Exponents and trailing zeroes - Anubhav Sehgal, NMIMS Mumbai


  • NMIMS, Mumbai (Marketing)


    Finding exponent of one number in another number

    1. Linear format
    2. Factorial format

    Linear format

    Q. Find the maximum power of 5 that will divide (25 * 30 * 35 * 40 * 42 * 45)?

    You take a term by term manual count and try to form a pattern if possible. 42 = 2 * 3 * 7.
    So no 5s. Rest are consecutive multiples of 5. So, (1 + 1 + 1 + 1 + 1) = 5 5s for 5 consecutive multiples.
    Add 1 more for an extra 5 for multiples of 25 (=5^2).
    You move ahead like this if there were higher multiples involved.
    Here, total 5s = 5 (for the 5 multiples of 5) + 1(additional from 25) = 6
    So, highest power of 5 that divides the above expression = 6.

    Q. Find the maximum power of 10 that will divide (25 * 30 * 35 * 40 * 42 * 45)?

    10 = 2 * 5
    So you look for power of 2 and 5 in the given expression and take the minimum out of those since we need a pair of 2 and 5 to make a 10. Any extra powers of one prime are irrelevant for us to get a 10.
    Check for divisibility by highest power of 2: 25 ( = 5^2) gives zero twos, 30 ( = 2 * 3 * 5) gives 1 two, 40 ( = 2^3 * 5) gives 3 twos, 42 gives 1 two. Total = 5 2s
    From previous question, we know number of 5s is 6.
    So five 2s and 5s combine to give you a maximum of five 10s while one 5 is extra and is irrelevant for us without another two.
    Answer: 5

    Factorial format

    Exponent of any prime P in n! = [n/p] + [n/p^2] + [n/p^2] + … until the last positive term where [.] is the greatest integer function which gives the greatest integer less than equal to the expression taken within the brackets.

    Example : Highest Power of 5 in 100! = 100/5 + 100/25 + 100/125 = 20 + 4 + 0 = 24.

    Exponent of any composite number can be seen in three categories

    1. Composite number is a product of primes with all primes having unit power.
      Example: 15 (= 3 * 5), 21(= 3 * 7), 30(= 2 * 3 * 5) and so on.
    2. Composite number is just (prime)^n where n > 1
      Example: 8(=2^3), 9(=3^2) and so on.
    3. Composite number if product of primes with not all having power as 1.
      Example: 12(=2^2*3), 72(2^3 * 3^2) and so on.

    Example : Highest power of 15, 8 and 12 in 100!

    15 = 3 * 5
    Check for powers of 3 and 5 individually.
    Power of 15 in 100! is the lesser of the two.
    Power of 3 in 100! = [100/3] + [100/9] + [100/27] + [100/81] = 33 + 11 + 3 + 1 + 0 = 48
    Power of 5 in 100! = 24 from previous example question.
    So power of 15 in 100! is minimum(24, 48) = 24.

    8 = 2^3
    Find power of 2 in 100! And divide it by 3 (as 8 = 2^3)
    Power of 2 in 100! = [100/2] + … + [100/64] = 50 + 25 + 12 + 6 + 3 + 1 = 97
    Power of 8 in 100! = [97/3] = 32

    12 = 2^2 * 3
    Find highest power of 2^2 and 3 in 100! separately by above methods and take the minimum of the two.
    Power of 2 in 100! = 97.
    Therefore, power of 2^2 in 100! = [97/2] = 48
    Power of 3 in 100! = 48.
    Power of 12(=2^2*3) in 100! = minimum (Power of 2^2, Power of 3) = minimum (48, 48) = 48.
    Does not affect in this case but that’s how it will be done.

    Finding number of trailing zeroes in n!

    1. In base 10 or decimal system
    2. In bases other than 10

    In base 10:
    Trailing Zeroes in base 10 are basically the number of 10s in the expression. 10 = 2 * 5. So you can easily find highest power of 10 in any expression (linear or factorial format) which will give you the number of trailing zeroes in that expression.

    In bases other than 10:
    Base system is nothing but a set of defined digits used for basic counting. From school days, we learn the counting as 1, 2, 3, 4, 5, 6, 7, 8, 9 followed by a 10 and further. This is the decimal or the base 10. A base-n is nothing but a system of digits 0 to (n – 1) which does the complete counting.

    Trailing zeroes are introduced in a base 10(or decimal system) whenever we reach a ten or 10(Read as one-zero in base 10). For any generic base n, a zero will be introduced whenever we reach a multiple of n.
    Example : Base 8 counting
    Zero is 0
    One is 1
    Two is 2
    Three is 3
    Four is 4
    Five is 5
    Six is 6
    Seven is 7
    Eight is 10
    ….
    Sixteen is 20

    Sixty four (=8^2) is 100 and so on.

    Hence, the number of trailing zeroes in k! in base n would be given as the highest power of n in k! calculated normally by the methods described above.

    Example : Number of trailing zeroes in 20! when written in base 8
    Power of 2 in 20! = 10 + 5 + 2 + 1 = 18.
    Power of 8 in 20! = 18/3 = 6.
    So, number of trailing zeroes in 20! When written in base 8 will be 6.


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