Question

If 6k a factor of 25!, what is the largest possible value of k? a.6 b.4 c.9. d.10​

Answer 1

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 .

Answer 2

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.

Learn more :-

let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46

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