Question

What is the maximum value of k for which 105!/12k is an integer?​

Answer 1

Answer:

50

Step-by-step explanation:

Highest power of 3 in 105! =

[105/3^1]+[105/3^2]+[105/3^3]+[105/3^4]

= 35+11+3+1 => 50

where [ ] this bracket signifies greatest integer function.

As heighest power of 2 will be greater in 105! We don't need to calculate this for all the prime factors of 12 to get the maximum value of k for 105!/12^k to be an integer

Answer 2

The max value of k is 50

Step-by-step explanation:

We know that for n!, and a given prime number p, the highest power of p in n! is given by

$\boxed{[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+.....}$

Where [ ] is the greatest integer function

In the given question

$\frac{105!}{12^k}$

$=\frac{105!}{(3\times 2^2)^k}$

$=\frac{105!}{3^k\times 2^{2k}}$

Power of 3 in 105!

$=[\frac{105}{3}]+[\frac{105}{3^2}]+[\frac{105}{3^3}]+[\frac{105}{3^4}+[\frac{105}{3^5}+...$

$=[35]+[11.67]+[3.89]+[1.3]+[0.43]$

$=35+11+3+1+0$

$=50$

Power of 2 in 105!

$=[\frac{105}{2}]+[\frac{105}{2^2}]+[\frac{105}{2^3}]+[\frac{105}{2^4}+[\frac{105}{2^5}+\frac{105}{2^6}+\frac{105}{2^7}+...$

$=[52.5]+[26.25]+[13.125]+[6.5625]+[3.28125]+[1.640625]+[0.8203125]$

$=52+26+13+6+3+1+0$

$=101$

Thus, if we take $k=50$, $\frac{105!}{12^k}$ will be an integer

Hope this answer is helpful.

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