Smallest Value of N!; Factorial TableDate: 11/07/2001 at 19:46:26 From: Melissa Subject: How to find N The symbol 3! means 3*2*1, which equals 6. Similarly, 5! means 5*4*3*2*1, which equals 120, and so on. Suppose that N! ends in exactly 3 zeros after fully multiplying it out. What is the smallest value that N can have? I have tried doing this on a calculator but it would have taken forever by hand. Is there any pattern or way to make this faster? Date: 11/07/2001 at 20:20:38 From: Doctor Paul Subject: Re: How to find N Think of what generates a zero at the end of a number. Pick a number - say 3. What do you have to do to 3 to add a zero to the end of it? I think you'll agree that the answer is: multiply it by ten. So certainly 30! will have at least three zeros at the end of it because you multiply by 10 three times in the computation of 30! But are there any other ways to get 10? Yes = 10 = 2*5 Every even number (every other number) contains a 2 - but the fives are much more rare. So it seems that 5! will end in one zero, 10! will end in 2 zeros and 15! will be the first number to end in 3 zeros. Similarly, 20! ends in 4 zeros but 25! ends in *six* zeros! Why the jump? Because 25 contains two fives! Here's a factorial table from 1! to 30! : 01! = 1 02! = 2 03! = 6 04! = 24 05! = 120 06! = 720 07! = 5040 08! = 40320 09! = 362880 10! = 3628800 11! = 39916800 12! = 479001600 13! = 6227020800 14! = 87178291200 15! = 1307674368000 16! = 20922789888000 17! = 355687428096000 18! = 6402373705728000 19! = 121645100408832000 20! = 2432902008176640000 21! = 51090942171709440000 22! = 1124000727777607680000 23! = 25852016738884976640000 24! = 620448401733239439360000 25! = 15511210043330985984000000 26! = 403291461126605635584000000 27! = 10888869450418352160768000000 28! = 304888344611713860501504000000 29! = 8841761993739701954543616000000 30! = 265252859812191058636308480000000 - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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