Number System- 2 - GATE CSE (CSE) Digital Logic Free MCQ Test with solutions

Given:
3240
Concept:
If k = a^x × b^y, then
a, and b must be prime number 
Sum of all factors = (a⁰ + a¹ + a² + ….. + a^x) (b⁰ + b¹ + b² + ….. + b^y)
Solution:
3240 = 2³ × 3⁴ × 5¹
Sum of factors = (2⁰ + 2¹ + 2² + 2³) (3⁰ + 3¹ + 3² + 3³ + 3⁴) (5⁰ + 5¹)
⇒ (1 + 2 + 4 + 8) (1 + 3 + 9 + 27 + 81) (1 + 5)
⇒ 15 × 121 × 6
⇒ 10890
∴ required sum is 10890

Concept used:
A twin prime is a prime number that is either 2 less or 2 more than another prime number.
The difference between the twin prime number is always two.
In twin prime number, both the number should be the prime number.

Calculation:
Twin primes are pairs of successive primes that differ by two. 
The primes from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Options:
(37, 41) - Difference between them is 4.
(3, 7) - The difference between them is 4.
(43, 47) - Difference between them is 4.
(71, 73) - Difference between them is 2.
Here, in the given option (71 and 73) are prime numbers and their difference is '2'.

Given:
Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.
Concept:
LCM: It is a number which is a multiple of two or more numbers.
Calculation:
LCM of (6, 12, 15, 20) = 60
All 4 bells ring together again after every 60 seconds
Now,
In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting) = 121 times
∴ In 2 hours they ring together for 121 times

Mistake Points
In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

Given:
HCF = 24
LCM = 168
Ratio of numbers = 1 ∶ 7.
Formula:
Product of numbers = LCM × HCF
Calculation:
Let numbers be x and 7x.
x × 7x = 24 × 168
⇒ x² = 24 × 24
⇒ x = 24
∴ Larger number = 7x = 24 × 7 = 168.

Given:
676xy is divisible by 3, 7 & 11
Concept:
When 676xy is divisible by 3, 7 &11, it will also be divisible by the LCM of 3, 7 &11. 
Dividend = Divisor × Quotient + Remainder
Calculation:
LCM (3, 7, 11) = 231
By taking the largest 5-digit number 67699 and divide it by 231.
∵ 67699 = 231 × 293 + 16
⇒ 67699 = 67683 + 16 
⇒ 67699 - 16 = 67683 (completely divisible by 231)
∴ 67683 = 676xy (where x = 8, y = 3)
(3x - 5y) = 3 × 8 - 5 × 3
⇒ 24 - 15 = 9 
∴ The required result = 9

A number is divisible by 11 if the difference of the sum of digits on odd places and the sum of digits on even places is zero or divisible by 11. So, we have the number A381.
Hence, (A + 8) - (3 + 1) = either 0 or multiple of 11
If we put A = 7, then we get
(7 + 8) - (3 + 1) = 15 - 4 = 11 (Which is divisible by 11).

According to the question when 'n' is divided by 5, we will get the remainder 2
n/5 = remainder 2
On putting n =7, and on dividing it with 5 we will get remainder 2
So, n = 7, and n² = 49
n²/5 = 49/5 = remainder 4

Given, 5^{(x + 3)} = 25^{(3x - 4)}
We can write it as -
5^{(x + 3)} = 5^{2 x (3x - 4)}
Or,
x + 3 = 2(3x - 4)
x + 3 = 6x - 8
or, 5x = 11
So, x = 11/5

Given a = -4 and b = -2
On putting the values in the given equation, we will get -
(-4)³ - 3 x (-4)² + 3 x (-4) + 3 x (-2) + 3 x (-2)² + (-2)³
= -64 - 48 - 12 - 6 + 12 - 8
= -126

Given ration 7 : 9
So, let the numbers be 7a and 9a. According to the question -
7a * 9a = 1575
63a² = 1575
a² = 1575/63
Or a² = 25
And a =5
So, the two numbers will be -
7a = 7 * 5 = 35
And 9a = 9 * 5 = 45
Hence, the greater number is 45 between the given numbers.