We have already learnt in the previous chapter how to find the number of zeros in a factorial. Finding number of zeros in any factorial is equivalent to finding the highest power of 5 in that given factorial since highest power of 2 will be definitely more than highest power of 5.
Now, we will look at some new aspects of number of zeros in a given factorial.
From which factorial onwards we will start getting trailing zeros and why?
Answer should be pretty simple and that is 5!. From 5! Onwards, we will start getting trailing zeros. We would not get any zero in 1!, 2!, 3! And 4! since these factorials do not have a 5.
5! – one trailing zero
6!, 7!, 8! and 9!, all of them have one trailing zeros as only one 5 is present in them.
Number of zeros will increase by 1 in 10!. Because 10! has got one extra 5, so it will have two trailing zeros. (10! = 10 × 9!)- 9! has one trailing zero and new number which has come in 10! is 10 which has one 5., that’s why number of zeros will increase by 1.
So, (10! to 14!)- All of them have got 2 trailing zeros.
We can conclude one thing: Any particular number of zero repeats in exactly five factorials.
Similarly, (15! to 19!) has 3 trailing zeros.
(20! to 24!) has 4 trailing zeros.
(25! to 29!) has 6 trailing zeros and not 5 zeros. Because 25! = 25 × 24!. New addition to 25! is 25 which has two 5’s, that’s why the number of zeros will increase by 2.
(30! to 34!) has 7 trailing zeros.
(35! to 39!) has 8 trailing zeros.
(40! to 44!) has 9 trailing zeros.
(45! to 49!) has 10 trailing zeros.
(50! to 54!) has 12 trailing zeros. Because 50! = 50 × 49!. New number is 50 which is a multiple of 25, so number of zeros in 50! will be 2 more than 49!.
We can conclude that all the factorials which are multiples of 25 will have a gap or difference of two trailing zeros as compared to the preceding factorials.
For e.g., 24! – 4 zeros, 25! – 6 zeros.
49! – 10 zeros, 50! – 12 zeros.
Next factorial where a jump of 2 trailing zeros will occur is 74! – 16 zeros and 75! – 18 zeros.
Similarly, we can write all the factorials with the difference of two trailing zeros. We need to observe one more thing is that if we know number of trailing zeros in a particular factorial, we can find out the number of zeros in subsequent factorial by just observing whether that factorial is a multiple of 5 or 25 or 125 or so on. If the new factorial is a multiple of 5, number of trailing zeros will increase by 1. And if the new factorial is a multiple of 25, the number of zeros will increase by 2.
For e.g., if we know number of zeros in 124!, we don’t need to find number of zeros in (125! to 129!) using highest power of 5. Because number of zeros in 125! will be three more than 124!.
Similarly, it can also work vice-versa. If we know number of zeros in 150!, we can find out number of zeros in (149! to 145!) without using highest power of 5. Number of zeros in 149! will be two less than 150!, because 150! is a multiple of 25.
E.g. 1: If n! has 2 more trailing zeros than (n – 1)!, then find how many three digit values can n take?
On reading the question, the logic should be clear that question is talking about multiples of 25. We have to find the number of three digit-values which n can take.
If n takes the value of 100!, then (n – 1) will take 99!, and the difference in the number of zeros will be 2. So, like that we can find out all the multiples of 25 uptill 999!. The last factorial which is a multiple of 25 is 975!, so, starting from 100!, 125!, 150!,……….. 975!.
One method is to count all the multiples of 25 by writing them down. But, it will take a lot of time. We have learnt a better method of counting out number of terms in a series in chapter Cyclicity. We will use that method now.
So, number of terms in a series can be found out by =[ (Last term – first term)/common difference] + 1.
But we know that some of the multiples of 25 are also multiples of 125 and 625, and in those factorials we will get a difference of three or four trailing zeros. But we want only those factorials with a difference of two zeros.
So, just count out multiples of 125 and 625 uptill 999. They are 125, 250, 375, 500, 625, 750, 875.
So, we will have to subtract these 7 multiples from 36 terms which we got as multiples of 25.
