+0

# I need help :P

+1
1294
3
+279

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 9 exactly once in each integer. What is the sum of the integers on Camy's list?

Nov 8, 2018

#1
+23593
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Here is my 'shot' at it

there are

5 x 4 x 3 x 2 x 1   =  120 possible numbers,

so each didgit must appear in each position 120/5 = 24 times

So each position of the 5 digit numbers must sum up to

24(1) + 24(3) + 24(4) + 24(5) + 24(9) = 24(1+3+4+5+9) = 528

P5      P4     P3    P2     P1

528    528   528   528   528

adding these together   (P1 would be 8  carry 52 to P2    etc)

I come up with

5,866,608     (I hope it is corrrect !)

Nov 8, 2018
#2
+1

5! = 120 possible 5-digit numbers
The average value of each digit is =1+3+4+5+9=22/5=4.4
4.4 x 10,000th place x 120 permutations of 5 numbers=5,280,000 +528,000+52,800 + 5,280 +528 =5,866,608

Which is the same as multiplying: 528 x 11,111 =5,866,608.

Nov 8, 2018
edited by Guest  Nov 8, 2018
#3
+279
0

Thank you both, I really appreciate it.

Nov 8, 2018