Cyclicity and trailing digits (unit’s digit, ten’s digit and more) - Anubhav Sehgal, NMIMS Mumbai


  • NMIMS, Mumbai (Marketing)


    Finding unit’s digit

    Method 1: Find remainder with 10
    Unit digit of 68 = 8 - which is nothing but 68 mod 10 = 8
    Unit digit of 21 * 47 * 18
    = Remainder when 21 * 47 * 18 is divided by 10
    = 1 * 7 * 8 mod 10
    = 6

    example : Find the unit digit of 45^65.
    Unit digit = 45^65 mod 10 = 5^65 mod 10 = 25^32 * 5 mod 10 = 5^33 mod 10
    5^33 mod 10 = 25^16 * 5 mod 10 = 5^17 mod 10 = 25^8 * 5 mod 10 = 5^9 mod 10
    5^9 mod 10 = 5

    How do you shorten this process?
    Try finding for a larger factor of power on first occasion
    45^65 mod 10
    = 5^65 mod 10
    = 3125^13 mod 10
    = 5^13 mod 10
    = 3125 * 3125 * 125 mod 10
    = 5

    Method 2: Use Cyclicity of digits

    Digit^1^2^3^4^5^6^7^8Cycle Length
    0000000001
    1111111111
    2248624864
    3397139714
    4464646462
    5555555551
    6666666661
    7793179314
    8842684264
    9919191912

    example : Find the unit digit of 37^74
    Unit digit of 37^74 = Unit digit of 7^74 = Unit digit of (7^18)^4 * 7^2 = 1 * 9 = 9

    Finding ten’s digit

    Method 1 : Finding remainder by 100

    Either directly find remainder by 100 or use Chinese remainder theorem discussed before with 100 = 25 * 4.

    Example Find the ten’s place digit of 517^100
    Usually it is better to go by Chinese remainder theorem.
    517^100 mod 4 = 1^100 mod 4 = 1
    517^100 mod 25 = 17^100 mod 25
    Since 17 and 25 are co-primes, we may apply Euler’s theorem to reduce the exponent.
    E(25) = 20 => 17^20 mod 25 = 1 => (17^20)^5 mod 25 = 1 => 17^100 mod 25 = 1
    4a + 1 = 25b + 1
    Remainder = 1 or 01
    which means ten’s digit = 0 while unit’s digit = 1.

    Method 2 : Using Cyclicity

    DigitsCyclicity
    2, 3, 820
    4, 910
    51
    65
    74

    So for really high powers you can first reduce them all by factors of 20 since a cycle of 1, 4, 5 or 10 taken multiple times (20, 5, 4, 2 times respectively) is same as a cycle of 20.

    Example Find the ten’s place digit of 7^43
    Ten’s place digit of 7^43 = Ten’s digit of (7^4)^10 * 7^3 = 01 * 43 = 43.
    So, ten’s place digit = 4. [As 7^4 = 2401. So cycle of 7^1 = 07, 7^2 = 49, 7^3 = 3|43, 7^4 = 24|01 repeats itself 10 times.]

    For further trailing digits like hundredths place digit and so on we take remainders with next powers of 10 like 1000 for hundredths place digit with probably 125 and 8 used for Chinese Remainder Theorem

    Patterns and Generalizations for Cyclicity

    For odd numbers

    1. (Any odd number with unit digit 1, 3, 7, 9)^20N will have 01 as the last two digits
      example Last two digits of 37^20 = Last two digits of 37^40 = Last two digits of 37^60
    2. (Any odd number with unit digit 5)^N where N >1 will have 25 as the last two digits
    3. Last two digits are same for 21^5 = 41^5 = 61^5 = 81^5 = 01^5 = 01(last two digits)
      11^5 = 31^5 = 51^5 = 71^5 = 91^5 = 51(last two digits)
      9^10 = 19^10 = 29^10 = …. = 99^10 = 01(last two digits)
    4. Last two digits for powers of 11 are as follows
      11^1 = 11, 11^2 = 21, 11^3 = 31, 11^4 = 41, 11^5 = 51, …. , 11^8 = 81, 11^9 = 91, 11^10 = 01

    For even numbers

    1. (Any even number with unit digit 2, 4, 6, 8)^20N will have 76 as the last two digits.
    2. (M6)^5N will have last two digits as 76 where M is any digit and power is any multiple of 5.
    3. (M4)^(odd powers of 5) OR (M8)^(odd powers of 5) will have last two digits as 24.
    4. (M4)^10N will have last two digits as 76.

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