+0

+1
103
5

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 $$n+S(n)+S(S(n))\equiv 0 \pmod{9}?$$

Oct 18, 2020

#1
+1

189  198  279  288  297  369  378  387  396  459  468  477  486  495  549  558  567  576  585  594  639  648  657  666  675  684  693  729  738  747  756  765  774  783  792  819  828  837  846  855  864  873  882  891  909  918  927  936  945  954  963  972  981  990  999  Total =  55 such integers.

Example: 189 + S(1+8+9) =18 + S(S(1+8)=9)=189 + 18 +9 = 216 mod 9 = 0

Oct 18, 2020
#2
0

doesn't seem right

Oct 18, 2020
#3
+1

Tell me, what is wrong with it? Every number in the list is congruent to: S + S(n) + S(S(n)) mod 9 =0. So, show me where the mistake is?

Guest Oct 18, 2020
#4
+1

111+3+3=117 = 0 (mod9)

I came up with the digtts must add to one of the following

3,6,9,12,15,18,21,24,27

The smallest is 111

the biggest is 999

and there are lots in between.

111

114

117

120

123

126

129

...

so there will be   (999-111)/3  +1 = 297 of them

That is what I get anyway.

Maybe you can find one of mine that does not work.  I certainly have not checked them.

Oct 18, 2020
#5
0

Oh, If you want to know how I came up with this answer then just ask. Melody  Oct 18, 2020