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  
 
-2
62
2
avatar+397 

deleted.

 Oct 29, 2019
edited by sinclairdragon428  Nov 20, 2019
 #1
avatar+23358 
+2

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?

 

I assume, we count the numbers of all permutations of 13459.

 

The sum is:

\(\begin{array}{|rcll|} \hline && \dfrac{5!}{5} \times (1+3+4+5+9)\times (10^0+10^1+10^2+10^3+10^4) \quad | \quad 5!=120 \\ &=& \dfrac{120}{5} \times 22\times 11111 \\ &=& 24 \times 22\times 11111 \\ &=&\mathbf{ 5 866 608 } \\ \hline \end{array} \)

 

The sum of the integers on Camy's list is 5866608

 

laugh

 Oct 29, 2019
 #2
avatar+397 
0

thank you!

sinclairdragon428  Oct 29, 2019

22 Online Users

avatar
avatar