+0  
 
0
58
1
avatar+85 

For each positive integer n, let S(n) denote the sum of the digits of n. How many three-digit n's are there such that  \(n+S(n)+S(S(n))\equiv 0 \pmod{9}\)?

 

Thanks for helping! Also- please explain your answer so I can better understand it!

 

(Btw I reposted this because the other one had an answer that wasn't explained; I had no idea how they got that)

 Sep 18, 2020
 #1
avatar
0

If I understand your question, then there are:

 

108  117  126  135  144  153  162  171  180  189  198  207  216  225  234  243  252  261  270  279  288  297  306  315  324  333  342  351  360  369  378  387  396  405  414  423  432  441  450  459  468  477  486  495  504  513  522  531  540  549  558  567  576  585  594  603  612  621  630  639  648  657  666  675  684  693  702  711  720  729  738  747  756  765  774  783  792  801  810  819  828  837  846  855  864  873  882  891  900  909  918  927  936  945  954  963  972  981  990  999  Total =  100 such numbers.

 

You simply start with a 3-digit number that is divisible by 9. Take the first one of 108. This is divisible by 9 because: 1 + 0 + 8 =9. Then you add this sum of 9 to 108: 108 +9 =117. Then add up the digits of 117 = 1 + 1+ 7 =9. Finally, you add this last 9 to 117: 9 + 117 =126 which is divisible by 9. Or 126 mod 9 =0.

 

Note: All the  100 numbers above meet your conditions. However, by the time you add their digit up 3 times, many of them will exceed 3 digits: Take 999. 9 + 9 + 9 =27. 27 + 999 =1026: and 1+0+2+6=9: 9 + 1026 =1035.........and so on. Whether, this is acceptable to your Math Teacher, that I don't know.

 Sep 18, 2020

58 Online Users

avatar