So, final answer is n can take 36 – 7 = 29 three-digit values.
E.g. 2: Find the number of trailing zeros in 1! × 2! × 3! × 4! × ………………… × 45!.
This is a very easy and logical question but most of the students get it wrong as they just don’t know to proceed or even if they do, they just count number of trailing zeros in 5!, 10!, 15!, 20! and so on. They miss out on trailing zeros which are also present at the end of 6!, 7!, 8! and so on.
Let us see how to solve it.
5! has one trailing zeros and we know that any particular number of trailing zero repeats itself in exactly five distinct factorials. So, number of trailing zeros in 5! to 9! when all of them are being multiplied will be 105.
Now, we got the logic, we can write and add up all the trailing zeros.
5! to 9! – 5 trailing zeros.
10! to 14! – 10 trailing zeros (each one having two trailing zeros)
15! to 19! – 15 trailing zeros (each one having three trailing zeros)
20! to 24! – 20 trailing zeros (each one having four trailing zeros)
25! to 29! – 30 trailing zeros (each one having six trailing zeros)
30! to 34! – 35 trailing zeros (each one having seven trailing zeros)
35! to 39! – 40 trailing zeros (each one having eight trailing zeros)
40! to 44! – 45 trailing zeros (each one having nine trailing zeros)
And lastly 45! – having 10 zeros.
Add them up, we will get the required number of trailing zeros which is 10210.
No. of trailing Zeros skipped in Factorials:
We have already seen that factorials which are multiples of 25 or 125 have a difference of two or three trailing zeros. Some trailing zeros will be skipped there.
For example, 24! has 4 zeros and 25! has 6 zeros. So, we can say that no factorial has five trailing zeros. Similarly, another trailing zero will be skipped between 49! and 50! which is 11. Another one will be skipped between 74! and 75! which is 17.
We can get a series for skipped trailing zeros like 5, 11, 17, 23, 29, 30, 36, 42, 48, 54, 60, 61, 67 and so on.
Just pay attention to the multiples of 125!, there will be a difference of 3 zeros between 124! And 125!, so, two trailing zeros will be skipped.(124! – 28 trailing zeros, 125! – 31 trailing zeros, so 29 and 30 will be skipped.)
E.g. 3: Which of the following cannot be the number of trailing zeros at the end of any one of these factorials?
a. 48 b. 49 c. 47 d. 50
On reading this problem, we can understand that the question is asking about skipped zeros. We know the series of skipped trailing zeros starting from 5, 11, 17, 23, 29, 30, 36, 42, 48, 54, and so on.
So, answer should be 48. All the other trailing zeros can be obtained at the end of any factorial, but 48 trailing zeros cannot be. So, if we know this funda, this problem can be solved in a jiffy.
Number of trailing zeros in a series of numbers except Factorials:
Questions related to number of trailing zeros in multiplication of different numbers except factorials are also asked in entrance exams. Let us see some examples.
E.g. 4: Find the number of trailing zeros in (10000)50 × (100)30.
This is a very easy question which can be solved orally.
(10000)50 × (100)30 can be written as (104)50 × (102)30.
On solving this, we will get 10200 × 1060 = 10260.
E.g. 5: Find the number of trailing zeros in 150! + 200!.
If we add up two numbers let’s say 10000 + 100, we can easily predict the answer. When we add up these two numbers, we will get 2 trailing zeros.
Similarly, when we add 10000 + 1000000, in this case answer will be 4 trailing zeros. So, logic should be that answer is dependent on the number having less number of zeros.
We can apply the same logic in the finding the number of trailing zeros in 150! + 200!. 150! will be a number with 37 trailing zeros and 200! will be a greater number (as compared to 150!) having 49 trailing zeros. So, answer should be 1037.
E.g. 6: Find the number of trailing zeros in 20 × 30 × 40 × 50 × 60 × 70 × 80 × 100.
