PE 3 - Numbers | Algebra - Number Theory

Answer: 190

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Explanation :

In the first 200 odd natural numbers the lowest is 1 and the highest is (2 × 200 - 1 =) 399.

All the natural numbers that have an odd number of factors are perfect squares and vice versa. 

Out of the first 200 odd natural numbers, there are 10 perfect squares i.e., 1², 3², 5², ..., 19². 
[19 is the 10^th odd natural number]

Hence, there are 200 – 10 = 190 numbers among the first 200 odd natural numbers that have an even number of factors.

Hence, 190.

Answer: 48

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Explanation :

Let A + B = 97, HCF of A and B be K and A = ka and B = kb, where a and b are coprime.

ka + kb = 97

or K(a + b) = 97

Since 97 is a prime number, it can be expressed as product of two natural numbers in only one way i.e., 1 × 97.

So, K has to be 1.

So whenever 97 is written as sum of two natural numbers, those natural numbers will always be coprime to each other. 

97 can be expressed as sum of two numbers in 48 ways starting from (1 + 96) and ending with (48 + 49).

Hence, 48.

Answer: 285

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Explanation :

Any such number can end with one of the digits out of 0, 1, 2, 3, 4, …, 8.

When the unit’s digit is 0: Ten’s and Hundred’s digit can be any digit out of 1, 2, 3, … 8 or 9. 
Number of such numbers = 9 × 9 = 81

When the unit’s digit is 1: Ten’s and Hundred’s digit can be any digit out of 2, 3, … 8 or 9. 
Number of such numbers = 8 × 8 = 64

…

When the unit’s digit is 7: Ten’s and Hundred’s digit can be any digit out of 9 or 8. 
Number of such numbers = 2 × 2 = 4 

When the unit’s digit is 8: Ten’s and Hundred’s digit can only be 9. 
Number of such numbers = 1 × 1 = 1

∴ Total such numbers will be 1 + 4 + 9 + … + 64 + 81 = 285.

Hence, 285.

Answer: 666

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Explanation :

Number of multiples of 2 till 1000 (excluding 1000) = 499

Number of multiples of 3 till 1000 (excluding 1000) = 333

Number of multiples of both 2 and 3 i.e., of 6 till 1000 = 166

Thus, total number of multiples of 2 or 3 = 499 + 333 – 166 = 666

These numbers will account for all multiples of 4, 6, 8 and 10 as well.

Hence, 666.

Answer: 398

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Explanation :

N has only four factors
∴ N is of the form p₁ × p₂ or p₂³, where p₁, p₂ and p₃ are prime numbers.

Since, N is not a perfect cube, hence N = p₁ × p₂, 

Factors of N less than N are 1, p₁ and p₂ and Sum of these factors

1 + p₁ + p₂ = 202
p₁ + p₂ = 201

So, the value of p₁ and p₂ are 2 and 199 (in any order)

(Sum of two prime number is odd so one of the numbers must be even i.e., 2)

∴ N = 2 × 199 = 398.

Hence, 398.

Answer: Option D

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Explanation :

x = 17⁵ + 17⁴ + 17 + 1
= 17⁴ (17 + 1) + (17 + 1)
= (17⁴ + 1) × (17 + 1)
Hence, x is a composite number.

y = 11⁸ – 11⁵ – 11³ – 1
= (11⁸ – 1) – (11⁵ + 11¹)
= (11⁴ – 1) (11⁴ + 1) – 11³ (11² + 1)
= (11² - 1) (11² + 1) (11⁴ + 1) – 11³ (11² + 1)
= (11² + 1)[(11² – 1)(11⁴ + 1) - 11³]
Hence, y is a composite number.

Hence, option (d).

Answer: Option B

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Explanation :

Case 1: p is of the form 6k + 1
p² – 1 = (6k + 1)² – 1
= (6k + 1 + 1) × (6k + 1 - 1)
= (6k + 2) × 6k
= 12 × k × (3k + 1)
Since, k(3k + 1) is always divisible by 2
12 × k × (3k + 1) is always divisible by 24.

Case 2: p is of the form 6k - 1
p² – 1 = (6k - 1)² – 1
= (6k - 1 + 1) × (6k - 1 - 1)
= 6k × (6k -  2)
= 12 × k × (3k - 1)
Since, k(3k - 1) is always divisible by 2
12 × k × (3k - 1) is always divisible by 24.

Hence, option (b).

Answer: 1

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Explanation :

Product of one 2 and one 5 gives one 0 at the end.

Out of all prime numbers there in only one even prime number i.e., 2 itself and only one prime number which is a multiple of 5 i.e., 5 itself.

Hence, product the elements of A consist of only one 0 at the end.

Hence, 1.

Answer: Option D

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Explanation :

Let the four-digit number contain digits a, b, c or d.

When we write it in expanded form the digits will have one of these 4 coefficients 1000, 100, 10 or 1.

Now the difference of any two of these coefficients will always be divisible by 9.

Hence, the difference of a 4-digit number and the number formed by permuting its digits would always be divisible by 9.

Hence, option (d).

Answer: 8

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Explanation :

Since N has 6 factors, 

6 can be factorized as 6 or 2 × 3.

∴ N must be of the form P₁⁵ or P₂ × P₃² where P₁, P₂, P₃ are prime numbers.

If N takes the form P₁⁵ then P₁ can only be 2. [10 ≤ N ≤ 50]

If N takes the form P₂ × P₃² then the possibilities are:
2 × 3², 2 × 5², 3 × 2², 5 × 2², 5 × 3², 7 × 2², 11 × 2²

So, N can take 1 + 7 = 8 values in all.

Hence, 8.