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

 Jan 5, 2022
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[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 ...........etc.

 

Since 5 is the largest prime factor of 60, it therefore follows that:

 

n ==28, so that: [942!  +  120!]  mod 60^28 ==0

 Jan 5, 2022

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