practice problems pending

1. Show that 10^(2n-1) + 1 is divisible by 11 for all natural number 'n'.
Solution:
10 = -1 mod 11
10^(2n-1) = (-1)^(2n-1) mod 11 = -1 mod 11
=> 10^(2n-1) + 1 = 0 mod 11
H.P.

2. Show that
11^(n+2) + 12^(2n+1) is divisible by 133 for all natural 'n'.

Solution:
133 = 7*19
So LHS should be div by both 7 and 19

Try 7 first:
11 mod 7 = 4, 12 mod 7 = 5
LHS = 4^(n+2) + 5^(2n+1)

5 mod 7 = -2 => 5^2 mod 7 = 4 mod 7
LHS = 16.4^n + 5.4^n = 21.4^n mod 7 = 0 mod 7

Now do the same with 19.
LHS = 121.11^n + 12.12^2n
12^2n = 144^n
144 = 11 mod 19
144^n = 11^n mod 19
121 = 7 mod 19

LHS = 7.11^n + 12.11^n = 19.11^n = 0 mod 19
H.P.

3. Find all integers a such that the quadratic expression (x+a)(x+1991) + 1 can be factored as (x+b)(x+c), where b and c are integers.

Solution:
Since -b is a root of RHS
=> 
(-b+a)(-b+1991) = -1

Case 1:
(-b+a) = 1
(-b+1991) = -1

=> b = 1992, a = 1993

Case 2:
(-b+a) = -1
(-b+1991) = 1
=> b = 1990, a = 1989

Answer: a = 1989, 1993.





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