Exponent Of Prime In N! Illustration Introduction
As we know formula to find exponent of
P
in
n
! is
p
, is a prime number let’s explore it fraction with examples.
Example 1:
Find exponent of 5 in 200!
Sol.
Exponent of 5 in 200! Would be
40 + 8 + 1 + 0 = 49
Example 2:
Find exponent of 4 in 98!
Sol.
Exponent of 4 in 98! Can be calculated with the help of exponent of 2 in 98! which is
49 + 24 + 12 + 6 + 3 + 1 + 0 = 95
Now exponent of 4 would be
Explanation:
Each 4 is a pair of 2 hence pair of 2 in 95 two’s is 47 with one 2 remaining hence 47 is the answer.
Example 3:
Find exponent of 7 in 343!?
Sol.
Exponent of 7 in 343! Would be
49 + 7 + 1 = 57
Example 4:
Find exponent of 11 in 541!?
Sol.
Exponent of 11 in 541! would be
49 + 4 + 0 = 53
Exponent Of Prime In N! Illustration Rapid Question
Rapid Question :
1.
Find exponent of 5 in 249!
2.
Find exponent of 7 in 198!
Illustration
:
1.
Find exponent of 8 in 165!
Sol.
Exponent of 8 in 165! Could be find with the help of exponent of 2 in 165! which is
82 + 41 + 20 + 10 + 5 + 2 + 1 + 0
= 161
Now exponent of 8 is
Explanation:
8 needs 3 two’s so set of 3 two’s in 161 two’s would be
2.
Find exponent of 9 in 401!
Sol.
Exponent of 9 in 401! could be find with the help of exponent of 3 in 401! hence
133 + 44 + 14 + 4 + 1 + 0
= 196
Now count for exponent of 9 in 401! is
Highest Power of a Prime Number p in nCr illustration
Highest Power of a Prime Number p in nCr illustration :
Prime number are either divisible by 1 or itself examples of prime numbers are 2, 3, 5, 7 etc.
represents combination or selection of r objects out of n and denoted as
, n! means
means
and
means
Exponent of prime number in
means finding exponents of all prime factors possible in
for example exponent of 3 in
Can be find by writing
as
=
Now we must find exponent of numerator and denominator separately and subtract them to get the final result Now 8!=
7
×
6
× 5 × 4 × 3 × 2 × 1 or
7 × 2 × 3 × 5 × 2 × 2 × 3 × 2 × 1 has exponent of 3 as 2, similarly in
6! = 6
× 5 × 4 × 3 × 2 × 1 exponent of 3 is 2 and in 2! =
exponent of 3 is 0, combined exponent of 3 in denominator is (2+0=2)
Resulting exponent of numerator and denominator is (2-2=0), hence exponent of
3 in
is 0.
Above process can be generalized for
as
Q =
,
R =
,
Exponent of 5 in
could be find as
Break
as
Exponent of 5 in 12! Would be
Exponent of 5 in 8! Would be
Exponent of 5 in 4! Would be
Now resulting exponent would be (2-1-0) = 1
Hence count of exponent for 5 in
would be 1
Prime Number p in nCr illustration Introduction
Introduction:
As discussed above exponent of prime p in
could be found by using above algorithm,
let’s explore it with examples.
Example 1:
Find exponent of 5 in
?
Solution:
could be written as
Exponent of 5 in 1000! Would be
Exponent of 5 in
850!
Would be
Exponent of 5 in
150!
Would be
Now resulting value would be (249-211-37) = 1)
Hence count of exponent for 5 in
would be 1
Example 2:
Find exponent of 19 in
?
Solution:
could be written as
Exponent of 19 in 1512! Would be
Exponent of 19 in 1483! Would be
Exponent of 19 in 29! Would be
Now resulting value would be (84-83-1 = 0)
Hence count of exponent for 19 in
is 0.
Example 3:
Find exponent of 23 in
?
Solution:
could be written as
Exponent of 23 in 52! Would be
2 + 0 = 2
Exponent of 23 in 46! Would be
2 + 0 = 2
Exponent of 23 in 5! Would be
0
Now resulting value would be (2-2-0 = 0)
Hence count of exponent for 23 in
is 0
Example 4:
Find exponent of 4 in
?
Solution:
4 as an exponent of prime factor would be
could be written as
Exponent of 2 in 100! would be
50 + 25 + 12+6+3+1+0=97
Exponent of
in 100! Would be
Exponent of 2 in 50! would be
25+12+6+3+1+0=47
Combined exponent of
is 47+47=94
Exponent of
in
Would be
Now resulting value would be (48-47 = 1)
Hence count of exponent for 4 in
is 1
Prime Number p in nCr illustration Rapid Question
1.
Find exponent of 29 in
in
?
2.
Find exponent of 31 in
?
Prime Number p in nCr illustration
1.
Find exponent of 15 in
?
Solution:
15 could be written to product of prime factor as
could be written as
Exponent of 3 in 130! Would be
Exponent of 3 in 92! Would be
Exponent of 3 in 38! Would be
Now resulting exponent of 3 would be (62-44-17 = 1)
Exponent of 5 in 130! Would be
Exponent of 5 in 92! Would be
Exponent of 5 in 38! Would be
Now resulting exponent of 5 would be (32-21-8 = 3)
Least of 3 and 1 is 1
Hence count of exponent for 15
in
would be 1
Prime Number p in nCr illustration Rapid Question
1.
Find exponent of 6 in
?
2.
Find exponent of 14 in
?
are same we must take care of arrangements while using fundamental rule of multiplication let’s explore it with examples.
