*In my previous post on divisibility, I had shared the logic applied behind the divisibility check for 3 and 9. Here, I share the working behind divisibility check by 11.*

**Divisibility by 11 : Method**

A number is said to be divisible by 11, if the difference of sums of its alternate digits is divisible by 11.

It simply means,

**Sum up the digits on odd place values and those on even place values separately** => **Find the difference between the sums** => if the resulting value is divisible by 11, then the number is divisible by 11, as indicated below:

For N = abcdef, |

sum1 = a + c + e |

sum2 = b + d + f |

calculate difference => sum1 – sum2 |

if difference value is divisible by 11, then N is divisible by 11 |

For example:- |

N = 5832, |

sum1 = 5 + 3 = 8 |

sum2 = 8 + 2 = 10 |

difference => sum1 – sum2 => -2 |

2 is not divisible by 11, thus N=5832 is not divisible by 11 |

**Divisibility by 11 : Logic**

Let’s see the logic and working behind why and how the method works!

For any N = abcdef, we rewrite it as: |

a x (10^{5}) + b x (10^{4}) + c x (10^{3}) + d x (10^{2}) + e x (10^{1}) + f x (10^{0}) |

which is same as: |

10^{5}a + 10^{4}b + 10^{3}c + 10^{2}d + 10^{1}e + 10^{0}f => 10 ^{5}a + 10^{4}b + 10^{3}c + 10^{2}d + 10e + f |

We write 10 = 11 – 1, and we get… |

(11-1)^{5}a + (11-1)^{4}b + (11-1)^{3}c + (11-1)^{2}d + (11-1)e + f |

Using the distributive laws and assuming all the multiples of 11 as m, we write |

[11m + (-1)^{5}]a +[11m + (-1)^{4}]b + [11m + (-1)^{3}]c + [11m + (-1)^{2}]d + [11m + (-1)]e + f |

Factoring out the common term 11m, we get |

N = 11m [a + b + c + d + e] – a + b – c + d – e + f |

Here, we can say:

[- a + b – c + d – e + f] is same as (b + d + f) – (a + c + e), which is nothing but the difference between sums of alternate digits, as described in the method earlier.

Through the above equation, we see that for N to be divisible by 11, the term [(b + d + f) – (a + c + e)] needs to be divisible by 11.

Let’s see few examples:

N= 45831 |

N = 4 x (10^{4}) + 5 x (10^{3}) + 8 x (10^{2}) + 3 x (10^{1}) + 1 x (10^{0}) = 4 x (11-1) ^{4} + 5 x (11-1)^{3} + 8 x (11-1)^{2} + 3 x (11-1) + 1 = 4 [11m + (-1) ^{4}] + 5 [11m + (-1)^{3}] + 8 [11m + (-1)^{2}] + 3 [11m + (-1)]d + 1 = 11m [4 + 5 + 8 + 3] + 4 – 5 + 8 – 3 + 1 = 11m [20] + 5 = 11m + 5 |

5 is not divisible by 11, hence 45831 is not divisible by 11 |

For N = 75933 |

N = 7 x (10^{4}) + 5 x (10^{3}) + 9 x (10^{2}) + 3 x (10^{1}) + 3 x (10^{0}) = 7 x (11-1) ^{4} + 5 x (11-1)^{3} + 9 x (11-1)^{2} + 3 x (11-1) + 3 = 7 [11m + (-1) ^{4}] + 5 [11m + (-1)^{3}] + 9 [11m + (-1)^{2}] + 3 [11m + (-1)]d + 3 = 11m [7 + 5 + 9 + 3] + 7 – 5 + 9 – 3 + 3 = 11m [24] + 11 = 11m + 11 |

Here, N = 45837 is divisible by 11 |

See below a very simple example without the assumption of multiples of 11 as m. |

For N = 121 |

= 1 x 10^{2} + 2 x 10 + 1 x 1= 1 [11-1] ^{2} + 2 [11-1] + 1= 1 [11 ^{2} – 11(2) + 1] + 2[11 + (-1)] + 1= 1 [11(11 -2) + 1] + 2 [11 + (-1)] + 1 = 1 [11(9) + 1] + 2 [11 + (-1)] + 1 = 11 [9+2] + 1 – 2 + 1 = 11[11] + 0 |

=> 0 is considered to be divisible by all numbers. Thus N=121 is seen as divisible by 11 |

Very Interesting! Thank you for sharing the logic and refreshing the facts.. keep sharing more maths logic ðŸ˜Š

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Thank you ðŸ™‚

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