# Factors | Algebra - Number Theory

**Factors | Algebra - Number Theory**

What is the number of factors of 151200?

Answer: 144

**Explanation** :

For a number n = a^{x} × b^{y} × c^{z}

Number of factors = (x +1)(y + 1)(z + 1)

151200 = 2^{5} × 3^{3} × 5^{2} × 7

∴ The number of factors of 70560 = (5 + 1)(3 + 1)(2 + 1)(1 + 1) = 6 × 4 × 3 × 2 = 144.

Hence, 144.

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**Factors | Algebra - Number Theory**

Find the sum of the factors of 360.

Answer: 1170

**Explanation** :

For a number n = a^{x} × b^{y} × c^{z}

Sum of factors = (a^{x} + a^{x-1} + ... + 1)(b^{y} + b^{y-1} + ... + 1)(c^{z} + c^{z-1} + ... + 1)

360 = 2^{3} × 3^{2} × 5

Sum of factors of 360 = (2^{3} + 2^{2} + 2^{1} + 1)(3^{2} + 3^{1} + 1)(5^{1} + 1)

∴ The sum of the factors of 360 = $\left[\frac{\left({2}^{\left(3+1\right)}-1\right)\left({3}^{\left(2+1\right)}-1\left)\right({5}^{(1+1)}-1)\right)}{\left(2-1\right)(3-1)(5-1)}\right]$ = 1170

Hence, 1170.

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**Factors | Algebra - Number Theory**

How many factors of 14400 are odd?

Answer: 9

**Explanation** :

14400 = 2^{6} × 3^{2} × 5^{2}

For factor to be odd, the power of 2 should be 0.

∴ Possible power of 2 in factor = 0, i.e., only 1 possibility

There is no restriction of powers of other prime factors.

⇒ The number of odd factors of 14400 = (1)(2 + 1)(2 + 1) = 9

**Note**: The odd factors of 14400 are 1, 3, 5, 9, 15, 25, 45, 75, 225.

Hence, 9.

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**Factors | Algebra - Number Theory**

How many factors of 14400 are even?

Answer: 54

**Explanation** :

14400 = 2^{6} × 3^{2} × 5^{2}

For factor to be even, the least power of 2 should be 1.

∴ Possible power of 2 in factor = 1 or 2 … or 6, i.e., 6 possibilities

There is no restriction of powers of other prime factors.

⇒ The number of even factors of 14400 = (6)(2 + 1)(2 + 1) = 54

Hence, 54.

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**Factors | Algebra - Number Theory**

In how many ways can 1925 be written as a product of two co-primes?

Answer: 4

**Explanation** :

For a number n = a^{x} × b^{y} × c^{z}

Number of ways of writing n as a product of two co-prime factors = 2^{m-1}, where m is the number of discting prime factor of n.

1925 = 5^{2} × 7 × 11

∴ The number of ways in which 1925 can be expressed as a product of two co-primes is 2^{3 – 1} = 4 (∵ there are 3 distinct prime factors i.e., 5, 7 and 11).

**Note**: The four ways in which 91091 can be expressed as a product of two co-primes is

25 × 77

11 × 175

7 × 275

1925 × 1

Hence, 4.

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**Factors | Algebra - Number Theory**

How many numbers lesser than 36 are co-prime to it?

Answer: 12

**Explanation** :

For a number n = a^{x} × b^{y} × c^{z}

Number of co-primes of n, less than n is = $\mathrm{n}\left(1-\frac{1}{\mathrm{a}}\right)\left(1-\frac{1}{\mathrm{b}}\right)\left(1-\frac{1}{\mathrm{c}}\right)$

36 = 2^{2} × 3^{2}

∴ The number of co-primes of 72, less than 72 is:

$=36\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=36\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)$ = 12

Note: The co-primes of 36, less than 36 are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35.

Hence, 12.

Workspace:

**Factors | Algebra - Number Theory**

Find the sum of co-primes of 36 which are less than 36.

Answer: 216

**Explanation** :

For a number n = a^{x} × b^{y} × c^{z}

Sum of co-primes of n, less than n is = $\frac{\mathrm{n}}{2}\times \mathrm{n}\left(1-\frac{1}{\mathrm{a}}\right)\left(1-\frac{1}{\mathrm{b}}\right)\left(1-\frac{1}{\mathrm{b}}\right)$

36 = 2^{2} × 3^{2}

∴ The sum of co-primes less than 36 = $\frac{36}{2}\times 36\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)$ = 216.

Note: 1 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 25 + 29 + 31 + 35 = 216.

Hence, 216.

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**Factors | Algebra - Number Theory**

Find the total number of factors of 104 divisible by 26.

Answer: 3

**Explanation** :

104 = 2^{3} × 13

26 = 2 × 13.

