Thursday, November 04, 2004
Number of 0's
13! = 13 × 12 × 11 × ... × 2 × 1 = 6227020800, and it can be seen that 13! contains two trailing zeros.
How many such trailing zeros does 1000! contain?
Answer:
Every time a number is multiplied by 10 an extra trailing zero is added, and as 10 = 2 × 5, we need only consider the number of factors of 5 present in 1000!; there are an abundance of factors of 2.
There are 1000/5=200 factors of 5, but we must also ensure that we take into account 25 = 5 × 5, which contains two factors of 5; hence there will be 1000/25=40 extra factors of 5. Similarly, 125 = 53, contains three factors of 5, and 625 = 54.
1000/5 = 200, 1000/25 = 40, 1000/125 = 8, and 1000/625 = 1.6.
Therefore, 1000! contains, 200 + 40 + 8 + 1 = 249 trailing zeros.
How many such trailing zeros does 1000! contain?
Answer:
Every time a number is multiplied by 10 an extra trailing zero is added, and as 10 = 2 × 5, we need only consider the number of factors of 5 present in 1000!; there are an abundance of factors of 2.
There are 1000/5=200 factors of 5, but we must also ensure that we take into account 25 = 5 × 5, which contains two factors of 5; hence there will be 1000/25=40 extra factors of 5. Similarly, 125 = 53, contains three factors of 5, and 625 = 54.
1000/5 = 200, 1000/25 = 40, 1000/125 = 8, and 1000/625 = 1.6.
Therefore, 1000! contains, 200 + 40 + 8 + 1 = 249 trailing zeros.
