Question

the highest exponent of 24 in 20! is​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1. The highest exponent of the prime number p in n! is

$= \displaystyle \sf{\sum\limits_{k=1}^{m} \: \: \frac{n}{ {p}^{k} } \: }$

$\sf{\: where \: k \: is \: the \: smallest \: integer \: such \: that \: \: {p}^{k +1} > n}$

2. BOX FUNCTION :

$\sf{ \: [ x ] = the \: greatest \: integer \: but \: not \: greater \: than \: x }$

TO DETERMINE

The highest exponent of 24 in 20!

CALCULATION

Here

$24 = 2 \times 2 \times 2 \times 3 = {2}^{3} \times 3$

So in the prime factorisation of 24 only two prime numbers 2 & 3 are present

So we will proceed to find the power of 2 & 3 in 20!

Now $\sf{ \: {2}^{5} > 20 \: }$

So the largest power of the prime number in 20! is

$= \displaystyle \: \sf{\bigg[\: \frac{20}{2} \bigg]} + \: \sf{\bigg[\: \frac{20}{ {2}^{2} } \bigg]} + \: \sf{\bigg[\: \frac{20}{ {2}^{3} } \bigg]} + \: \sf{\bigg[\: \frac{20}{ {2}^{4} } \bigg]}$

$= \displaystyle \: \sf{\bigg[\: \frac{20}{2} \bigg]} + \: \sf{\bigg[\: \frac{20}{ 4 } \bigg]} + \: \sf{\bigg[\: \frac{20}{ 8 } \bigg]} + \: \sf{\bigg[\: \frac{20}{ 16 } \bigg]}$

$= \displaystyle \sf{10 + 5 + 2 + 1}$

$= \displaystyle \sf{18}$

$\sf{ \: Again \: \: {3}^{3} > 20}$

So the largest power of the prime number in 20! is

$= \displaystyle \: \sf{\bigg[\: \frac{20}{3} \bigg]} + \: \sf{\bigg[\: \frac{20}{ {3}^{2} } \bigg]}$

$= \displaystyle \: \sf{\bigg[\: \frac{20}{3} \bigg]} + \: \sf{\bigg[\: \frac{20}{9} \bigg]}$

$= \displaystyle \: \sf{6 + 2}$

$= \displaystyle \: \sf{8}$

Hence the highest exponent of 24 in 20! is

= min ( 18/3, 8)

= min ( 6, 8 )

= 6

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