Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
Find the number of trailing zeros in 50!
10
12
14
16
12
Divide the number repeatedly by 5 and add the quotients, ignoring remainders. For 50, do 50/5 = 10 and 10/5 = 2, then add 10+2 = 12.
We are given the factorial of a number, specifically 50!.
Trailing Zeros=∑k=1∞⌊5kn⌋
Divide the number repeatedly by 5 and add the quotients, ignoring remainders. For 50, do 50/5 = 10 and 10/5 = 2, then add 10+2 = 12.
Students often make the mistake of dividing only by 5 once, forgetting that higher powers of 5 (like 25, 125, etc.) also contribute additional factors of 5.
Identify the power of 5
To find trailing zeros in n!, we need the count of factors of 10, which is determined by the number of pairs of prime factors (2imes5). Since 5 is the limiting factor, we count the total occurrences of 5 in the prime factorization of 50!.
n=50
Apply Legendre's Formula
Using the formula, divide 50 by increasing powers of 5: first by 51 and then by 52.
⌊550⌋+⌊2550⌋
Calculate the quotients
Divide 50 by 5 to get 10, and divide 50 by 25 to get 2.
10+2=12
Final Result
Summing the quotients gives the total number of trailing zeros in 50!.
12
B is correct because the sum of the quotients ⌊550⌋+⌊2550⌋ equals 10+2=12.
This logic is essential for number theory problems involving trailing zeros and is also useful in finding the highest power of any prime p that divides n!.
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download • 100K+ Aspirants