Number of zeroes concept is one of the most important and frequently tested topics in quantitative aptitude and number theory for competitive exams. It is asked in almost every major exam including SSC CGL, SSC CHSL, Bank PO, Bank Clerk, Railway RRB, CAT and UPSC CSAT. This topic is also important for AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. A strong understanding of trailing zeroes concept, highest power of a prime number in factorial and highest power of composite numbers in factorial is essential for scoring well in these exams. In this post we cover everything from how zeroes are formed in a product, number of trailing zeroes in factorial of any number, highest power of any prime number in a factorial using both formula method and successive division method, and highest power of composite numbers in factorial — all explained with solved examples.
Number of zeroes in an expression:-
Zero will be formed when we multiply 2 and 5
10 = 2×5
100 = 2²×5²
1000 = 2³×5³
10000 = 2⁴×5⁴
So we can say that to have ‘n’ zeroes at the end of a product we need exactly ‘n’ combinations of 2 and 5.
Number of zeroes in the factorial of a number:-
Number of zeroes in the factorial of a number will be the highest power of 10 in the factorial & 10 can be written as 2×5 in its prime factor form and we know that in any factorial number power of 2 is always greater than power of 5. Hence to find the number of trailing zeroes we only need to find the highest power of 5 in the factorial & that will be the number of trailing zeroes in the factorial of the number.
Ex:- Find the highest power of 10 or number of trailing zeroes in 160!
Sol:-
i.e. highest power of 5 in 160! = 32 + 6 + 1
= 39
∴ Number of trailing zeroes in 160! = 39 Answer
Ex:- Find the highest power of 1000 in 1000!
Sol:-
i.e. = 200 + 40 + 8 + 1 = 249
∴ highest power of 5 in 1000! is 249 & that of 5³ will be \(\frac{{249}}{3}\) = 83
∴ Highest power of 1000 in 1000! will be 83 Answer
Highest power of a number in N!:-
method (1):-
Highest power of prime number p that divides n! exactly i.e. without leaving any remainder is given by:-
where [a] represents greatest integer less than or equal to a.
Example:- Find the highest power of 3 in 100!
Solution:-
= 33 + 11 + 3 + 1
= 48 Answer
method (2):-
Quotient form → Successive division
(Q). Find the highest power of 6 in 150!
Solution:- First, since 6 is a composite number so convert it into its prime factors.
6 = 2 × 3
now
So highest power of 2 in 150! = 146
& highest power of 3 in 150! = 72
∴ Highest power of 6 in 150! = minimum of above two values = 72 Answer