# PE 3 - Numbers | Algebra - Number Theory

**PE 3 - Numbers | Algebra - Number Theory**

For how many of the first 200 odd natural numbers is the total number of factors even?

Answer: 193

**Explanation** :

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

Out of the first 300 odd natural numbers, there are 7 perfect squares i.e., 1, 9, 25, 49, 81, 121 and 169.

Hence, there are 200 – 7 = 193 numbers among the first 200 odd natural numbers that have an even number of factors.

Hence, 193.

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

In how many ways can 97 be expressed as the sum of two natural numbers that are coprime to each other?

Answer: 48

**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.

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

How many three-digit natural numbers are there such that unit’s digit of each of which is always less than tens as well as hundreds digit?

Answer: 285

**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.

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

How many natural numbers below 1000 are divisible by at least one of 2, 3, 4, 6, 8 and 10?

Answer: 666

**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.

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

A natural number N which is not a perfect cube, has exactly four factors. The sum of the positive factors of N that are less than N is 202. Find the value of N.

Answer: 398

**Explanation** :

N has only four factors

∴ N is of the form p_{1} × p_{2} or p_{2}^{3}, where p_{1}, p_{2} and p_{3} are prime numbers.

Since, N is not a perfect cube, hence N = p_{1} × p_{2},

Factors of N less than N are 1, p_{1} and p_{2} and Sum of these factors

1 + p_{1} + p_{2} = 202

p_{1} + p_{2} = 201

So, the value of p_{1} and p_{2} 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.

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

If x = 17^{5} + 17^{4} + 17 + 1 and y = 11^{8} – 11^{5} – 11^{3} – 1, then which of the following statements is true?

- A.
x and y are both prime

- B.
x is prime but y is composite

- C.
y is prime but x is composite

- D.
x and y are both composite

Answer: Option D

**Explanation** :

x = 17^{5} + 17^{4} + 17 + 1

= 17^{4} (17 + 1) + (17 + 1)

= (17^{4} + 1) × (17 + 1)

Hence, x is a composite number.

y = 11^{8} – 11^{5} – 11^{3} – 1

= (11^{8} – 1) – (11^{5} + 11^{1})

= (11^{4} – 1) (11^{4} + 1) – 11^{3} (11^{2} + 1)

= (11^{2} - 1) (11^{2} + 1) (11^{4} + 1) – 11^{3} (11^{2} + 1)

= (11^{2} + 1)[(11^{2} – 1)(11^{4} + 1) - 11^{3}]

Hence, y is a composite number.

Hence, option (d).

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

If p is a prime number greater than 3, (p^{2} – 1) is divisible by which of the following?

- A.
6

- B.
24

- C.
2

- D.
All of these

Answer: Option B

**Explanation** :

**Case 1**: p is of the form 6k + 1

p^{2} – 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

p^{2} – 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).

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

A denotes the set of all the prime numbers less than 100. How many zeroes does the product of the elements of A end with?

Answer: 1

**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.

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

The difference of a four-digit number and any number formed by permuting its digits would always be divisible by

- A.
18

- B.
11

- C.
10

- D.
9

Answer: Option D

**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).

Workspace:

**PE 3 - Numbers | Algebra - Number Theory**

N is a natural number which has 6 factors. If 10 ≤ N ≤ 50, then how many values are possible for N?

Answer: 8

**Explanation** :

Since N has 6 factors,

6 can be factorized as 6 or 2 × 3.

∴ N must be of the form P_{1}^{5} or P_{2} × P_{3}^{2} where P_{1}, P_{2}, P_{3} are prime numbers.

If N takes the form P_{1}^{5} then P_{1} can only be 2. [10 ≤ N ≤ 50]

If N takes the form P_{2} × P_{3}^{2} then the possibilities are:

2 × 3^{2}, 2 × 5^{2}, 3 × 2^{2}, 5 × 2^{2}, 5 × 3^{2}, 7 × 2^{2}, 11 × 2^{2}

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

Hence, 8.

Workspace:

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