+0  
 
+1
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13
avatar+187 

How many five digit even integers have a digit sum of 13?

 

edit: i dont want them listed I want a way to do it

 Jul 15, 2019
edited by Mathgenius  Jul 15, 2019
 #1
avatar
0

There are 968 such numbers that begin with:

10048  10066  10084  10138  10156  10174  10192  10228  .............etc.

Note: If you want them all listed, just let us know.

 Jul 15, 2019
 #2
avatar+23557 
0

Never mind !    I see it says 'even' numbers    !!!!!!

ElectricPavlov  Jul 15, 2019
edited by ElectricPavlov  Jul 15, 2019
 #3
avatar+187 
0

its even numbers

Mathgenius  Jul 15, 2019
 #4
avatar+23557 
0

Sorry, M-G.....saw that a little too latecheeky

ElectricPavlov  Jul 15, 2019
 #5
avatar+30021 
+2

Here is a "brute force and ignorance" piece of pseudo code to count the number of five-digit even integers that have a digit sum of 13:

 

Set n = 0                                       n is the counter

for k = 10000 : 2 : 99998              loop from 10000 to 99998 in steps of 2

       a = floor(k/10^4)

       t = k – a*10^4

       b = floor(t/10^3)

       t = t – b*10^3

       c = floor(t/10^2)

       t = t – c*10^2

       d = floor(t/10)

       t = t – d*10

       sum = a + b + c + d + t

       if sum ==13 then n = n+1

end for loop

display n

 Jul 15, 2019
 #6
avatar
0

Alan: Here is a "brute force and ignorance" REAL code that lists 968 numbers with a sum total of 13. But, that is not what the questioner wants! He/she, I believe, wants a solution using combinations and permutations, and I haven't the faintest idea how to approach it!.

 


n=0;p=0;cycle:a(10000+n);b=int(a/10000);c=int(a/1000);d=c%10;e=int(a/100);f=e%10;g=int(a/10);h=g%10;i=int(a/10);j=a%10;n=n+1;if(a%2==0 and b+d+f+h+j==13, goto loop,goto cycle);loop:p=p+1;printa," ",;if(n<84001, goto cycle, 0);print"Total = ",p

 Jul 15, 2019
 #7
avatar+30021 
0

I have no idea how to do it using permutations and combinations either - that's why I listed the pseudo code!

Alan  Jul 15, 2019
 #8
avatar
0

OK, young person, here is a solution to your problem courtesy of Wolfram/Alpha:
expand | (x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9) (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9)^4:


x^45 + 5 x^44 + 15 x^43 + 35 x^42 + 70 x^41 + 126 x^40 + 210 x^39 + 330 x^38 + 495 x^37 + 714 x^36 + 992 x^35 + 1330 x^34 + 1725 x^33 + 2170 x^32 + 2654 x^31 + 3162 x^30 + 3675 x^29 + 4170 x^28 + 4620 x^27 + 4998 x^26 + 5283 x^25 + 5460 x^24 + 5520 x^23 + 5460 x^22 + 5283 x^21 + 4998 x^20 + 4620 x^19 + 4170 x^18 + 3675 x^17 + 3162 x^16 + 2654 x^15 + 2170 x^14 + 1725 x^13 + 1330 x^12 + 992 x^11 + 714 x^10 + 495 x^9 + 330 x^8 + 210 x^7 + 126 x^6 + 70 x^5 + 35 x^4 + 15 x^3 + 5 x^2 + x.


So, the answer is the coefficient x^13, which is 1725 minus 757, which is number that sums up to 13 for all ODD numbers.
Therefore, the total number of all EVEN 5-digit numbers that sum up to 13 is =1,725 - 757 =968 such numbers.

 Jul 15, 2019
 #9
avatar+187 
0

how did you get how many odd numbers there are?

Mathgenius  Jul 16, 2019
 #10
avatar
0

By the above computer code.

