Updated On : August 7, 2023

Trailing Zeroes is one of the essential concepts under mathematics, and you can expect 2-3 questions from this topic in the Integrated Program in Management Aptitude Test.

As we all know, the quantitative aptitude section holds more weight in the IPMAT entrance exam. Hence, it is vital to focus on each and every concept to score good marks in the exam.

This post shall guide you through the list of important IPM Aptitude Trailing zeroes Questions with solutions in this post.

As said above, trailing zeroes is one of the** important topics for the IPMAT Exam**. Before moving on to conceptual questions, it is imperative to understand, what is a trailing zero?.

In simple words, it is a zero digit with no non-zero numbers to the right of it. Let us understand this with an example:

**9100340560000**

In the above example, there are four trailing zeros. Most of you might think that there are three other zeros in the given number, but they do not count as trailing zeros because there are non-other zero digits that are less significant.

**How many trailing zeros are in the number 10000?**

The number 10000 has four trailing zeros as 1 has four zero digits and no non-zero digit to the right.

**How many trailing zeros are in the number 10500?**

The number 10500 has only two 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 vital 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

Most of you you might be worried about what type of questions asked from trailing zeroes and the difficulty level of questions.

To help you get an idea about the type of trailing zeroes questions asked in the quant section, we have curated important questions that are collected from **previous year's IPMAT Question Papers**.

Let us check out the IPM trailing zeroes questions with solutions from the post below.

**Question 1**

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

- 10!
- 40!
- 100!

**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, as shown below.

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

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

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

10! = (5 x 2) x 9 x (2 x 3) x 7 x (3 x 2) x 5 x (2 x 2) 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.

**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 divide 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

- 10!

Now solving for 10! using the trick mentioned above,

divide 10/2 = 5 (as the answer is more significant than 2, we will not stop at this stage), now dividing five by the given prime number we get (5/2 = ) 2 (as we have only to 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 accurate for every prime number available in the factorial.

For calculating the number of 5’s present in 10!

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

Ans = 2

- 40!

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)

* Read more*:

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

- 100!

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

* Read more*:

**Question 2**

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

- 10!

- 100!

- 150!

**Solution**

- 10!

Here we need to find the power of 2 that will completely divide 10! using the concept discussed above, we calculate the maximum capacity 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.

- 100!

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

* Read more*:

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

- 150!

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.

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.

The following are some of the sample questions based on trailing zeroes for the upcoming exam.

**Question 1**

Find the highest value of ‘n’ that so that 10n completely divides

- 100!

- 250!

**Solution**

- 100!

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

- 250!

* Read more*:

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

* Read more*:

**Question 2**

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 3**

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.

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