+0  
 
0
38
1
avatar

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)) (mod  9)?

 May 27, 2021
 #1
avatar+113616 
+1

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)) (mod  9)?

 

I am having a hard time getting my head around this question.

 

I think    n(mod9) = s(n) (mod9)

So that would mean that  S(S(n)) mod 9 must equal 0

 

\(100\le n \le 999\\ 1\le S(n) \le 27\\ 1 \le S(S(n))\le 10\\ so\\ S(S(n)) \;\;must\;=9\)

 

if

\(S(S(n))=9\;\;then\\ S(n)\; \; \text{must be 9 or 18 or 29} \)

 

So the digits of n must add to 9, 18 or 27

this means that n must be dividable by 9

So how many numbers between 100 and 999 are divisible by 9

111-11 = 100

 

I think there is 100 but I might not be thinking very clearly.

 May 27, 2021

49 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar
avatar
avatar
avatar