Three got mad!

For every real positive integer n=akak1ak2...a0n = \overline{a_ka_{k-1}a_{k-2}...a_0} can be written as n=a0×100+a1×101+...+ak×10k n = a_0 \times 10^0 + a_1 \times 10^1 + ... + a_k \times 10^k . For example, 576=5×102+7×101+6×100 576 = 5 \times 10^2 + 7 \times 10^1 + 6 \times 10^0 .

For every real positive integer AA , let S(A) S(A) denotes the sum of the digits of A A. For example, S(576)=5+7+6=18 S(576) = 5 + 7 + 6 = 18 .

Question 1. For real positive integer N=akak1ak2...a0 N = \overline{a_ka_{k-1}a_{k-2}...a_0} , prove that S(N)N S(N) - N is divisible by 33.

Question 2. From Q1, please explain why is N N divisible by 33 if and only if S(N) S(N) is divisible by 33

Question 3. Let mm and nn be two positive real integer. And let xx be a number that is obtained by concatenating mm and nn. For example, if m=23 m = 23 and n=546 n = 546, then x=23546 x = 23546 . Prove that S(x)S(m)S(n) S(x) - S(m) - S(n) is divisible by 33. And also, prove that x=m+n (mod 3) x = m + n \space (\text{mod} \space 3) .

Question 4. A sequence b1,b2,b3,... b_1 , b_2 , b_3, ... defined as written below.

b1=1;b2=12;b3=123;b4=1234;b5=12345; b6=123456 etc b_1 = 1; \quad b_2 = 12; \quad b_3 = 123; \quad b_4 = 1234; b_5 = 12345; \space b_6 = 123456 \space \text{etc}.

For every positive integer k k, prove that b3k+1 b_{3k + 1} is not divisible by 33.

Question 5. Based on the sequence from Q4, among b1,b2,b3,...,b2017 b_1 , b_2 , b_3 , ... , b_{2017} , how many number are there that are divisible by 33 ?

Note by Fidel Simanjuntak
2 years, 4 months ago

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