N is a 7-digit positive integer whose product of the digits is 20160.
Determine the probability that the sum of the digits of N is a perfect square.
20160 = 2^6*3^2*5*7 so one of the digits is a 5 and another is a 7.
The other five digits contain all the 2 and 3 factors.
The 3 factors could be alone, together making a 9, or with a 2 making a 6. Here is a table with every possibility of the seven digits along with the digit sum and the divisor in the permutation count.
7598811 39 4
7598421 36 1
7598222 35 6
7594441 34 6
7594422 33 4
7533881 35 4
7533842 32 2
7533444 30 12
7563841 34 1
7563822 33 2
7563444 31 6
7566821 35 2
7566441 33 4
7566422 32 4
The only perfect square is the 36. So of all the possibilities for N, 7!
are perfect squares.
The reciprocals of the last column sum to 17/3. So N has 17/3*7! members.
So the solution is 3/17.
(If I counted wrong, I'll try to fix it.)
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Posted by Jer
on 2022-08-19 10:10:34 |
Solution if I counted right
