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Seven digits from the digits 1,2,3,4,5,6,7,8,9 are written in random order. the probability that this seven digit no. is divisible by 9 is ?




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Recall the fact that an integer is divisible by 9 if the sum of its digits is divisible by 9.

Since the sum of first 9 natural numbers is \frac{9(9+1)}{2}=45, which is divisible by 9, it must be the case that the sum of the two integers that we don't pick to form the seven digit number is divisible by 9.

Number of ways of choosing two integers from 9 integers :  ^9C_2=36

Number of two digit pairs whose sum is divisible by 9 : \{(1,8),(2,7),(3,6),(4,5)\}=4

Simply take the ratio to get the probability that the seven digit number so formed is divisible by 9
1 5 1
:) basically we're using this :
"9m - n" is divisible by 9 only if "n" is divisible by 9
because, 9m - n = 9t implies n = 9m-9t = 9(m-t)
which simply means that "n" is a multiple of 9
right. tha'ts interesting ! thanks a lot, AGAIN !
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