Guest Jul 16, 2019
 #11
avatar+24948 
+5

How many five digit even integers sums up to 13

 

\(\begin{array}{|l|l|r|r|r|} \hline \text{5 digit even integers} & \text{partition} & \text{permutation} & - \text{partition} &- \text{permutation} \\ \hline 9\{4,0,0\}0 & P(4,1), P(4,2), P(4,3) & \\ &\{4,0,0\},\{3,1,0\},\{2,1,1\} & \binom{6}{2} \\ & \qquad\qquad \{2,2,0\} & \\ 9\{2,0,0\}2 & P(2,1), P(2,2), P(2,3) & \\ &\{2,0,0\},\{1,1,0\} & \binom{4}{2} \\ 9\{0,0,0\}4 & 1 & \frac{3!}{3!}=1= \binom{2}{2} \\ \hline 8\{5,0,0\}0 & P(5,1), P(5,2), P(5,3) & \\ &\{5,0,0\},\{4,1,0\},\{3,1,1\} & \binom{7}{2} \\ & \qquad\qquad \{3,2,0\},\{2,2,1\} & \\ 8\{3,0,0\}2 & P(3,1), P(3,2), P(3,3) & \\ &\{3,0,0\},\{2,1,0\},\{1,1,1\} & \binom{5}{2} \\ 8\{1,0,0\}4 &P(1,1), P(1,2), P(1,3) & \\ &\{1,0,0\} & \binom{3}{2} \\ \hline 7\{6,0,0\}0 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 7\{4,0,0\}2 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ 7\{2,0,0\}4 & P(2,1), P(2,2), P(2,3) & \binom{4}{2} \\ 7\{0,0,0\}6 & 1 & \frac{3!}{3!}=1= \binom{2}{2} \\ \hline 6\{7,0,0\}0 & P(7,1), P(7,2), P(7,3) & \binom{9}{2} \\ 6\{5,0,0\}2 & P(5,1), P(5,2), P(5,3) & \binom{7}{2} \\ 6\{3,0,0\}4 & P(3,1), P(3,2), P(3,3) & \binom{5}{2} \\ 6\{1,0,0\}6 & P(1,1), P(1,2), P(1,3) & \binom{3}{2} \\ \hline 5\{8,0,0\}0 & P(8,1), P(8,2), P(8,3) & \binom{10}{2} \\ 5\{6,0,0\}2 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 5\{4,0,0\}4 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ 5\{2,0,0\}6 & P(2,1), P(2,2), P(2,3) & \binom{4}{2} \\ 5\{0,0,0\}8 & 1 & \frac{3!}{3!}=1= \binom{2}{2} \\ \hline 4\{9,0,0\}2 & P(9,1), P(9,2), P(9,3) & \binom{11}{2} \\ 4\{7,0,0\}2 & P(7,1), P(7,2), P(7,3) & \binom{9}{2} \\ 4\{5,0,0\}4 & P(5,1), P(5,2), P(5,3) & \binom{7}{2} \\ 4\{3,0,0\}6 & P(3,1), P(3,2), P(3,3) & \binom{5}{2} \\ 4\{1,0,0\}8 & P(1,1), P(1,2), P(1,3) & \binom{3}{2} \\ \hline 3\{10,0,0\}0 & P(10,1), P(10,2), P(10,3) & \binom{12}{2} & \{10,0,0\} & - \frac{3!}{1!2!} \\ 3\{8,0,0\}2 & P(8,1), P(8,2), P(8,3) & \binom{10}{2} \\ 3\{6,0,0\}4 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 3\{4,0,0\}6 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ 3\{2,0,0\}8 & P(2,1), P(2,2), P(2,3) & \binom{4}{2} \\ \hline 2\{11,0,0\}0 & P(11,1), P(11,2), P(11,3) & \binom{13}{2} & \{11,0,0\} & - \frac{3!}{1!2!} \\ & & & \{10,1,0\} & - \frac{3!}{1!1!1!} \\ 2\{9,0,0\}2 & P(9,1), P(9,2), P(9,3) & \binom{11}{2} \\ 2\{7,0,0\}4 & P(7,1), P(7,2), P(7,3) & \binom{9}{2} \\ 2\{5,0,0\}6 & P(5,1), P(5,2), P(5,3) & \binom{7}{2} \\ 2\{3,0,0\}8 & P(3,1), P(3,2), P(3,3) & \binom{5}{2} \\ \hline 1\{12,0,0\}0 & P(12,1), P(12,2), P(12,3) & \binom{14}{2} & \{12,0,0\} & - \frac{3!}{1!2!} \\ & & & \{11,1,0\} & - \frac{3!}{1!1!1!} \\ & & & \{10,2,0\} & - \frac{3!}{1!1!1!} \\ & & & \{10,1,1\} & - \frac{3!}{1!2!} \\ 1\{10,0,0\}2 & P(10,1), P(10,2), P(10,3) & \binom{12}{2} & \{10,0,0\} & - \frac{3!}{1!2!} \\ 1\{8,0,0\}4 & P(8,1), P(8,2), P(8,3) & \binom{10}{2} \\ 1\{6,0,0\}6 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 1\{4,0,0\}8 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ \hline \end{array} \)

