+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)
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.
