Factorials | Algebra - Number Theory

Answer: Option D

Explanation :

0! = 1

Hence, option (d).

Answer: 336

Explanation :

$\frac{8!}{5!}$ = $\frac{8\times 7\times 6\times 5!}{5!}$ = 8 × 7 × 6 = 336

Hence, 336.

Answer: Option D

Explanation :

All the terms in the series from 5! onwards are divisible by 5.

∴ Remainder will be the same as the remainder when (1! + 2! + 3! + 4!) is divided by 5 

⇒ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33.

∴ The remainder will be R(33/5) = 3.

Hence, option (d).

Answer: Option E

Explanation :

The product of any n consecutive numbers would always be divisible by the product of the 1^st n consecutive natural numbers. i.e., n!.

Alternately,

For such questions, quickly check for some values of n e.g., 1, 2, 3 etc. and look for which options satisfy

Hence, option (e).
 

Answer: Option A

Explanation :

Let the even numbers be 2a, 2a + 2, 2a + 4, …, 2a + 38

∴ Product of these numbers (P) = 2a × (2a + 2) × (2a + 4) × …, × (2a + 38).

⇒ P = 2²⁰ × [a × (a + 1) × (a + 2) × (a + 3) × … (a + 19)]

⇒ P = 2²⁰ × product of 20 consecutive natural numbers.

We know the product of 20 consecutive natural numbers is always divisible by 20! 

∴ P is always divisible by 2²⁰ × 20!.

Hence, option (a).

Answer: Option D

Explanation :

To find the highest power of a prime number (p) in N!, we successively divide N by p and add all the quotients.

∴ Highest power of 2 in 50! = $\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+\left[\frac{50}{2^3}\right]+\left[\frac{50}{2^4}\right]+\left[\frac{50}{2^5}\right]$ = 25 + 12 + 6 + 3 + 1 = 47.

Hence, option (d).

Answer: 23

Explanation :

4 = 2²

∴ Highest power of 2 in 50! = $\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+\left[\frac{50}{2^3}\right]+\left[\frac{50}{2^4}\right]+\left[\frac{50}{2^5}\right]$ = 25 + 12 + 6 + 3 + 1 = 47.

⇒ Highest power of 2² (i.e., 4) in 50! = 23.

Hence, 23.

Answer: 15

Explanation :

24 = 2³ × 3

∴ Highest power of 2 in 50! = $\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+\left[\frac{50}{2^3}\right]+\left[\frac{50}{2^4}\right]+\left[\frac{50}{2^5}\right]$ = 25 + 12 + 6 + 3 + 1 = 47.

⇒ Highest power of 2³ (i.e., 8) in 50! = 15.

∴ Highest power of 3 in 50! = $\left[\frac{50}{3}\right]+\left[\frac{50}{3^2}\right]+\left[\frac{50}{3^3}\right]$ = 16 + 5 + 1 = 22.

Hence, 50! = (8)¹⁵ × 3²² × k (k is an integer and does not contain any power of 2 or 3)

⇒ 50! = (24)¹⁵ × 3⁷ × k

∴ Highest power of 24 in 50! Is 15.

Hence, 15.

Answer: 24

Explanation :

100! + 200! + 300! = 100! × (1 + 101 × 102 × … × 200 + 101 × 102 × … × 300) = 100! × (a number which ends in 1)

∴ 100! + 200! + 300! will have as many zeroes at the end as 100! does.

∴ Number of zeros at the end of 100! = highest power of 5 in 100! = $⌊\frac{100}{5}⌋+⌊\frac{100}{25}⌋$ = 20 + 4 = 24

⇒ 100! has 24 zeroes.

Hence, 24.