The Integrated Programme in Management Aptitude Test popularly known as IPMAT is a national level entrance exam conducted by the Indian Institute of Management Indore (IIM-I) every year for the students seeking admission in the Five-year Integrated Management Programme.

The course is currently being offered by only 5 IIMs- IIM Indore, IIM Rohtak, IIM Jammu, IIM Bodh Gaya, and IIM Ranchi.

Due to the sectional time limit of 40 minutes in the IPMAT exam, Quantitative Aptitude becomes even harder to score. Therefore, it is important to devote time carefully to each question.

Mathematics questions are lengthy and time-consuming, therefore you are advised to follow **important tips and tricks for the IPMAT Quant** section to enhance your preparation. One of the important concepts to save time in trailing zero-related questions is provided below to help students increase their efficiency.

## Conceptual questions based on Trailing Zeroes

Before moving on to conceptual questions it is really important to understand, What is a trailing zero?. In simple words, it is a zero digit with no non-zero digits to the right of it. Let’s understand this using an example.

How many trailing zeros are in the number 10000?

The number 10000 has 4 trailing zeros as 1 has four zero digits having no non-zero digit to the right of them.

How many trailing zeros are in the number 10500?

The number 10500 has only 2 trailing zeros. Note that there is another zero in the representation of the number, but it doesn't count as trailing zeros because there are non-zero digits to the right of it.

Another important concept required to understand trailing zero conceptual questions is the factorial of a number. The factorial of a whole number 'n' is defined as the product of that number with every whole number till 1. For example,

2! = 2 x 1 = 2

3! = 3 x 2 x 1 = 6

4! = 4 x 3 x 2 x 1 = 24

## IPM Aptitude Questions & Answers based on Trailing Zeroes

**Question 1. Find the Maximum no. of 2’s and 5’s present in the following**

**Solution.**

Here in this question, we have to find the number of factors of 2 and 5 present in the given factorial. Calculating the number of 2’s and 5’s can be done manually but it will be a cumbersome procedure like shown below.

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800

Factoring terms of 10! In the form of 2’s and 5’s,

10! = (5 x 2) x 9 x (23) x 7 x (3 x 2) x 5 x 22 x 3 x 2 x 1

Counting from the above we have, number of 2’s = 8 and number of 5’s = 2

This method can be used to solve small factorials but wouldn't be convenient for calculating large numbers like 40!

To solve these questions quickly and conveniently you can use the following trick.

**Read more:** **Know Important Formulas for IPMAT Quantitative Aptitude**

**Trick-1:- **to find the maximum power of a prime number which is a factor of the given factorial number, we have to divide the factorial by the prime number, then dividing the integer part of the answer again and again by the prime number up till the answer is not smaller than the prime number.

**Better Solution**

Now solving for 10! using the trick mentioned above, divide 10/2 = 5(as the answer is bigger than 2 we will not stop at this stage), now dividing 5 by the given prime number we get (5/2 = ) 2 (as we have to only take integer values of the answer), finally 2/2 = 1(as 1 < 2, we will stop at this stage)

Ans = 5+2+1 = 8

This method will be true for every prime number available in the factorial.

For calculating no. of 5’s present in 10!

⇒ 10/5 = 2 (as 2 < 5 , we will stop at this stage)

Ans = 2

For calculating no.of 2’s present in 40!

⇒ 40/2 = 20

⇒ 20/2 = 10

⇒ 10/2 = 5

⇒ 5/2 = 2

⇒ 2/2 = 1 (as 1 < 2 , we will stop at this stage)

Ans = 20 +10 +5 +2 +1 = 38

For calculating no.of 5’s present in 40!

⇒ 40/5 = 8

⇒ 8/5 = 1 (as 1 < 5, we will stop at this stage)

Ans = 8 +1 = 9

For calculating no.of 2’s present in 100!

