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nuox95
let 2xyz7 be a five-digit number such that the product of the digits is zero and the sum of the digits is divisible by 9. Find how many such numbers
- the product will be zero when one from those will be zero and one number from indifferent how many digits is divisible with 9 if the sum of this numbers is divisible with 9 like - for example 26307 those sum is 6+3=9 and 7+2=9 and if you assumes again 9+9=18 -- so 1+8=9 - so if you have thought it the every digits can be using only one time from 0 till 9 hence you get this numbers :21087,23067,24057,25047,26037,28017,20367,20457, 20187,20817,23607,24507,21807, - i think those are all
i think you might have missedd some. for example 22077
the 2 and 7 are fixed so we are only looking at xyz. it is possible that all three are zero one way to do this it is possible that 2 are zero in which case the other must be a 9. three ways to do this:009, 900,090 now it is possibel that only one is zero. three ways to do this and for erach one you have the following possibilities for the other two 18 81 27 72 36 63 45 54 99 for a grand total of 9*3=27 possiblilites in thi scase. so i think there are 1+3+27=31 possibilities. i could be wrong