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