How to Find Number of Trailing Zeros in a Factorial or Product
Under the topic of Number of Zeros, it is expected to find out the number of trailing zeros at the end of the number. In simple words, it can be said that to calculate the No. of Zeros at the right side of the number.
To make the things more clear, let us take a simple example to understand the concept of Number of Zeros
Example : 1234057000 → No. of Trailing zeros → 3
1050500000 → No. of Trailing zeros → 5
1.5 x 10⁵ → No. of Trailing zeros → 4
Such Zeros that are represented at the end of the numbers are actually the Trailing Zeros.
Now the next thing arises about the formation of Zeros in the Numbers.
So, the fundamental regarding the formation of a Zero is the presence of a Pair of 2 & 5.
i.e. 10 → 2¹ x 5¹
100 → 2² x 5²
1000 → 2³ x 5³
10000 → 2⁴ x 5⁴
KEY HIGHLIGHTS :
Number of trailing zeroes is going to be the power of 2 or 5, whichever is lesser.
For a number to be divisible by 10, it should be divisible by 2 & 5.
Practice Material :
Q 1 .Calculate Number of Zeros in 100!.
23
24
20
22
Sol : To calculate Number of Zeros in 100! , it is required to calculate Number of 2’s and Number of 5’s in 100!.
Calculating No. of 5’s
100
━━ = 20
5
20
━━ = 4
5
No. of 5’s in 100! = 20 + 4 = 24
Now calculate No. of 2’s
100
━━ = 50
2
50
━━ = 25
2
25
━━ = 12
2
12
━━ = 6
2
6
━━ = 3
2
3
━━ = 1
2
Number of 2’s in 100! = 50 + 25 + 12 + 6 + 3 + 1 = 97
{ Number of trailing zeroes is going to be the power of 2 or 5, whichever is lesser. }
Now the Number of Zeros is the Number of Pairs of 2’s and 5’s form = 24
Q 2. Which of the following has 23 trailing zeros ?
99!
95!
No such factorial exist
None of these
Sol : No such factorial exist with 23 Trailing Zeros. Because upto 99! there are 22 Trailing Zeros.
No. of Trailing Zeros in 100 ! = 24 Zeros
No. of Trailing Zeros in 951 to 99! = 22 Zeros
At 100! , 2 Zeros add to these 22 zeros; making 24 Zeros in 100 Factorial.
There is no valid value for which there are 23 trailing zeros.
Q 3. What is the number of trailing in 125000 ?
3
4
7
6
Sol : Prime Factorization of 125000 = 2³ x 5⁶
Since, Number of trailing zeroes is going to be the power of 2 or 5, whichever is lesser.
Hence, the number of trailing in 125000 is 3
Q 4. Find the number of trailing zeros in
1¹ x 2² x 3³ x 4⁴ x ……………………...100¹⁰⁰ .
1300
1305
1050
1500
Sol : Since , the number of trailing zeros depends upon the number of 2’s and 5’s.
We can clearly say the power of 5 is the limiting factor in the expression.
1¹, 2² , 3³, 4⁴ and so on are such numbers which have no 5’s
Number of 5’s lies only in the multiple of 5.
5⁵ → No of 5’s → 5
10¹⁰ → No of 5’s → 10 (2 x 5)
15¹⁵ → No of 5’s → 15 (3 x 5)
20²⁰ → No of 5’s → 20 (4 x 5)
25²⁵ → No of 5’s → 50 (5 x 5)
30³⁰ → No of 5’s → 30 (5 x 6)
35³⁵ → No of 5’s → 35 (5 x 7)
40⁴⁰ → No of 5’s → 40 (5 x 8)
45⁴⁵ → No of 5’s → 45 (5 x 9)
50⁵⁰ → No of 5’s → 100 (2 x 5 x 5)
55⁵⁵ → No of 5’s → 55 (5 x 11)
60⁶⁰ → No of 5’s → 60 (5 x 12)
65⁶⁵ → No of 5’s → 65 (5 x 13)
70⁷⁰ → No of 5’s → 70 (5 x 14)
75⁷⁵ → No of 5’s → 150 (3 x 5 x 5)
80⁸⁰ → No of 5’s → 80 (5 x 16)
85⁸⁵ → No of 5’s → 85 (5 x 17)
90⁹⁰ → No of 5’s → 90 (5 x 18)
95⁹⁵ → No of 5’s → 95 (5 x 19)
100¹⁰⁰ → No of 5’s → 200 (2 x 2 x 5 x 5)
No. of 5’s = 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45 + 100 + 55 + 60 + 65 + 70 + 150 + 80 + 85 + 90 + 95 + 200 = 1300
No. of Trailing Zeros = 1300
Q 5. What is the number of trailing zeros in 1123! ?
224
250
277
296
Sol : Power of 5 is the limiting factor in the expression
Calculating No. of 5’s
1123
━━ = 224
5
224
━━ = 44
5
44
━━ = 8
5
8
━━ = 1
5
No. of 5’s in 1123! = 224 + 44 + 8 + 1
Hence , the number of trailing zeros in 1123! = 224 + 44 + 8 + 1 = 277
Very Informative
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