Number Theory Questions - Factorial

Have given below one very different question on factorial

Question

1. Given N is a positive integer less than 31, how many values can n take if (n+1) is a factor of n! ?

Correct Answer:

18 values

Explanatory Answer:

The best starting point for this question is to do some trial and error.

3 is not a factor of 2!

4 is not a factor of 3!

5 is not a factor of 4!

6 is a factor of 5!

7 is not a factor of 6!

8 is a factor of 7!

The first thing we see is that n+1 cannot be prime. If (n +1) were prime, it cannot be a factor of n!.

So, we can eliminate all primes.

Now, let us think of all numbers where (n+1) is not prime. In this instance, we should be able to write (n+1) as a * b where a,b are not 1 and (n+1). So, (a,b) will lie in the set { 1, 2, 3, .....n} or, a* b will be a factor of (n + 1)!

So, for any composite (n+1), (n+1) will always be a factor of n! {Is there any exception?}

For any prime number (n +1), (n+1) will never be a factor of n!

The above rule works well even for all the examples we have seen, except when (n+1) = 4. 4 = 2 * 2; So, 4 is not a factor of 3!. But this is the only exception.

Counting on from here, we can see that n can take values 5, 7, 8, 9, 11, 13, 14,15, 17, 19, 20, 21, 23, 24, 25, 26, 27, 29. Essentially, all numbers where N+1 is greater than 4 and is not prime will feature in this list. If N+1 is not prime, we should be able to write is as a product of 2 numbers less than N+1. This will feature in n!. This question is just a different way of asking one to count primes (and then account for the exception of 4)

n can 18 different values.

Now, if we know that there are 25 prime numbers less than 100, can you answer the following question - Given N is a positive integer less than 101, how many values can n take if (n+1) is a factor of n!?

CAT Number Theory Questions - Factorial

Have given below two questions on Number Theory, focusing on factorial.

Questions

1. How many trailing zeroes (zeroes at the end of the number) does 60! have?
2. What is the highest power of 12 that divides 54! ?

Correct Answers

Question 1: 14 zeroes
Question 2: 25

Explanatory Answers

1. How many trailing zeroes (zeroes at the end of the number) does 60! have?

To start with, the number of trailing zeroes in the decimal representation of a number = highest power of 10 that can divide the number.

For instance,

3600 = 36 * 10²

45000 =  45 * 10³ 

In order to approach this question, let us first see the smallest factorial that ends in a zero.

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

Now, 5! ends in a zero as we have get a product of 10 when we compute 1 * 2 * 3 * 4 * 5

10 is 2*5, so we get a factor of 10 every time we get a 2 and a 5 in the factorial.

So, 5! has 1 zero. The factorial that ends with 2 zeroes is 10!

15! has 3 zeroes.

20! has 4 zeroes and so on.

An extra zero is created every time a 2 and 5 combine. Every even number gives a two, while every fifth number gives us a 5.

Now, the critical point here is that since every even number contributes at least a 2 to the factorial, 2 occurs way more frequently than 5. So, in order to find the highest power of 10 that can divide a number, we need to count the highest power of 5 that can divide that number. We do not need to count the number of 2’s in the system as there will be more than 2’s than 5’s in any factorial.

Now, every multiple of 5 will add a zero to the factorial. 1 * 2 * 3 *…..59 * 60 has twelve multiples of 5. So, it looks like 60! will end in 12 zeroes. But we need to make one more adjustment here.

25 is 5², so 25 alone will contribute two 5’s, and therefore add two zeroes to the system. Likewise, any multiple of 25 will contribute an additional zero. 

So, 20! has 4 zeroes, 25! has 6 zeroes.

60! will have [60/5] zeroes arising due to the  multiples of  and an additional  [60/25] due to the presence of 25 and 50. {We retain only the integer component of 60/25 as the decimal part has no value}

So, 60! will end with 12 + 2 zeros. = 14 zeros.

In general, any n! will end with [n/5] + [n/25] + [n/125] + [n/625] …… zeroes.

Generalizing further, in case we want to find the highest power of 3 that divides n!, this is nothing but [n/3] + [n/9] + [n/27] + [n/81] ……

The highest power of 7 that divides n! is [n/7] + [n/49] + [n/343] ……

In case of a composite number, we need to break into the constituent primes and compute the highest power that divides the number.

For instance, if we want to find the largest power of 15 that divides n!, this will be driven by the highest powers of 3 and 5 that divide n!. Similar to the scenario we saw with trailing zeroes, we can observe that there will definitely be at least as many 3’s than 5’s in any factorial. So, the highest power of 15 that divides n! is simply [n/5] + [n/25] + [n/125] + [n/625] ……

 

2. What is the highest power of 12 that divides 54! ?

12 = 2² * 3, so we need to count the highest power of 2 and highest power of 3 that will divide 54! and then we can use this to find the highest power of 12.

The method to find highest powers of 2 and 3 are similar to the one outlined in the previous question.

Highest power of 2 that divides 54! = [54/2] + [54/4] + [54/8] + [54/16] + [54/32] = 27 + 13 + 6 + 3 + 1 = 50

Highest power of 3 that divides 54! = [54/3] + [54/9] + [54/27] = 18 + 6 + 2 = 26

Or 54! is a multiple of 2⁵⁰ * 3²⁶. Importantly, these are the highest powers of 2 and 3 that divide 54!.  

2² * 3 = 12. We need to see what is the highest power of 22 * 3 that we can accommodate within 54!

In other words, what is the highest n such that (2²*3)ⁿ can be accommodated within 2⁵⁰* 3²⁶

Let us try some numbers, say, 10, 20, 30

(2² *3)¹⁰ = 2²⁰ * 3¹⁰, this is within 2⁵⁰ * 3²⁶

(2² *3)²⁰ = 2⁴⁰ * 3²⁰, this is within 2⁵⁰ * 3²⁶

(2² *3)³⁰ = 2⁶⁰ * 3³⁰, this is not within 2⁵⁰ * 3²⁶

The highest number possible for n is 25.

(2² *3)²⁵ = 2⁵⁰ * 3²⁵, this is within 2⁵⁰ * 3²⁶, but (2² *3)²⁶ = 2⁵² * 3²⁶, this is not within 2⁵⁰ * 3²⁶

So, 54! can be said to be a multiple of (2² * 3)²⁵. Or, the highest power of 12 that can divide 54! is 25.

Note: For most numbers, we should be able to find the limiting prime. As in, to find the highest power of 10, we need to count 5s. For the highest power of 6, we count 3s. For 15, we count 5s. For 21, we count 7’s. However, for 12, the limiting prime could be 2 or 3, so we need to check both primes and then verify this.

Rajesh

Director, 2IIM

CAT classroom course in Chennai, Bangalore and Mumbai