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# Using divisibility tests, determine which of the following numbers are divisible by 11 :

**(a)** 5445

**(b)** 10824

**(c)** 7138965

**(d)** 70169308

**(e)** 10000001

**(f)** 901153

To do :

We have to find whether the given numbers are divisible by 11.

Solution :

Divisibility rule of 11:

A positive integer N is divisible by $11$ if the difference of the alternating sum of digits of N is a multiple of $11$.

(a) 5445

Sum of the digits at odd places $= 5 + 4$

$= 9$

Sum of the digits at even places $= 4 + 5$

$= 9$

Difference $= 9 - 9$

$= 0$

0 is a multiple of $11$. (Zero is a multiple of every integer)

Hence, 5445 is divisible by 11.

(b) 10824

Sum of the digits at odd places $= 4 + 8 + 1$

$= 13$

Sum of the digits at even places $= 2 + 0$

$= 2$

Difference $= 13 - 2$

$= 11$

11 is a multiple of $11$.

Hence, 10824 is divisible by 11.

(c) 7138965

Sum of the digits at odd places $= 5 + 9 + 3 + 7$

$= 24$

Sum of the digits at even places $= 6 + 8 + 1$

$= 15$

Difference $= 24 - 15$

$= 9$

9 is not a multiple of $11$.

Hence, 7138965 is not divisible by 11.

(d) 70169308

Sum of the digits at odd places $= 8 + 3 + 6 + 0$

$= 17$

Sum of the digits at even places $= 0 + 9 + 1 + 7$

$= 17$

Difference $= 17 - 17$

$= 0$

0 is a multiple of $11$.

Hence, 70169308 is divisible by 11.

(e) 10000001

Sum of the digits at odd places $= 1 + 0 + 0 + 0$

$= 1$

Sum of the digits at even places $= 0 + 0 + 0 + 1$

$= 1$

Difference $= 1 - 1$

$= 0$

0 is a multiple of $11$.

Hence, 10000001 is divisible by 11.

(f) 901153

Sum of the digits at odd places $= 3 + 1 + 0 $

$=4$

Sum of the digits at even places $= 5 + 1 + 9$

$= 15$

Difference $= 15 - 4$

$= 11$

11 is a multiple of $11$.

Hence, 901153 is divisible by 11.

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