If 6k a factor of 25!, what is the largest possible value of k? a.6 b.4 c.9. d.10
Answers
Answer:
Option d
Step-by-step explanation:
Correct Question :-
If 6^k a factor of 25!, what is the largest possible value of k?
Solution :-
Given that 6^ k is a factor of 25! then
We have to find the number of 6's in 25!
So
6 can be written as 2×3
So we have to find powers of 2 and 3 in 25!
Number of 2 's in 25!
=> 25/2 + 25/4 + 25/8 + 25/16
=> 12+6+3+1
=> 22
Number of powers of 2 = 22
Number of 3's
=> 25/3 + 25/9
=> 8+2
=> 10
Number of powers of 3's = 10
=> 25! = 2²²×3¹⁰
=> 25! = 2¹⁰×3¹⁰×2¹²
=> 25! = 6¹⁰× 2¹²
Since the powers of 2 > The powers of 3 then we have to calculate only powers of 3
6^k = 6¹⁰
K = 10
The possible value = 10
Answer:-
The largest possible value of k for the given problem is 10
Used formulae:-
n! = n(n-1)(n-2)...1
Definition of n!:-
The product of all positive integers less than or equal to n is called factorial n and it is denoted by n! .Where n is a non-negative integer .
Given :- If 6^k a factor of 25!, what is the largest possible value of k ?
a.6
b.4
c.9.
d.10
Solution :-
→ 25 ! = 6^k
→ 25 ! = (2 * 3)^k
→ 25 ! = 2^k * 3^k .
so, number of 2's in 25! ,
→ (25/2¹) + (25/2²) + (25/2³) + (25/2⁴)
→ (25/2) + (25/4) + (25/8) + (25/16)
→ 12 + 6 + 3 + 1
→ 22
and, number of 3's in 25!, { since 3³ > 25 .}
→ (25/3¹) + (25/3²)
→ (25/3) + (25/9)
→ 8 + 2
→ 10
then,
→ 25 ! = 2^22 * 3^10
→ 25 ! = 2^(10 + 12) * 3^10
→ 25 ! = 2^(10) * 2^(12) * 3^10
→ 25 ! = (2 * 3)^10 * 2^12
→ 25 ! = (6)^10 * 2^12 = 6^k .
therefore, the largest possible value of k will be (D) 10.
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