Number of trailing zeroes in a factorial (n!)


  • QA/DILR Mentor | Be Legend


    Number of trailing zeroes in n!
    = Number of times n! is divisible by 10
    = Highest power of 10 which divides n!
    = Highest power of 5 in n!

    The question can be put in any of the above ways but it can be answered using the simple formula given below:

    [n/5] + [n/25] + [n/125] +......

    { [x] is the greatest integer function. [4.99] = 4, [4.01] = 4, [- 4.99] = 5, [-4.01] = 5}

    The above formula gives us the exact number of 5s in n! because it will take care of all multiples of 5 which are less than n. Not only that it will take care of all multiples of 25, 125, etc. (higher powers of 5).

    Note : Instead of dividing by 25, 125, etc. (higher powers of 5); it would be much faster if you divided by 5 recursively.

    Let us use this to solve a few examples:

    Q1) What is the number of trailing zeroes in 23! ?

    [23/5 ] = 4. It is less than 5, so we stop here.
    The answer is 4.

    Q2) What is the number of trailing zeroes in 123! ?

    [123/5] = 24
    Now we can either divided 123 by 25 or the result in the above step i.e. 24 by 5.
    [24/5 ]= 4. It is less than 5, so we stop here.
    The answer is = 24 + 4 = 28

    Q3): What is the number of trailing zeroes in 1123!?

    [1123/5] = 224
    [224/5] = 44
    [44/5] = 8
    [8/5] = 1.
    It is less than 5, so we stop here.
    The answer is = 224 + 44 + 8 + 1 =277


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