Question 705380
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If the number is divisible by 2 and 5 then it is divisible by 10.  If it is divisible by 10, then the one's digit must be zero.  The 100s digit of a 3 digit number cannot be zero, so there are 5 ways to choose the 100s digit.  The 10s digit can either be any of the given digits or any of the given digits less the one you chose for the 100s digit and the zero that must be used for the last digit -- which it is is dependent on whether or not digits can be repeated in the number. So there are either 6 or 4 ways to choose the 10s digit.  And then, since the 1s digit must be zero, there is only one way to choose the last digit.


In summary, for each of the 5 ways to choose the 1st digit, there are either 4 or 6 ways to choose the second digit making a total of either 20 or 30 ways to select the first two digits.  Then there is only one way to select the last digit.  Altogether, there are eithere 20 or 30 possibilities, depending on repeatability of digits.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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