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Q.

The largest non - negative number k such that $24^{k}$ divides 12! is

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a

2

b

3

c

4

d

5

answer is B.

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Detailed Solution

We know that if there are two numbers p and q which are relative prime (it means greatest common factor (GCD) between p and q is 1) and they exactly divide any number n if and only if their product pq also exactly divides the number n. It is one of the divisibility rules. We know that the largest power on any prime p is r such that pr divides n! then we have,
r =

[\frac{n}{p}]

+

[\frac{n}{p^{2}}]

+

[\frac{n}{p^{3}}]

+ ...
Here [x] is the greatest integer function which returns greatest integer less than equal to x. Now, let us consider the prime factorization of 24. We have
24 = 2×2×2×3 =

2^{3}

×3 = 8 × 3
If a number is divisible by the relative primes 8 and 3 then the number is also divisible by 24. Let us consider some natural numbers λ

24^{k}

=

2^{m} 3^{l}

λ......(1)
Here m and l are the largest power on 2 and 3 respectively. If

24^{k}

divide 12! then

2^{m}

and

3^{l}

also divide 12! . We use largest power formula for n=12 and p=2 to get m as,
m =

[\frac{12}{2}]

+

[\frac{12}{2^{2}}]

+

[\frac{12}{2^{3}}]

+

[\frac{12}{2^{4}}]

...
⇒ m = [6] + [3] + [1.5] + [0.75] + ...
⇒ m = 6 + 3 + 1 + 0 + 0...
⇒ m = 10
We use largest power formula for n=12 and p=3 to get ll as,
l =

[\frac{12}{3}]

+

[\frac{12}{3^{2}}]

+

[\frac{12}{3^{3}}]

+ ...
⇒ l = [4] + [1.

\underset{̲}{3}

] + [0.

\underset{̲}{4}

] +...
⇒ l = 4 + 1 + 0...
⇒ l = 5
We put the value of m,l in equation (1) to have,

24^{k}

=

2^{10} 3^{5}

λ
⇒

24^{k}

= (2×2×2×2×2×2×2×2×2×2×2×3×3×3×3×3)λ
We consider the divisibility rule of 24 and arrange the product 24 = 8×3 =

2^{3}

× 3 will be close to each other. We have,
⇒

24^{k}

=

2^{3}

× 3 ×

2^{3}

× 3 ×

2^{3}

× 3 ×

2^{3}

× 3 × λ
⇒

24^{k}

=

{(2^{3} \times 3)}^{3}

× 18λ
⇒

24^{k}

=

24^{3}

× 18λ We compare power on 24 on both sides of the equation and find the largest power on k such that

24^{k}

divides 12! as 3.
So, the correct answer is “Option 2”.

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