How many zeroes are at the end of $42! + 10!$ ($42$ factorial plus $10$ factorial)? (Reminder: The number $n!$ is the product of the integers from $1$ to $n$. For example, $5!=5\cdot 4\cdot3\cdot2\cdot 1= 120$.)
Since 10! is much much smaller than 42!, then no matter how zeros 42! has it will end having the same number of zeros as 10!, since you are adding 10! + 42!.
Therefore, you will have:
10! = 3,628,800 - or 2 zeros at the end of 42! + 10!
Since 10! is much much smaller than 42!, then no matter how zeros 42! has it will end having the same number of zeros as 10!, since you are adding 10! + 42!.
Therefore, you will have:
10! = 3,628,800 - or 2 zeros at the end of 42! + 10!
+1348 
