Determine the largest possible integer n such that 942! + 120! is divisible by 60^n.
60 = 2^2 * 3 * 5
[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.....etc.
Since 5 is the largest prime factor of 60, therefore 28 would be the largest exponent of 60, or 60^28.
