;

y = $x_{0}$ * 10⁰ + $x_1$ * 10¹ + $x_{2}$ * 10² + $x_{3}$ * 10³ + $x_{4}$ * 10⁴ + $x_5$ * 10⁵ + . . . + $x_{n}$ * 10ⁿ

Since 10⁰ = 1, let's rewrite y as follows;

y = $x_{0}$  + $x_1$ * 10¹ + $x_{2}$ * 10² + $x_{3}$ * 10³ + $x_{4}$ * 10⁴ + $x_5$ * 10⁵ + . . . + $x_{n}$ * 10ⁿ

Now, to test if y is divisible by 11, replace 10 in the equation above by -1. Since 10 divided by 11 gives -1 (mod 11)     [mod means modulus]

y = $x_{0}$  + $x_1$ * (-1)¹ + $x_{2}$ * (-1)² + $x_{3}$ * (-1)³ + $x_{4}$ * (-1)⁴ + $x_5$ * (-1)⁵ + . . . + $x_{n}$ * (-1)ⁿ

=> y =  $x_{0}$  - $x_1$ + $x_{2}$ - $x_{3}$ + $x_{4}$ - $x_5$ + . . . + $x_{n}$ (-1)ⁿ (mod 11)

Therefore, it can be seen that, y is divisible by 11 if and only if alternating sum of its digits $x_{0}$  - $x_1$ + $x_{2}$ - $x_{3}$ + $x_{4}$ - $x_5$ + . . . + $x_{n}$ (-1)ⁿ is divisible by 11

Let's take an example

Check if the following is divisible by 11.

i. 1859

Solution

1859 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (1 - 8 + 5 - 9) is divisible by 11.

1 - 8 + 5 - 9 = -11.

Since -11 is divisible by 11 so is 1859

ii. 31415

Solution

31415 is divisible by 11 if and only if the alternating sum of its digit is divisible by 11. i.e if (3 - 1 + 4 - 1 + 5) is divisible by 11.

3 - 1 + 4 - 1 + 5 = 10.

Since 10 is not divisible by 11 so is 31415 not divisible.