
How many five digit multiples of 11 are there, if the five digits are 3,4,5,6 and 7?
A. 12
B. 13
C. 10
D. 16
Answer
587.4k+ views
According to the divisibility rule of 11 , A number can be divisible by 11 if the difference between sum of digits at odd place and sum of digits at even place should be zero or multiple of 11.
Also n numbers can be arranged in n! ways.
Complete step-by-step answer:
As given digits are 3,4,5,6 and 7.
First we will arrange given digits in the form of that number in which the difference between sum of digits at odd place and sum of digits at even place should be zero or multiple of 11.
Required number can be formed from given digits is 63547.
Sum of digit at odd places is 6+5+7=18
Sum of digits at even places is 3+4=7
Difference between sum of digit at odd places and sum of digit at even places is 18-7=11 , which is a multiple of 11.
Now we can try to find possible numbers which can be formed in this way.
Number of digits at odd places is 3.
So we can arrange it in 3! Ways.
Number of digits at even places is 2.
So we can arrange it in 2! Ways.
So total possible number which can be formed in this way is
$3!\times 2!=3\times 2\times 1\times 2\times 1=12$
Hence option A is correct.
Note: First we need to arrange given digits in a way to get a number multiple of 11 according to the divisibility rule of 11.
Also we can expand n! as
$n!=n\times (n-1)\times (n-2)\times (n-3)\times ...............\times 1$
Also n numbers can be arranged in n! ways.
Complete step-by-step answer:
As given digits are 3,4,5,6 and 7.
First we will arrange given digits in the form of that number in which the difference between sum of digits at odd place and sum of digits at even place should be zero or multiple of 11.
Required number can be formed from given digits is 63547.
Sum of digit at odd places is 6+5+7=18
Sum of digits at even places is 3+4=7
Difference between sum of digit at odd places and sum of digit at even places is 18-7=11 , which is a multiple of 11.
Now we can try to find possible numbers which can be formed in this way.
Number of digits at odd places is 3.
So we can arrange it in 3! Ways.
Number of digits at even places is 2.
So we can arrange it in 2! Ways.
So total possible number which can be formed in this way is
$3!\times 2!=3\times 2\times 1\times 2\times 1=12$
Hence option A is correct.
Note: First we need to arrange given digits in a way to get a number multiple of 11 according to the divisibility rule of 11.
Also we can expand n! as
$n!=n\times (n-1)\times (n-2)\times (n-3)\times ...............\times 1$
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