This set of Aptitude Questions and Answers (MCQs) focuses on “Prime and Co-prime Numbers”.
1. If l, m and n are prime numbers, such that l<m<n<13. In how many cases (l + m + n) is also a prime number?
a) 0
b) 1
c) 2
d) 3
View Answer
Explanation: The prime numbers less than 13 are 2, 3, 5, 7 and 11.
Also, l<m<n<13 and (l+m+n) is a prime number.
Hence, only two possible pairs exist i.e., (3, 5, 11) and (5, 7, 11).
2. If A, A+2 and A+4 are prime numbers, then how many possible solutions do A have?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: When A is a prime number, there is only one possible case that A, A+2 and A+4 are prime numbers, i.e., 3,5,7. Therefore, number of possible solutions is 1.
3. Every prime number of the form 3x + 1 can be represented in the form 6y + 1 where, x and y are integers, then what should be the values of x?
a) Odd number
b) Perfect square
c) Natural number
d) Even number
View Answer
Explanation: Every prime number of the form 3x + 1 can be represented in the form 6y + 1, only when x is even.
For example, when x = 2, 3x + 1 i.e., 7 which is a prime number can also be represented by 6y + 1, where y=1.
When x = 4, 3x + 1 i.e., 13 which is a prime number can also be represented by 6y + 1, where y=2.
4. In how many ways the number 3234 can be written as the product of two numbers which are co-prime to each other?
a) 8
b) 4
c) 16
d) 32
View Answer
Explanation: Formula is 2n-1, where n is the distinct primes in the factorized form of the number.
3234 = 2*3*72*11.
Here, n=4.
So, 3234 can be written as the product of 2 co-prime numbers in 24-1 = 8 ways.
5. How many integers between 201 and 401 are co-prime to 20?
a) 70
b) 60
c) 80
d) 90
View Answer
Explanation: 20 = 22 * 5
So, we need to find the numbers which do not have 2 or 5 as factors.
Numbers between 201 and 401 which are multiple of 2 = 200/2 = 100.
Numbers between 201 and 401 which are multiple of 5 = 200/5 = 40.
Now we have to find numbers which are multiples of both 2 and 5, i.e., multiple of 10, as these numbers would be counted twice while counting the numbers divisible by 2 and 5.
Numbers between 201 and 401 which are multiple of 10 = 200/10 = 20.
So, total number of integers between 201 and 401 are co-prime to 20 is = 200 – (100 + 40 – 20) = 80.
6. Find the number of co-primes to the number 144, that are less than the number itself?
a) 48
b) 32
c) 64
d) 56
View Answer
Explanation: If N is the number that can be written as ap, bq, cr, ……, then the number of co-primes of N, which are less than N is = N*(1-\(\frac{1}{a}\))*(1-\(\frac{1}{b}\))*(1-\(\frac{1}{c}\))……
144 = 24 * 32
The number of co-primes to the number 144, that are less than the number itself is
= 144*(1-\(\frac{1}{2}\))*(1-\(\frac{1}{3}\)) = 48.
7. Find the sum of co-primes to 96, that is less than the number itself?
a) 1536
b) 3072
c) 768
d) 2304
View Answer
Explanation: The sum of co-primes of N, that is less than N is
= \(\frac{N}{2}\)*number of co primes of N,which are less than N
If N is the number that can be written as ap, bq, cr, ……, then the number of co-primes of N, which are less than N is = N*(1-\(\frac{1}{a}\))*(1-\(\frac{1}{b}\))*(1-\(\frac{1}{c}\))……
96 = 25 * 3
The number of co-primes to the number 96, that are less than the number itself is
= 96*(1-\(\frac{1}{2}\))*(1-\(\frac{1}{3}\)) = 32.
The sum of co-primes to 96, that is less than the number itself is = \(\frac{96}{2}\)*32 = 1536.
8. How many positive integers from 100 to 200 are co-prime to 630?
a) 25
b) 50
c) 40
d) 76
View Answer
Explanation: 630 = 2 * 32 * 5 * 7.
Total numbers from 100 to 200 = 101
Numbers from 100 to 200 which are divisible by 2 = 51
Numbers from 100 to 200 which are divisible by 3 = 33
Numbers from 100 to 200 which are divisible by 5 = 21
Numbers from 100 to 200 which are divisible by 7 = 14
Total = 119
Now, we have to find out double counting cases and remove them.
Numbers from 100 to 200 which are divisible by 2
by 2*3 = 17
by 2*5 = 11
by 2*7 = 7
by 3*5 = 7
by 3*7 = 5
by 5*7 = 3
by 2*3*5 = 3
by 2*3*7 = 2
by 2*5*7 = 1
by 3*5*7 = 1
by 2*3*5*7 = 0
Total to be removed = (17+11+7+7+5+3) – (3+2+1+1) = 43.
So, positive integers from 100 to 200 are co-prime to 630 is = 101 – (119 – 43) = 25.
9. If X is a natural number less than 100, then for how what values of N are the numbers 6X+1 and 15X+2 are relatively co-prime?
a) For all values of X less than 100
b) For all values of X less than 50
c) X = 100
d) For all values of X less than 10
View Answer
Explanation: Suppose 6X+1 and 15X+2 are not relatively co-prime.
Let P be a prime number dividing both of them.
Then P also divides 5(6X+1) and 2(15X+2), i.e., P divides 30X+5 and 30X+4.
Hence, P divides the difference (30X+5) – (30X+4) = 1.
Now 1 is not divisible by any other number. Hence it contradicts our assumption.
So, the pair 6X+1 and 15X+2 are relatively co-prime for all the values of X less than 100.
10. Select the set of numbers which are not mutually co-prime.
a) 31, 37, 65
b) 31, 32, 33
c) 72, 73, 74
d) 44, 45, 47
View Answer
Explanation: (31, 37, 65), (31, 32, 33) and (44, 45, 47) have no common factors in between them. Hence, they are mutually co-prime.
(72, 73, 74) has a common factor of 2 between first and third term. Hence, it is not mutually co-prime.
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