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- Mon Dec 04, 2017 1:09 pm
- Forum: Junior Level
- Topic: The Chinese Remainder Theorem
- Replies: 1
- Views: 462
Consider the following simultaneous congruence. x=3 (mod II), x = 5 (mod 6). It is easy to find a solution, x = 47, by inspection. Here's another method. Since 61-II, we can find a linear combination of 6 and II that equals one, for example, ( -I) . II + 2·6 = I. Now compute 5·(-1)·11 +3·2·6= -19. T...
Let f(n) denote the sum of the digits of n. (a) For any integer n, prove that eventually the se quence f(n),f(f(n) ),f(f(f(n))), . . . will become constant. This constant value is called the digital sum of n. (b) Prove that the digital sum of the product of any two twin primes, other than 3 and 5, i...