This series is not a factorial, its a series of numbers. In a factorial, its hundred percent sure that number of trailing zeros will depend on the highest power of 5. But, if its not a series of factorial, we will have to find number of 2’s and 5’s individually in the series. And then the least power of 2 or 5 will give the number of zeros.
So, in the question, we can easily identify number of 5’s and 2’s.
22 × 2 × 23 × 2 × 22 × 2 × 24 × 22 × 5 × 5 × 5 × 52 × 5 × 5 × 5 × 52.
So, on simplifying, we get 216 × 510. So, number of trailing zeros will be 1010.
E.g. 7: Find the number of trailing zeros in the multiplication of 1 × 4 × 7 × 10 × 13 × 16 × 19 ×…………..× 100.
In this series, if we observe carefully, we will see that number of 2’s are present in alternate terms, but 5’s are present in every 5th term. So, the frequency of occurrence of 5’s is less than that of 2’s, so number of trailing zeros will depend on the number of 5’s in the series.
So, our objective is to count the number of terms having 5’s in the series.
We get the first term having 5 is 10 and then next term will be 25. The difference between the two terms involving 5 is 15, so the next term which will contain 5 will be 40, then 55, 70, 85 and 100.
We count the number of terms involving 5 are 7 (10, 25, 40, 55, 70, 85, 100). So, 107 should be the number of trailing zeros. But we know that all the multiples of 25 have one extra 5 which will combine with 2(which is present in abundance in the series) to form one more trailing zero.
So, we will find the terms which are a multiple of 25. Only two terms 25 and 100 are multiples of 25.
Finally, we can say that total number of trailing zeros at the end of the given series will not be 107 but 109.
What if the range is large?
Let’s say if the previous question asked to find number of zeros uptill 481. The original question was just uptill 100, so we could count easily, but if the range is large, if we start counting, it will take a lot of time. So, we need to revisit a simple method of counting, which we have already learnt in cyclicity.
E.g. 8: Find the number of trailing zeros in the multiplication of 1 × 4 × 7 × 10 × 13 × 16 × 19 ×…………..× 481.
The logic is still the same which we used in the last example. The difference is just that the range is greater than the previous question.
We just need to find how many terms will have 5. The first term having 5 is 10 and if we can find the last term having 5, we can use the logic of finding out number of terms in a sequence.
Number of terms in a series can be found out by =[ (Last term – first term)/common difference] + 1.
How to find the last term?
The series is increasing in steps of 3. So, if we have to find the last term having 5, we should start from the last term which is 481 and start subtracting 3 untill we get a multiple of 5.
If we subtract 3 from 481, we get 478 which is not a multiple of 5. Again we subtract 3 from 478, we get 475 which is a multiple of 5. Now, if we know the first term containing 5 is 10 , then 25, 40 and till 475. We can find the number of terms using the formula which we have learnt.
All these terms will make trailing zeros. So, we got 1032.
Now, we also have to find terms which are multiples of 25 because all the multiples of 25 will give us additional zeros.
First multiple of 25 is 25 and we are getting 25 in the series. Now, we just need to find the second multiple of 25, our purpose will be solved.
First we got was 25, next multiple is 50. The difference between 25 and 50 is of 25 and the jump is of 15. (Jump means next term which we are getting is coming after an interval of 15).
Can we cover a distance of 25 with a jump of 15? Answer should be no. We can cover a distance of 15, 30, 45, 60 and so on. So, we would not get 50 in the series.
Next multiple of 25 is 75. The gap between 25 and 75 is 50. Can we cover a distance of 50 with a jump of 15? No, that means 75 will also not occur in the series.
Next multiple of 25 is 100. The gap between 25 and 100 is of 75. Can we cover a distance of 75 with a jump of 15. Yes, we can.
So, first multiple of 25 which we got in the series was 25 and next is 100.
Our purpose is solved. The gap between 25 and 100 is 75, so next multiple of 25 which will occur in the series will also be at the gap of 75. Now, we can write all the multiples of 25 which are going to be found in the series.
So, 25, 100, 175, 250, 325, 400, 475. All these seven terms will have one extra zero each So, we will get 107 extra.
