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The largest integer value of x such that 6^x divides exactly into 100! is?

 Nov 28, 2020
 #1
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x = [100 / 3 + 100 / 3^2  + 100 / 3^3  + 100/ 3^4] = 48

 Nov 28, 2020
 #2
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Since 6 = 2 x 3, then the greatest power of 3 when factoring 100! =48. And since 2 has power of 97, then 6^x =[2^48 x 3^48] = 6^48.

 

100!= 2^97 * 3^48 * 5^24 * 7^16 * 11^9 * 13^7 * 17^5 * 19^5 * 23^4 * 29^3 * 31^3 * 37^2 * 41^2 * 43^2 * 47^2 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97

Guest Nov 28, 2020

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