# English Maths

## Divisibility rule by 11

If the sum of the digits in the odd places and the sum of the digits in the even places difference is a multiple of 11 or zero then we can say that the given number is divisible by 11.

Example: Is 1023 divisible by 11?

Solution:

Sum of the digits in the odd places (Black Color) = 1 + 2 = 3

Sum of the digits in the even place (Red Color) = 0 + 3 = 3

Difference between the two sums = 3 - 3 = 0

Hence, 1023 is divisible by 11.

Example: Is 1749 divisible by 11?

Solution:

Sum of the digits in the odd places (Black Color) = 1 + 4 = 5

Sum of the digits in the even place (Red Color) = 7 + 9 = 16

Difference between the two sums = 16 - 5 = 11

Hence, 1749 is divisible by 11.

Example: Is 592845 divisible by 11?

Solution:

Sum of the digits in the odd places (Black Color) = 5 + 2 + 4 = 11

Sum of the digits in the even place (Red Color) = 9 + 8 + 5 = 22

Difference between the two sums = 22 - 11 = 11

Hence, 592845 is divisible by 11.

Example: Is 694201 divisible by 11?

Solution:

Sum of the digits in the odd places (Black Color) = 6 + 4 + 0 = 10

Sum of the digits in the even place (Red Color) = 9 + 2 + 1 = 12

Difference between the two sums = 12 - 10 = 2

Hence, 694201 is not divisible by 11.

### Let’s have practice of divisibility rule by 11:

Which of the following numbers are divisible by 11?

852346, 45982, 326117, 85932, 6987751, 5676

Solution:

852346 = [8 + 2 + 4 =14 & 5 + 3 + 6 = 14] 14 – 14 = 0 Hence, divisible by 11

45982 = [4 + 9 + 2 =15 & 5 + 8 = 13] 15 – 13 = 2 Hence, not divisible by 11

326117 = [3 + 6 + 1 =10 & 2 + 1 + 7 = 10] 10 – 10 = 0 Hence, divisible by 11

85932 = [8 + 9+ 2 = 19 & 5 + 3 = 8] 19 – 8 = 11 Hence, divisible by 11

698775 = [6 + 8 + 7 =21 & 9 + 7 + 5 = 21] 21 – 21 = 0 Hence, divisible by 11

5676 = [5 + 7 =12 & 6 + 6 = 12] 12 – 12 = 0 Hence, divisible by 11

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Divisibility Rule by 10