We also know that multiples of 125 give us one more trailing zero. All Multiples of 125 are also multiples of 25, so we do not need to know any extra work. Just check from the list of multiples of 25, some of them might be multiples of 125.
Only one term 250, a multiple of 125 is found. So, we will get one more zero.
So, the final answer: Multiples of 5 gave us: 1032 zeros.
Multiples of 25 gave us 107 additional zeros.
Multiples of 125 gave us 101.
So, total number of trailing zeros are 1032 × 107 × 101 = 1040.
E.g. 9: Find the number of trailing zeros at the end of the multiplication of 2 × 6 × 10 × 14 × 18 × 22 × ………………. × 502.
Same logic has to be used while solving since the range is pretty long.
First of all, we will just find out highest power of 5, since number of 2’s are present in every term and number of 5’s are present in every fifth term. So, number of 5’s will give us the trailing zeros.
Starting with finding the number of 5’s first. First 5 is obtained at 10 and if we go ahead in the series, we will get next 5 in 30. We get the common difference between first term containing 5 and second term containing 5, the remaining terms will also occur after the same gap.
To find the last term containing 5, we will start subtracting 4 from the last term which is 502. If we subtact 4, we get 498 which is not a multiple of 5. Again we subtract 4 from 498, we get 494 which is not a multiple of 5. Again we subtract 4 from 494, we get 490 which is a multiple of 5.
Now, we need to check how many multiples of 25 will occur in the series.
We got the first term as 10 and then 30 and then we will get 50 which is the first multiple of 25.
Now we need to find the second multiple of 25 so that we can establish the common difference between two multiples of 25. Also, note series given in the question is of even number, so we need not check odd multiples of 25. Next even multiple of 25 is 100.
Gap between 50 and 100 is 50 and there is a jump of 20, can we cover a distance of 50 with a jump of 20?
Now, we cannot.
Next even multiple of 25 is 150, so the gap between 50 and 150 is of 100. Can we cover a distance of 100 with a jump of 20? Yes we can. So, we will get 150 in the series.
Now, we got the first and second multiple of 25, now the remaining multiples of 25 which are going to appear in the series will also come after same frequency.
We can write those: 50, 150, 250, 350, 450. We got 5 terms i.e. we will get 5 more trailing zeros.
And lastly we need to check for multiples of 125. Only one of them is present in the multiples of 25 which is 250 which will give us one more trailing zero.
So, number of trailing zeros at the end of product of those numbers will be 1025 × 105 × 101 = 1031.
So, after solving these many examples, all of us should have got a fair idea to deal with the number of trailing zeros. All these questions are pretty simple, if we practice them 3-4 times, we can solve each one of these problem in less than a minute.
Now, we will solve some more questions for practice.
Exercise:
1. Find the number of trailing zeros in the multiplication of 201 × 202 × 203 × 204 ×………………..× 300.
2. Find the number of zeros in the multiplication of (1!)2 × (2!)2 × (3!)2 × (4!)2 × ……….. × (35!)2.
3. If (n +1)! has 3 more trailing zeros than n!, then how many three-digit values can n take?
4. Which of the following can be the number of trailing zeros at the end of any factorial?
a. 63 b. 68 c. 74 d. 80 e. All of these
5. Which of the following cannot be the number of trailing zeros at the end of any factorial?
a. 28 b. 36 c. 19 d. 47
6. Find the number of trailing zeros in the product of 11 × 55 × 1010 × 1515 × 2020 × 2525………. × 5050.
7. Find the number of trailing zeros in the following series: 1! + 2! + 3! + 4! +………..+ 18!.
8. Find the number of trailing zeros in the expansion of (20! × 21! × 22! × ……………… × 33!)3!.
9. If P is a natural number and P! ends with y trailing zeros, then the number of zeros that (5P)! ends with will be
Answer Key:
1. 25 2. 10246 3. 7 4. (e) 5. (b) 6. 10250 7. No trailing zeros 8. 10468 9. (P + Y) trailing zeros
Explanations:
Question1.