⇒ 100/2 = 50

⇒ 50/2 = 25

⇒ 25/2 = 12

⇒ 12/2 = 6

⇒ 6/2 = 3

⇒ 3/2 = 1 (as 1 < 2 , we will stop at this stage)

Ans = 50 + 25 + 12 + 6 + 3 + 1 = 97

For calculating no.of 5’s present in 1000!

⇒ 100/5 = 20

⇒ 20/5 = 4 (as 4 < 5, we will stop at this stage)

Ans = 20 +4 = 24

**Question 2. Find the highest value of ‘n’ so that 2****n**** completely divides **

**Solution:**

Here we need to find the power of 2 that will completely divide 10! using the concept discussed above we calculate the maximum power of the prime number available in the factors of the factorial number.

Writing 10! = 2n x N(other factors)

Using the above method

⇒ 10/2 = 5

⇒ 5/2 = 2

⇒ 2/2 = 1

Ans = 5 + 2 + 1

Therefore 28 is the maximum factor of 2 and the highest value of n would be 8.

Writing 100! = 2n x N(other factors)

Using the above method

⇒ 100/2 = 50

⇒ 50/2 = 25

⇒ 25/2 = 12

⇒ 12/2 = 6

⇒ 6/2 = 3

⇒ 3/2 = 1

Ans = 50 + 25 + 12 + 6 + 3 + 1 = 97

Therefore 297 is the maximum factor of 2 and the highest value of n would be 97.

Writing 150! = 2n x N(other factors)

Using the above method

⇒ 150/2 = 75

⇒ 75/2 = 37

⇒ 37/2 = 18

⇒ 18/2 = 9

⇒ 9/2 = 4

⇒ 4/2 = 2

⇒ 2/2 = 1

Ans = 75 + 37 + 18 + 9 + 4 +2 + 1 = 146

Therefore 2146 is the maximum factor of 2 and the highest value of n would be 146.

**Question 3. Find the highest value of ‘n’ that so that 10****n**** completely divides **

Solution:

As 10 is not a prime number we wouldn't be able to use the method used above in question number two. For these questions, factorize the term into prime factors form.

Like 10 = 21 x 51

We now need to find the maximum number of 2 x 5 present in 100! Which can be found using the trick mentioned above.

Maximum no. of 2’s present in 100! = 97

Maximum no. of 5’s present in 100! = 24

Therefore the maximum pairs of 2 x 5 are 24.

Hence answer = 24

We need to find the maximum number of 2 x 5 present in 250! Which can be found using the trick mentioned above.

Maximum no. of 2’s present in 250! = 244

⇒ 250/2 = 125

⇒ 125/2 = 62

⇒ 62/2 = 31

⇒ 31/2 = 15

⇒ 15/2 = 7

⇒ 7/2 = 3

⇒ 3/2 = 1

Maximum no. of 5’s present in 250! = 62

⇒ 250/5 = 50

⇒ 50/5 = 10

⇒ 10/5 = 2

Therefore the maximum pairs of 2 x 5 are 62.

Hence answer = 62

**Question 4. Find the trailing zeros of 250!.**

**Solution:**

The number of trailing zeroes of a number is the highest power of n for which we can completely divide 10n.

Note

**Trailing zeros = highest power = highest power of 5**

We don't need to calculate the highest power of 2 which we have done above as the deciding factor is the term having minimum power. This is true only in the case of pure factorials, not in cases like 100! x 5200.

Hence answer = 62.

**Question 5. Find the number of trailing zeros at the end of the first hundred multiples of 10**

**Solution:**

Before moving to the solution try to solve this question on your own.

Here we need to find the number of trailing zeroes in 10 x 20 x 30 x 40 ……x 1000

⇒ (10 x 1 ) x (10 x 2) x (10 x 3 )…… (10 x 100)

⇒ 10100 x (1 x 2 x 3 x 4 ….. x 100)

⇒ 10100 x 100!

⇒ 10100 x 1024 x N (writing 100! = 1024 x N )

⇒ 10124 x N

Hence answer = 124.