 

Sum off all permutations:

\(\begin{array}{|rcll|} \hline && \binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2} \quad &|\quad 9\ldots , \text{ and } 8\ldots \\ &+& \binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2} \quad &|\quad 7\ldots ,\ \text{ and } 6\ldots \\ &+& \binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2} \quad &|\quad 5\ldots ,\ \text{ and } 4\ldots \\ &+& \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2}+ \binom{12}{2}+ \binom{13}{2}- 2\times\frac{3!}{1!2!}- 1\times \frac{3!}{1!1!1!} \quad &|\quad 3\ldots ,\ \text{ and } 2\ldots \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 3\times\frac{3!}{1!2!}- 2\times \frac{3!}{1!1!1!} \quad &|\quad 1\ldots \\\\ &=& \underbrace{\binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2} }_{= \binom{8}{3}\text{( hockey stick identity)} } \quad &|\quad 9\ldots , \text{ and } 8\ldots \\ &+& \underbrace{\binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}}_{=\binom{10}{3}\text{( hockey stick identity)} } \quad &|\quad 7\ldots ,\ \text{ and } 6\ldots \\ &+& \underbrace{\binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2} }_{=\binom{12}{3}\text{( hockey stick identity)} } \quad &|\quad 5\ldots ,\ \text{ and } 4\ldots \\ &+& \underbrace{\binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2}+ \binom{12}{2}+ \binom{13}{2} }_{=\binom{14}{3} -\binom{3}{2} -\binom{2}{2} \text{( hockey stick identity)} }- 2\times\frac{3!}{1!2!}- 1\times \frac{3!}{1!1!1!} \quad &|\quad 3\ldots ,\ \text{ and } 2\ldots \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 3\times\frac{3!}{1!2!}- 2\times \frac{3!}{1!1!1!} \quad &|\quad 1\ldots \\\\ &=& \binom{8}{3} + \binom{10}{3} + \binom{12}{3} + \binom{14}{3} -\underbrace{\left(\binom{2}{2} +\binom{3}{2}\right)}_{=\binom{4}{3}\text{( hockey stick identity)} } \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 5\times\frac{3!}{1!2!}- 3\times \frac{3!}{1!1!1!} \\\\ &=& \binom{8}{3} + \binom{10}{3} + \binom{12}{3} + \binom{14}{3} - \binom{4}{3} \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 5\times\frac{3!}{1!2!}- 3\times \frac{3!}{1!1!1!} \\\\ && \boxed{\binom{8}{2}+\binom{8}{3} = \binom{9}{3} \\ \binom{10}{2}+\binom{10}{3} = \binom{11}{3} \\ \binom{12}{2}+\binom{12}{3} = \binom{13}{3} \\ \binom{14}{2}+\binom{14}{3} = \binom{15}{3} } \\\\ &=& \mathbf{\binom{9}{3} + \binom{11}{3} + \binom{13}{3} + \binom{15}{3} - \binom{4}{3} + \binom{6}{2} - 5\times\frac{3!}{1!2!}- 3\times \frac{3!}{1!1!1!}} \\\\ &=& \binom{9}{3} + \binom{11}{3} + \binom{13}{3} + \binom{15}{3} - \binom{4}{3} + \binom{6}{2} - 5\times 3- 3\times 6 \\\\ &=& \binom{9}{3} + \binom{11}{3} + \binom{13}{3} + \binom{15}{3} - \binom{4}{3} + \binom{6}{2} -33 \\\\ &=& 84 + 165 + 286 + 455 - 4 + 15 -33 \\ &=& \mathbf{968} \\ \hline \end{array}\)

 

laugh

 Jul 16, 2019
edited by heureka  Jul 16, 2019
 #12
avatar+1904 
+3

An Amazing and Very COOL Presentation, Heureka! smiley

GingerAle  Jul 16, 2019
 #13
avatar+24948 
+3

Thank you, GingerAle !

 

laugh

heureka  Jul 16, 2019

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