We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
343
3
avatar+233 

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
avatar+18324 
+1

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
avatar
+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
avatar+233 
0

Thank you both, I really appreciate it.

 Nov 8, 2018

7 Online Users

avatar