There are multiple ways of solving this question. To find the trailing zeros in 201 × 202 × 203 × 204 × 205 × 206 × 207 × ………… × 300, we can find out trailing zeros in 300! and 200!, and subtracting those trailing zeros will give us the trailing zeros for (201 × 202 × 203 × 204 × 205 × 206 × 207 × ………… × 300).
We found out no. of trailing zeros in (300!) and (200!), subtracted them which gave us the answer for trailing zeros of (201 × 202 × 203 × 204 × 205 × 206 × 207 × ………… × 300).
Alternative Method: To find out trailing zeros in (201× 202 × 203 × 204 × 205 × 206 × 207 × ………… × 300), we can find out number of 2’s and 5’s, depending on the lesser power of 2 or 5, we can get the number of trailing zeros.
We observe the sequence given in the question, number of 2’s are occurring every alternate digits but number of 5’s are occurring after every 4th number. So we can say that number of 5’s in the sequence will be less than number of 2’s. So, our number of trailing zeros will be dependent on the power of 5. So, if we can just find out number of 5’s in the sequence, we can get the number of trailing zeros.
We will get 5 in terms like 205, 210, 215, 220,………………, 300 at an interval of 5. We can find out number of terms in the sequence with the help of formula learnt earlier.
So, number of terms having 5 are [(300 – 205) ÷ 5] + 1 = 20 terms. So, we will get 1020.
But, we will get some multiples of 25 in the series which have an extra 5, so we will count multiples of 25. They are 225, 250, 275 and 300. So, we will get 4 more zeros.
Also, in this sequence, there in one multiple of 125, this will give us one more extra zero.
So, total number of zeros in this series are = 1020 × 104 × 101 = 1025.
Intuitive Method: There is one more method of solving this question orally. In any set of product of 100 natural numbers, we will get a minimum of 24 trailing zeros. Number of trailing zeros will increase by 1 if that sequence has one multiple of 125. Number of trailing zeros will increase by 2 if that sequence has a multiple of 625. If there are no multiples of 125 or 625, then number of trailing zeros will remain as 24.
In the sequence given in the question, we will get a minimum of 1024, and since the series has a multiple of 125 i.e. 250, number of zeros will increase by 1, resulting in total number of 1025 trailing zeros.
For e.g. (1 × 2 × 3 × 4 × 5 ×…………..× 100) is also a sequence of product of 100 natural numbers, this series will have 24 trailing zeros since no multiples of 125 is present in the sequence.
Similarly, (101 × 102 × 103 × 104 × 105 ×…………………× 200) will also have minimum of 24 trailing zeros, but since there is a multiple of 125 present in it, total number of trailing zeros will be 25.
And , (601 × 602 × 603 × 604 × ……………… × 700) will have minimum of 24 trailing zeros + 2 more trailing zeros as multiple of 625 is present in it. (625 has 2 more zeros than multiples of 25).
Question 2.
To find the number of zeros in the series of product of factorials, we should start from (5!) as factorials preceding it will not have any zeros. We also know that a particular number of zero repeats itself in exactly five factorials.
[(5!)2 × (6!)2 × (7!)2 × (8!)2 × (9!)2] = 102 × 102 × 102 × 102 × 102 = 1010 .
[(10!)2 × (11!)2 × (12!)2 × (13!)2 × (14!)2] = 104 × 104 × 104 × 104 × 104 = 1020 .(All the factorials starting from (10!) to (14!) have two trailing zeros each, and when we square them, the number of trailing zeros in each of them becomes 4.
[(15!)2 × (16!)2 × (17!)2 × (18!)2 × (19!)2] = 106 × 106 × 106 × 106 × 106 = 1030 .
[(20!)2 × (21!)2 × (22!)2 × (23!)2 × (24!)2] = 108 × 108 × 108 × 108 × 108 = 1040 .
[(25!)2 × (26!)2 × (27!)2 × (28!)2 × (29!)2] = 1012 × 1012 × 1012 × 1012 × 1012 = 1060 .
