Computer Science Question and Answer 2022

COMPUTER SCIENCE
STUDENT ID:
[Pick the date]

Question 1
(a) Here a, b, m, n € Z and also, m>0 and n>0
Let
a = b (mod n)
From definition a- b = k n (for some random integer k)
Now, m | n
n = m p (for some random integer p)
Thus,
a-b = k (m p)
After using associative and commutative properties of the product of the integers.
a-b = m (k p)
Here, as k and p are the integers and k p is also an integer thus, a -b would be integral
multiple of the m. In mathematical expression m | a-b
Therefore,
a = b (mod n)
(b) Prove / Disprove of statements
(i) If a = b mod m, a2=b2 mod m
Here,
a = b mod m → a-b =0 mod m
Such that
a2 – b2 = (a-b) (a+b) = 0 mod m → a2 =b2 mod m
Hence, it can be said that the above statement is true.
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(ii) If a2= b2 mod m, a = b mod m
As, m| (a2 – b2) and a2 – b2 = (a-b) (a+b)
As a result of this, m | (a-b) or m |(a+b)
Thus,
a = ±b hence, it can be that the above statement is false.
(iii) If a = b mod m, a2 = b2 mod m2
As, m| (a2 – b2) and a2 – b2 = (a-b) (a+b)
As a result of this, m | (a-b) or m |(a+b)
Further squaring a = b mod m, it is apparent that the given statement is true.
(c) The given numbers in the numerator can be represented as follows.
100 = (99+1)
101 = (99+2)
102 = (99+3)
103 = (99+4)
Hence, the given expression is [(99+1)*(99+2)*(99+3)*(99+4)]/99
On multiplying the above four terms in the numerator, all the resultant terms would be
multiple of 99 except the term 1*2*3*4 or 24. Since all the other terms would be multiple of
99, hence these would be perfectly divisible by 99 leaving 24 as the remainder.
Thus, the requisite remainder is 24.
(d) (10011001)/3 = (999+2)1001/3
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