SOLUTION: The prime factorization of "k!" is {{{ 2^25 }}} x {{{ 3^13 }}} x {{{ 5^6 }}} x {{{ 7^4 }}} x {{{ 11^2 }}} x {{{ 13^2 }}} x 17 x 19 x 23. The value for "k" is

Algebra ->  Divisibility and Prime Numbers  -> Lessons -> SOLUTION: The prime factorization of "k!" is {{{ 2^25 }}} x {{{ 3^13 }}} x {{{ 5^6 }}} x {{{ 7^4 }}} x {{{ 11^2 }}} x {{{ 13^2 }}} x 17 x 19 x 23. The value for "k" is      Log On


   



Question 1188320: The prime factorization of "k!" is +2%5E25+ x +3%5E13+ x +5%5E6+ x +7%5E4+ x +11%5E2+ x +13%5E2+ x 17 x 19 x 23. The value for "k" is
Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
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The prime factorization of "k!" is 2%5E25%2A3%5E13%2A5%5E6%2A7%5E4%2A11%5E2%2A13%5E2%2A17%2A19%2A23. The value for "k" is
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Looking at the factors (multipliers)  13%5E2  and  17, we can conclude that  26<= k < 34.


Looking at the factor 11%5E2,  we can exacerbate this inequality  26 <= k < 33.


Looking at the factor 7%5E4,  we can exacerbate this inequality  28 <= k < 33.


Looking at the factor 5%5E6,  we can exacerbate this inequality  28 <= k < 30.


Looking at the factor 3%5E13,  we can confirm this inequality  28 <= k <= 29.


Since the prime number 29 is not presented in the k! factorization, we conclude that 28 <= k < 29,

             which means that k = 28.


CHECK.  Check it on your own that k = 28 provides the degree of 2 equal to 24 in the factorization of k!


ANSWER.  (a) The condition has a highlight%28MISTAKE%29 stating that the index of 2 is 25: it is 24.


         (b)  With the corrected prime factorization of k! = 2%5E24%2A3%5E13%2A5%5E6%2A7%5E4%2A11%5E2%2A13%5E2%2A17%2A19%2A23,  the value of k is 28.


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Again:   the given formulation is  INCORRECT.

There is  NO  integer number of  " k "  with the given factorization of  k!

With the  corrected factorization,  the answer is   k = 28.