[(30!)2 × (31!)2 × (32!)2 × (33!)2 × (34!)2] = 1014 × 1014 × 1014 × 1014 × 1014 = 1070 .
[(35!)2 = 1016 .
So, total number of trailing zeros in the sequence will be (10 + 20 + 30 + 40 +60 + 70 + 16) = 10246.
Question 3.
On reading the question, it should be quite obvious that we need to find out all the multiples of 125 which are of three-digits because all the multiples of (125!) have 3 more zeros than the preceding factorials.
If n is 124!, then (n + 1)! = 125!, a difference of 3 zeros in these two factorials.
Similarly, n can take values 249, 374, 499, 749, 875 and 999. Correspondingly, (n+ 1) will take 250, 375, 500, 750, 875 and 1000.
Don’t get confused about why we are counting 1000 also because (n + 1) is taking the value of 1000, but n is taking the value 999 which is a three-digit number. So, n can take 7 values.
Also, we did not count (625!) as there will be difference of 4 zeros between (624!) and (625!).
Question 4.
To find the number of trailing zeros which will be at the end of any factorial, we can find out the trailing zeros which will be skipped. We have already learnt about the series of skipped trailing zeros in factorials.
Trailing zeros skipped will be: 5, 11, 17, 23, 29, 30, 36, 42, 48, 54, 60, 61, 67, 73, 79, 85, 91 and so on.
And in the question is asking which one of the options will be trailing zero at the end of any factorials
a. 63 b. 68 c. 74 d. 80 e. All of these.
All the options given will be the trailing zeros at the end of any factorial. So, answer should be “all of these”.
Question 5.
This question is asking which one of the options cannot be number of zeros at the end of any factorial, which means we have to find out which particular zero given in the question is skipped. Options are
a. 28 b. 36 c. 19 d. 47.
Comparing the values given in the options with the series of skipped zero in the previous question, answer should be 36.
Question 6.
To find the trailing zeros in of (11 × 55 × 1010 × 1515 × 2020 × 2525.......... × 5050), we should just count number of 2’s in the sequence as number of 5’s are in every term except the first term, but number of 2’s are occurring every alternate term. So, number of 2’s will be less than number of 5’s, that’s why number of trailing zeros will depend on number of 2’s.
Now, we will count 2’s which will be found in (1010 × 2020 × 3030 × 4040 × 5050). Number of 2’s are [210 × (22)20 × 230 × (23)40 × 250] = [210 × 240 × 230 × 2120 × 250] = 2250.
If number of 2’s are 250, then number of trailing zeros will also be 250 as number of 5’s in the series are going to be more than 250.
So, answer is 10250.
Question 7.
To find trailing zeros in the sum of (1! + 2! + 3! + 4! +………..+ 18!), trailing zeros will start occurring from (5!) onwards. But the first four terms do not have any zero and when we add a non-zero number with a number ending with zero, we get non-zero digit at the unit’s place.
For e.g., if we add 2 + 4 + 7 + 100 = we would not any zero at the unit’s place, rather we will get a non-zero digit. Same thing happens in the question given.
If we add, we get ( 1 + 2 + 6 +24 + 120 + 720 + higher values of factorials ending with one or more trailing zeros). When we add up these numbers, unit’s place will not have any zero.
That’s why the answer is no trailing zero.
Question 8.
This is the same question based on the same lines of question 2.
Number of zeros in (20! × 21! × 22! × ……………… × 33!)3! can be calculated like:
(20! to 24!): Each has 4 trailing zeros resulting in total of 20 trailing zeros.
(25! to 29!): Each has 6 trailing zeros resulting in total of 30 trailing zeros.
(30! to 33!): Each has 7 trailing zeros resulting in total of 28 trailing zeros.
Number of trailing zeros in the series (20! × 21! × 22! × ……………… × 33!) is (1020 × 1030 × 1028) = 1078.
But the question is asking trailing zeros in (20! × 21! × 22! × ……………… × 33!)3! = (1078)6 = 10468.
