Camy made a list of every possible distinct five-digit positive integer that can be formed using each of the digits 1, 3, 4, 5 and 8 exactly once in each integer. What is the sum of the integers on Camy's list?
+128407 Let's look at two simpler examples and see if we can find an answer
Consider the digits 2 , 4 , 6
We have :
246
264
426
462
624
642
Now....consider the digits 1,2,3 and 4
1234 2134 3124 4123
1243 2143 3142 4132
1324 2314 3214 4213
1342 2341 3241 4231
1423 2413 3412 4312
1432 2431 3421 4321
It appears that each digit will appear n! / n times in each column where n is the number of digits we are considering
So the sum of any column of the 5 digits we are given must be
( 1 + 3 + 4 + 5 + 8) ( 5! / 5) = (21) (24) = 504
So......the sum must be
504 ( 10^3 + 10^2 + 10 + 1) = 559944
