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
Stacey Warren - Expert brainly.com
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- 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
for a grand total of 9*3=27 possiblilites in thi scase. so i think there are
i could be wrong