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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 S(n) = n (mod 9) ?

 Sep 21, 2021
 #1
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Hey guest,

Try forming cases, if n == {0, 1, 2, ... ,8} mod 9

 Sep 21, 2021
 #2
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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 S(n) = n (mod 9) ?

 

I think this last bit: S(9)==n (mod 9) should read: n==S(n) mod 9, which means n mod 9==sum of digits of n.

Example: n=313 and 313 mod 9 ==7 which is the sum of digits of 313.

 

If that is correct, then you have 120 such n as follows:

(100, 101, 102, 103, 104, 105, 106, 107, 110, 111, 112, 113, 114, 115, 116, 120, 121, 122, 123, 124, 125, 130, 131, 132, 133, 134, 140, 141, 142, 143, 150, 151, 152, 160, 161, 170, 200, 201, 202, 203, 204, 205, 206, 210, 211, 212, 213, 214, 215, 220, 221, 222, 223, 224, 230, 231, 232, 233, 240, 241, 242, 250, 251, 260, 300, 301, 302, 303, 304, 305, 310, 311, 312, 313, 314, 320, 321, 322, 323, 330, 331, 332, 340, 341, 350, 400, 401, 402, 403, 404, 410, 411, 412, 413, 420, 421, 422, 430, 431, 440, 500, 501, 502, 503, 510, 511, 512, 520, 521, 530, 600, 601, 602, 610, 611, 620, 700, 701, 710, 800) >>Total No = 120 such poitive integers.

 Sep 21, 2021

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