Date: Thu, 20 Jul 2000 20:53:01 EDT Subject: Factorials Hi I was just wondering if you could tell me how many zeros are in 100,000! (factorial.) I am a college sophomore but i believe this could be considered a highschool level question. I would appreciate your help. Thank you. sincerely Lauren Hi Lauren, Let's try 25! instead. This is the general idea. 25! = 1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x...x20...x25 What power of 10 divides 25!? 10 = 2x5 thus every time 10 divides 25! we need a 2 and a 5 to divide 25! How many 5 divide 25!? 5 divides every 5th number (5, 10, 15, 20, 25) and 52 divides 25 thus there are 6 5's in 25! How many 2's? 2 divides every second number, 22 divides every 4th number, 23 divides every 8th number and 24 divides every 16th number, thus the number of 2 that appear is much larger than the number of 5's. Since there were only 25/5 + 25/52 = 5 + 1 = 6 5s in 25! we only get 10 to divide 25! 6 times in spite of the presence of all those 2s. In fact 25! = 15511210043330985984000000 For 200!, 10 divides it 200/5 + 200/52 + 200/53 = 40 + 8 + 1 = 49 times. (Notice here that 200/53 = 200/125 = 1.6 but we want only the integer part, 1.) For 100,000 the argument is the same. 5 divides every 5th number in the sequence 1, 2, ..., 100,000 52 divides every 25th number in the sequence 1, 2, ..., 100,000 53 divides every 125th number in the sequence 1, 2, ..., 100,000 54 divides every 54th number in the sequence 1, 2, ..., 100,000 etc. In total 100,000/5 + 100,000/52 + 100,000/53 + ... times. (Again in each division you want only the integer part.) Cheers, Denis Go to Math Central