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Find the number of 10 digit numbers, where the sum of the digits is divisible by 10
 

 
 Nov 24, 2021
 #1
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I wrote a compute program, and got an answer of 250000000.

 
 Nov 24, 2021
 #2
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There are: 999,999,999 ten-digit numbers that are divisible by 10 as follows in C++:

 

z=10;a=matrix(11,z);
for(i,0,9,a[1][i]=1);
i=2;
loopi: j=0;
loopj: a[i][j]=sumfor(k,0,9,a[i-1][(j - k + z) % z]);
j++; if(j print 10^(i-1),"-",10^i-1,":", a[i][0] - 1;
i++; if(i

 

OUTPUT: Between 1,000,000,000 - 9,999,999,999 = 999,999,999 ten-digit numbers that are divisible by 10

 
 Nov 24, 2021
 #3
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From 1 000 000 000     to  9 999 999 999    is   9 000 000 000    ten digit numbers

    only the ones that end in a zero are divisible by 10

       that is 1/10 of the total

          900, 000,000

 
 Nov 25, 2021
 #4
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The question is about the "SUM" of the digits that are divisible by 10 !! Example: 1,000,000,009, 1,000,000,o18, 1,000,000,027.......Just add up the digits: [1 + 9], [1 + 1 + 8], [1 + 2 + 7]....and so on.

 
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