Let P be 1000 times N, where N is some 2-digit positive integer. If P is a multiple of every digit 1 through 9, then what is the value of N?
If N, in the previous problem, is a 3-digit positive integer, there is no longer a unique solution. How many possible values of N are there if P is 1000 times N and a multiple of every digit 1 through 9?
Three 10-sided dice are rolled. If each die has a different digit from 0 to 9, inclusive, on each face, what is the probability that the three digits rolled can be arranged to create one of the possible 3digit values of N from the previous problem? Express your answer as a common fraction.
Good Luck
Sincerely,
John Morse
+129845 I'll try my hand at this one........
For the first one......we are looking for a 2 digit number that is divisible by both 9 and 7.......since we are multipying this by 1000, the result will be divisible by the other integers 1-9.
This number, N, is 63
For the second one.......we need all the multiples of 63 between 100 - 1000.....these are
126, 189, 252, 315, 378, 441, 504, 567, 630, 693, 756, 819, 882, 945
For the third, there are 1000 possible outcomes for each dice roll.
For each of the above numbers - except 252, 441 and 882 - there are 3! = 6 outcomes that could produce that particular number......for example, the outcome 126 plus the rearrangement of the outcomes 162, 216, 261, 612, 621 would all produce 126 ......for 252, 441 and 882, there are just 3 identifiable outcomes associated with each of these numbers......for example, 252 , 225 and 552
So the probability that the dice could be rearranged to make one of the numbers is :
[11 * 6 + 3 *3] / 1000 = [66 + 9] / 1000 = 75/1000 = 3/40
That's my best shot.....!!!!!
+129845 Edited answer for the third part :
126, 189, 252, 315, 378, 441, 504, 567, 630, 693, 756, 819, 882, 945
For the third one....I also noticed that 189, 819 and 567 and 756 actually have the same digits
So......my answer should have been
[9 * 6 + 3 *3] / 1000 = 63/1000
Sorry!!!!
