practice problems

Q2. For any arbitrary integer a, 3 | a (2a^2 + 7) True or false
S2.
a(2a^2 + 7) mod 3 = a(2a^2 + 1) = 2a^3 + a
Case 1: a = 0 mod 3 => True
Case 2: a = 1 mod 3 => 2 + 1 True
Case 3: a = 2 mod 3 => 4 + 2 = 6 True
True = answer

Q3. If a,b are both odd integers then a^4 + b^4 - 2 | 16 True or false.
S3.
For an even integer a^4 mod 16 = 0. Clearly.
For an odd integer a^4 mod 16 = 1. Why?
a = 2m + 1
a^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1
m(m+1) is even = 2k
=> a^2 = 8k + 1
a^4 = 64k^2 + 16k + 1
H.P.
So a^4 + b^4 - 2 = 0 mod 16 H.P.
True

Q4. If a is an odd integer then 24 | a ( a^2 - 1)
S4.
True

Method 1:
a(a^2 - 1) = (a-1).a.(a+1)
In 3 consecutive integers, one is always div by 3.
(a-1) and (a+1) are 2 consecutive even integers.
In 2 consecutive even integers, one is div by 2, other by 4.
So in total, div by 8.
So div by both 8,3 H.P.

Method 2:
a(a^2 - 1) = (a-1).a.(a+1) = (2m)(2m+1)(2m+2) = 4m(2m+1)(m+1) = 4m(m+1)(2m+1)
We need to show that m(m+1)(2m+1) is divisible by 6.
Clearly m(m+1) is div by 2.
Now show div by 3.
Case 1: m = 0 mod 3 => 0.1.(1) = 0 mod 3
Case 2: m = 1 mod 3 => 1.2.(3) = 0 mod 3
Case 3: m = 2 mod 3 => 2.3.(7) = 0 mod 3
H.P.

Q5. The expression (3n)!/(3!) yields a non-integral real number. True/False?
S5.
False.
Any multiplication of k consecutive integers is divisible by k!.
We can rewrite it as:
1.2.3/3! * 4.5.6/3! * 7.8.9/3!... (3n-2)(3n-1).3n/3!
So it is indeed an integer.