For factor to be divisible by 26, the least power of 2 should be 1 and that of 13 also should be 1.

∴ Possible power of 2 in factor = 1 or 2 or 3, i.e. 3 possibilities

Similarly, possible power of 13 in factor = 1, i.e. 1 possibility

⇒ The total number of factors of 104 divisible by 26 = (3) × (1) = 3.

Hence, 3.

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**Factors | Algebra - Number Theory**

How many factors of 5625 are not perfect squares?

Answer: 9

**Explanation** :

5625 = 3^{2} × 5^{4}

For factor to be a perfect square, the power of prime factors should be even.

Let us first calculare number of factors which can be perfect squares

∴ Possible powers of 3 in factor = 0 or 2 i.e., 2 possibilities

Similarly, possible powers of 5 in factor = 0 or 2 or 4 i.e., 3 possibilities

⇒ Number of factors which are perfect squares = 2 × 3 = 6 factors.

Also, total factors of 5625 = (2 + 1)(4 + 1) = 3 × 5 = 15 factors.

∴ Number of factors which are not perfect squares = 15 – 6 = 9.

Hence, 9.

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**Factors | Algebra - Number Theory**

Find the total number of sets of factors of 264 which are co-prime to each other.

Answer: 31

**Explanation** :

For a number n = a^{x} × b^{y} × c^{z}

The number of sets of factors that are co-prime to each other = (x + 1)(y + 1)(z + 1) - 1 + xy + yz + xz + 3xyz

⇒ 264 = 2^{3} × 3^{1} × 11^{1}

∴ x = 3; y = 1; z =1

∴ The number of sets of factors that are co-prime to each other = (3 + 1)(1 + 1)(1 + 1) – 1 + 3 + 1 + 3 + 3 (3 × 1 × 1) = (4 × 2 × 2) + 6 + 9 = 31

Hence, 31.

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**Factors | Algebra - Number Theory**

In how many ways can 360 be written as product of two of its factors?

Answer: 12

**Explanation** :

Number of ways of writing a non-perfect square as a product of two of its factors is equal to half the number of its factors.

360 = 2^{3 }× 3^{2} × 5

Number of factors of 360 = (3 + 1)(2 + 1)(1 + 1) = 24

∴ Number of ways of writing 360 as a product of two of its factors = ½ × 24 = 12 ways.

[1×360, 2×180, 3×120, 4×90, 5×72, 6×60, 8×45, 9×40, 10×36, 12×30, 15×24, 18×20]

Hence, 12.

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**Factors | Algebra - Number Theory**

In how many ways can 144 be written as product of two of its distinct factors?

Answer: 7

**Explanation** :

Number of ways of writing a perfect square as a product of two of its distinct factors is equal to half the (number of its factors - 1). Here 144 is a perfect square.

144 = 2^{4} × 3^{2}

Number of factors of 144 = (4 + 1)(2 + 1) = 15

Number of ways of writing a perfect square as a product of two of its distinct factors = ½ × (15 - 1) = 7

[1×144, 2×72, 3×48, 4×36, 6×24, 8×18, 9×16]

Hence, 7.

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**Factors | Algebra - Number Theory**

In how many ways can 144 be written as product of two of its factors?

Answer: 8

**Explanation** :

Number of ways of writing a perfect square as a product of two of its factors (same or distinct) is equal to half the (number of its factors + 1). Here 144 is a perfect square.

144 = 2^{4} × 3^{2}

Number of factors of 144 = (4 + 1)(2 + 1) = 15

Number of ways of writing a perfect square as a product of two of its distinct factors = ½ × (15 + 1) = 8

[Note: 12 × 12 is one additional way of writing 144 as product of two of its same factors]

Hence, 8.

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**Factors | Algebra - Number Theory**

How many factors of 344 are divisible by 4?

Answer: 4

**Explanation** :

344 = 2^{3} × 43

For a factor to be divisible by 4 (= 2^{2}), the minimum power of 2 in it should be 2.

∴ Possible powers of 2 in the factor = 2 or 3, i.e., 2 possibilities.

Their is no restriction on powers of other prime numbers

∴ Possible powers of 43 = 0 or 1, i.e., 2 possibilities.

Total number of such factors = 2 × 2 = 4.

Hence, 4.

Workspace:

**Factors | Algebra - Number Theory**

What is the product of all the factors of 360?

- (a)
360

^{12} - (b)
360

^{24} - (c)
360

^{11} - (d)
None of these

Answer: Option A

**Explanation** :

360 = 2^{3} × 3^{2} × 5

Product of all the factors of N = N^{(number of factors)/2}

Number of factors of 360 = (3 + 1)(2 + 1)(1 + 1) = 24

∴ Product of all factors of 360 = 360^{24/2} = 360^{12}

Hence, option (a).

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