The number of $6$ digit numbers that can be formed using digits $0, 1, 2, 5, 7 and 9$ which are divisible by $11$ and no digit is repeated are

060

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60.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

060.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

060.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

Given digits: $0, 1, 2, 5, 7 and 9$
Let the six digit number be: $d_{1}, d_{2}, d_{3}, d_{4}, d_{5}, d_{6}$

A number is divisible by $11,$ if the difference of alternating sum of digits of it is divisible by $11.$
$d_{1} d_{2} d_{3} d_{4} d_{5} d_{6}$ is divisible by $11$
$\Rightarrow [(d_{1} + d_{3} + d_{5}) - (d_{2} + d_{4} + d_{6})]$ is divisible by $11.$
Write $d_{1} + d_{3} + d_{5} = k_{1} and d_{2} + d_{4} + d_{6} = k_{2}$

$∵$ Sum of given digits $0 + 1 + 2 + 5 + 7 + 9 = 24$

$\Rightarrow k_{1} + k_{2} = 24 and ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} k_{1} - k_{2} = 22 k_{1} - k_{2} = 11 k_{1} - k_{2} = 0 \end{matrix}$

Now we will check weather these cases are possible or not
$C A S E 1 : k_{1} - k_{2} = 22$
Use $k_{1} + k_{2} = 24$ to find the values of $k_{1}, k_{2}$
$k_{2} = 1$ which is not possible as the sum of three natural numbers can’t be $1$

$C A S E 2 : k_{1} - k_{2} = 11$
Use $k_{1} + k_{2} = 24$ to find the values of $k_{1}, k_{2}$
$k_{1} = 17.5$ which is not possible as the sum of three natural numbers can’t be fraction.

$C A S E 3 : k_{1} - k_{2} = 0$

Use $k_{1} + k_{2} = 24$ to find the values of $k_{1}, k_{2}$
$k_{1} = k_{2} = 12$ this case is possible.

Now we have to check what values of $d_{1} + d_{3} + d_{5} = d_{2} + d_{4} + d_{6} = 12$

So there are two possible triplets ${0, 5, 7}, {1, 2, 9}$ , therefore we will have two cases
$C A S E 1 : {d_{1}, d_{3}, d_{5}} = {0, 5, 7}$ and ${d_{2}, d_{4}, d_{6}} = {1, 2, 9}$
Number of ways to fill $d_{1}$ are $2$ as $0$ can not comes at first place, the number of ways to fill the $d_{3}$ are $2$ as $0$ can be filled here and repetiotion is not allowed and the number of ways to fill $d_{5}$ are $1$ .
Similarly, the number of ways to fill $d_{2}, d_{4}, d_{6}$ are $3!$
$∴$ total number of ways are $= 2 \times 2 \times 3! = 24$

$C A S E 2 : {d_{1}, d_{3}, d_{5}} = {1, 2, 9}$ and ${d_{2}, d_{4}, d_{6}} = {0, 5, 7}$
$∴$ Number of ways to arrange = Number of ways to arrange ${d_{1}, d_{3}, d_{5}} \times$ Number of ways to arrange ${d_{2}, d_{4}, d_{6}}$

$\Rightarrow$ Number of ways to arrange $= 3! \times 3! = 36$

Hence, total number $6$ digit numbers divisible by $11$ are $= 24 + 36 = 60$

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