Lyndon chose this as one of his favourite problems. It is
accessible but needs some careful analysis of what is included and
what is not. A systematic approach is really helpful.
Find the highest power of 11 that will divide into 1000! exactly.
How many zeros are there at the end of the number which is the
product of first hundred positive integers?
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides
exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest
power of two that divides exactly into 100!?
I have forgotten the number of the combination of the lock on my
briefcase. I did have a method for remembering it...
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . .