Practice Questions
Trailing Zeroes in n!
Trailing zeroes come from factors 10 = 2×5. Count powers of 5 in n! (always fewer than powers of 2).
Formula
Trailing zeroes = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …
Examples
- 10! → 10/5 = 2 trailing zeroes
- 25! → 25/5 + 25/25 = 5 + 1 = 6
Highest power of p in n! (prime p): ⌊n/p⌋ + ⌊n/p²⌋ + …
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Q1 EasyHow many zeros will be there at the end of the product 15! ?ExplanationTrailing zeroes in 15! = ⌊15/5⌋ = 3.
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Q2 EasyHow many zeros will be there at the end of the product 7! ?ExplanationTrailing zeroes in 7! = ⌊7/5⌋ = 1.
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Q3 EasyWhen 30 factorial is written out completely, how many trailing zeros appear at the end of the number?ExplanationTrailing zeroes in 30! = ⌊30/5⌋ + ⌊30/25⌋ = 6 + 1 = 7.
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Q4 EasyHow many trailing zeros are there at the end of 10 factorial?ExplanationTrailing zeroes in 10! = ⌊10/5⌋ = 2.
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Q5 EasyWhen 5 factorial is evaluated completely, how many zeros appear at the end of the result?Explanation5! = 120, so there is 1 trailing zero.
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Q6 MediumHow many zeros are there at the end of the product 25! ?ExplanationTrailing zeroes in 25! = ⌊25/5⌋ + ⌊25/25⌋ = 5 + 1 = 6.
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Q7 MediumHow many zeros are there at the end of the product 100! ?ExplanationTrailing zeroes in 100! = ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24.
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Q8 MediumFind the number of zeroes at the end of the product 50!.ExplanationTrailing zeroes in 50! = ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12.
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Q9 MediumWhat is the minimum value of n for which n! will contain at least 3 trailing zeros?ExplanationAt least 3 trailing zeros need ⌊n/5⌋ ≥ 3, so n must be at least 15.
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Q10 MediumWhat is the largest value of n below 20 for which n! ends with exactly one trailing zero?ExplanationExactly one trailing zero means ⌊n/5⌋ = 1. This holds for n = 5 to 9 and again for n = 10 to 14. The largest below 20 is 14.
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Q11 MediumHow many times does the factor 5 appear in the prime factorisation of 40!?ExplanationPower of 5 in 40! = ⌊40/5⌋ + ⌊40/25⌋ = 8 + 1 = 9.
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Q12 HardHow many zeros will be there at the end of the product 125! ?ExplanationTrailing zeroes in 125! = ⌊125/5⌋ + ⌊125/25⌋ + ⌊125/125⌋ = 25 + 5 + 1 = 31.
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Q13 HardWhat is the highest power of 3 that divides 20! completely?ExplanationPower of 3 in 20! = ⌊20/3⌋ + ⌊20/9⌋ + ⌊20/27⌋ = 6 + 2 + 0 = 8.
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Q14 HardHow many trailing zeros appear at the end of 200 factorial?ExplanationTrailing zeroes in 200! = ⌊200/5⌋ + ⌊200/25⌋ + ⌊200/125⌋ = 40 + 8 + 1 = 49.
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DS1 Easy
How many trailing zeros are there at the end of n factorial?
Statement IThe value of n is 25.
Statement IIThe integer n is greater than 20.
Explanation25! has ⌊25/5⌋ + ⌊25/25⌋ = 6 trailing zeros. Statement I alone is sufficient. Statement II alone only tells us n > 20 and does not fix the count. -
DS2 Easy
How many trailing zeros are there at the end of 50 factorial?
Statement IThe number 50 factorial has exactly 12 trailing zeros.
Statement IIThe integer 50 is divisible by 10.
ExplanationStatement I directly gives the answer as 12. Statement II is true but does not by itself determine the number of trailing zeros.
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DS3 Medium
What is the highest power of 5 that divides 30 factorial completely?
Statement IThe prime factorisation of 30! can be written as 2^a × 3^b × 5^c × …
Statement IIThe exponent c in the prime factorisation of 30! is equal to 7.
ExplanationStatement II directly gives c = 7, which is the highest power of 5 in 30!. Statement I alone names the form of the factorisation but does not compute c without further work. -
DS4 Hard
What is the smallest positive integer n such that n! has at least 5 trailing zeros?
Statement IThe value of ⌊n/5⌋ is at least 5.
Statement IIThe integer n is a multiple of 5.
ExplanationAt least 5 trailing zeros require n ≥ 25. Statement I alone fixes the smallest such n as 25. Statement II alone does not determine the minimum value of n.