JEE Mathematics Prime and Composite Numbers MCQs Set 01

Answer: (b) 2

Explanation:
1. Every even number greater than 2 is divisible by 2 and so has a divisor other than 1 and itself.
2. Hence 2 is the only even prime.

Teacher's Note:
1. 2 is also the smallest prime.
2. 0 and 4 are not prime.

Answer: (a) 2 × 3 × 5²

Explanation:
1. 150 ÷ 2 = 75, 75 ÷ 3 = 25, 25 ÷ 5 = 5, 5 ÷ 5 = 1.
2. So 150 = 2 × 3 × 5 × 5 = 2 × 3 × 5².

Teacher's Note:
1. Divide by primes in increasing order.
2. 15 is not prime, so the last option is not a prime factorisation.

Answer: (d) 12

Explanation:
1. 150 = 2¹ × 3¹ × 5².
2. Number of divisors = (1 + 1)(1 + 1)(2 + 1) = 12.

Teacher's Note:
1. Add 1 to each exponent and multiply.
2. The divisors include 1 and 150.

Answer: (c) 3

Explanation:
1. Among p, p + 2, p + 4 the residues modulo 3 are all different, so one of them is a multiple of 3.
2. That number must be the prime 3 itself.
3. Only p = 3 works: 3, 5, 7 are all prime.

Teacher's Note:
1. Test small primes: 5, 7, 9 fails; 7, 9, 11 fails.
2. A prime multiple of 3 can only be 3.

Answer: (b) 6

Explanation:
1. Among any three consecutive integers one is divisible by 3 and at least one is even.
2. So the product is divisible by 2 × 3 = 6.
3. 12 fails for 1 × 2 × 3 = 6.

Teacher's Note:
1. Test 1 × 2 × 3 = 6 to rule out larger options.
2. The general rule is divisibility by r!.

Answer: (a) 24

Explanation:
1. The product of 4 consecutive integers is divisible by 4! = 24.
2. For n = 1 the product is 24, so no larger number works.

Teacher's Note:
1. Check n = 1 to find the upper limit.
2. For n = 2 the product is 120, divisible by 24.

Answer: (d) 7

Explanation:
1. The primes are 23, 29, 31, 37, 41, 43, 47.
2. That is 7 primes.

Teacher's Note:
1. Check divisibility only by 2, 3, 5 and 7 for numbers below 50.
2. 49 = 7² is not prime.

Answer: (c) 4

Explanation:
1. 1 is neither prime nor composite; 2 and 3 are prime.
2. The first number with a divisor other than 1 and itself is 4 = 2 × 2.

Teacher's Note:
1. Composite numbers have at least three divisors.
2. 6 is composite but not the smallest.

Answer: (b) 30

Explanation:
1. Use the three smallest primes: 2 × 3 × 5 = 30.
2. Any other choice of three distinct primes gives a larger product.

Teacher's Note:
1. Take the smallest primes to minimise the product.
2. 24 = 2³ × 3 has only two distinct prime factors.

Answer: (a) 22

Explanation:
1. x(a − 10) = a gives x = 1 + 10/(a − 10).
2. For positive integer x, a − 10 must be a positive divisor of 10: 1, 2, 5, 10.
3. Then x = 11, 6, 3, 2, and the sum is 22.

Teacher's Note:
1. Negative divisors of 10 give x ≤ 0, so they are rejected.
2. Do not count a − 10 = 2 twice.

Answer: (d) p is even

Explanation:
1. p² = 2q² is even, so p² is even.
2. The square of an odd number is odd, so p must be even.
3. Writing p = 2m leads to q even as well, a contradiction.

Teacher's Note:
1. An even square comes only from an even number.
2. The contradiction is that both p and q would have the common factor 2.

Answer: (c) \( \frac{a - b}{b} \)

Explanation:
1. 1 + √2 = a/b gives √2 = a/b − 1 = (a − b)/b.
2. This would be rational, contradicting the irrationality of √2.

Teacher's Note:
1. Subtract 1 from both sides.
2. A rational minus a rational is rational.

Answer: (b) 13

Explanation:
1. 1001 = 7 × 143 = 7 × 11 × 13.
2. The largest prime factor is 13.

Teacher's Note:
1. Try dividing by 7, 11, 13 in turn.
2. 143 is composite (11 × 13).

Answer: (a) 24

Explanation:
1. p is odd and not a multiple of 3, so p − 1 and p + 1 are consecutive even numbers, one a multiple of 4, and one of p − 1, p, p + 1 is a multiple of 3.
2. Hence 8 and 3 divide (p − 1)(p + 1), so 24 divides it.
3. For p = 5, p² − 1 = 24, so no larger number works for all p.

Teacher's Note:
1. Test p = 5, 7, 11: 24, 48, 120.
2. Their HCF is 24.

Answer: (d) (17, 19)

Explanation:
1. 17 and 19 are both prime and differ by 2.
2. 13 and 17 differ by 4, 27 = 3³ is not prime, and 7 and 11 differ by 4.

Teacher's Note:
1. Check both numbers for primality, then the difference.
2. Twin primes below 20: (3, 5), (5, 7), (11, 13), (17, 19).

Answer: (c) 91

Explanation:
1. 91 = 7 × 13.
2. 89, 97 and 83 have no divisor among 2, 3, 5, 7, so they are prime.

Teacher's Note:
1. Test divisibility by primes up to the square root.
2. 91 is a common trap because it is not divisible by 2, 3 or 5.

Answer: (b) 6

Explanation:
1. p < √200 ≈ 14.1, so p ∈ {2, 3, 5, 7, 11, 13}.
2. 17² = 289 is too large.

Teacher's Note:
1. Find the largest prime below 14.14.
2. Count 2 as a prime.

Answer: (a) 28

Explanation:
1. The first five primes are 2, 3, 5, 7, 11.
2. Sum = 28.

Teacher's Note:
1. Remember 2 is the first prime.
2. 1 is not prime.

Answer: (d) 8

Explanation:
1. 30 = 2 × 3 × 5, so we exclude multiples of 2, 3 or 5.
2. The numbers are 1, 7, 11, 13, 17, 19, 23, 29: 8 numbers.

Teacher's Note:
1. Coprime means no common prime factor.
2. Euler's function: 30 × (1/2) × (2/3) × (4/5) = 8.

Answer: (c) 6

Explanation:
1. n³ − n = (n − 1)n(n + 1), a product of three consecutive integers.
2. It is always divisible by 6, and n = 2 gives exactly 6.

Teacher's Note:
1. Factorise as (n − 1)n(n + 1).
2. Test n = 2: 8 − 2 = 6.