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Determine the largest possible integer n such that 942! + 120! is divisible by 60^n.

 Oct 19, 2021
 #1
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The largest integer n is 13.

 Oct 19, 2021
 #2
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Since 60 ==2^2 * 3 * 5 and  [942! + 120!] == 2^116 * 3^58 * 5^28 * 7^19 * 11^10 * 13^9 * 17^7 * 19^6 * 23^5 * 29^4 * 31^3 * 37^3 * 41^2 * 43^2 * 47^2 * 53^2 * 59^2 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 3923 * 363677 * .........etc.

 

Therefore, the exponent of 5, which is 28, is the largest integer n, or: 60^28

 Oct 19, 2021

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