if a nine digit no is formed using 1 to 9 without repetition and if the probability that they will be divisible by 11 is p/q then p+q=
can yu tell me the divisibility of 11
if a nine digit no is formed using 1 to 9 without repetition and if the probability that they will be divisible by 11 is p/q then p+q=
can yu tell me the divisibility of 11
A number is divisible by 11 if the difference between the sum of the digits in the odd positions counting from the left (the first, the third ...) and the sum of the remaining digits is divisible by 11.
For example:
1. 58564
The sum of the digits in the odd positions counting from the left is 5+5+4=14
The sum of the remaining digits is 8+6=14
14-14=0, and 0 is divisible by 11, so 58564 is divisible by 11 too.
2. 635047
The sum of the digits in the odd positions counting from the left is 6+5+4=15
The sum of the remaining digits is 3+0+7=10
15-10=5, 5 isn't divisible by 11, so 635047 isn't divisible by 11.