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