In order to be divisible by <st1:metricconverter w:st="on" ProductID="990, a">990, a</st1:metricconverter> pandigital number must be divisible by 9, 10, and 11. The sum of digits of all pandigital numbers is 45, therefore all of them are divisible by 9. Forcing the last digit to be 0 , takes care of the divisibility by 10, so all we must see to, is its divisibility by 11.
For being divisible by 11 the sum of the digits in the uneven places of the number, minus the sum of the even placed digits, must be divisible by 11 or be zero.
Denoting the sum of the 5 uneven placed digits as A, and the sum of the 4 even placed digits as B ( the10th digit is zero ), we get the requirement :
A+B = 45
A-B = 11 * N
where N is a positive or negative natural number. As A and B must be positive, natural numbers, and the max. of A is : A=9+8+7+6+5 = 35, the only solution will be (for N=1), A=28 and B=17, or, (for N=-1) , A=17 and B=28.
It is easily shown that dividing the 9 digits 1…9 into 2 groups answering the above conditions, gives only 6 possible groupings, as follows :
A B
1, 7, 5, 6, 9 2, 3, 4, 8
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2, 4, 5, 8, 9 1, 3, 6, 7
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5, 6, 8, 2, 7 1, 3, 4, 9
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6, 8, 9, 2, 3 5, 1, 4, 7
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1, 2, 3, 4, 7 5, 6, 8, 9
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1, 2, 3, 5, 6 4, 7, 8, 9
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The total different arrangements of those first 9 digits is found for each pair (A,B), by multiplying the number of possible permutations of A ( 5!), by the possible permutations of B (4!), times 6 (the number of the possible different groups) -
M, the total different numbers divisible by 990 will therefore be :
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M= 5! * 4! * 6 = 120 * 24 * 6 = 17280
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The total different pandigital numbers is 9! (no leading zero) , and the required percentage will be :
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(6 * 5! * 4! )/ 9! = 4.7619%
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Analytical Solution
