Problem Solving (PS) | Re: What is the largest value of x for which 12^x leaves zero remainder

50!

Number of 3s =\(\frac{50}{3}\) +\(\frac{50}{9}\) +\(\frac{50}{27}\) = 16 + 5+ 1 = 22

Number of 2s =\(\frac{50}{2}\) +\(\frac{ 50}{4}\) +\(\frac{50}{8}\) +\(\frac{50}{16}\) +\(\frac{50}{32}\) = 25 +12 + 6 + 3 + 1 = 47. Hence number of 4s = 46/2 = 23

Hence no. of 12s = 22

10!

Number of 3s =\(\frac{10}{3}\) +\(\frac{10}{9}\) = 3 + 1 = 4

Number of 2s =\(\frac{10}{2}\) +\(\frac{ 10}{4}\) +\(\frac{10}{8}\) = 5 +2 + 1 = 8. Hence number of 4s = 8/2 = 4

Hence no. of 12s = 4

Hence\(\frac{50!}{10!}\) will contain\( 12^{22 - 4}\) =\(12^{18}\)
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