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Thread: Unit digit-how to find out?

  1. #1
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    Unit digit-how to find out?

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    Can someone please tell me how to find of the unit digit of the following.

    Find out the unit digit of 4^27 * 5^27 * 3^27?

    Not only the above one,how to find unit digit of any value to any power?

    Please some one explain me this concept.

    Thx.

  2. #2
    moe
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    5, 15, 25.... to any power, always unit digit=5, and same goes for 6

    others follow easy patterns:

    the easiest are 4 and 9:

    4, 6, 4, 6.... and 9, 1...

    2^1=2
    2^2=4
    2^3=8
    2^4=16
    2^5=32 (back to the start, unit = 2)

    for 8 it's simmilar to 2:
    8, 4, 2, 6

    3 and 7 are simmilar too:

    3, 9, 7, 1 and 7, 9, 3, 1

    helps?!

  3. #3
    moe
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    and to find out the unit digit of 4^27 * 5^27 * 3^27... 4^27 will have unit = 4 and 5^27, unit = 5 ==> 4^27 * 5^27, unit = 0 ==> 4^27 * 5^27 * 3^27, unit = 0

    is that the answ?

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    Thank you moe,Yes.u r answer is correct.

    but I did not understand this part

    and anything to the power of 2, mean that first I should check where it is repeating,then, that reaping one is the unit digit or else ,I strucked up here.


    2^1=2
    2^2=4
    2^3=8
    2^4=16
    2^5=32 (back to the start, unit = 2)


    for 8 it's simmilar to 2:
    8, 4, 2, 6

    3 and 7 are simmilar too:

    3, 9, 7, 1 and 7, 9, 3, 1

    I think I have not understand the concept perfectly,can you please help me by saying bit clearly.

  5. #5
    moe
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    ok... lets see if i explain myself any better...

    2^1=2
    2^2=4
    2^3=8
    2^4=16
    2^5=32 ==> same unit digit as 2 to the power of 1 (2^1=2)
    2^6=64 ==> same unit digit as 2 to the power of 2 (2^2=4)
    .
    ..
    ...
    units digit can be grouped into a sequence of 4 numbers that repeat themselves 2/4/8/6==> 2 will be the unit number for 2 to the power of 1, 5, 9, 13... and, in the same way, 4 will be the unit number for 2 to the power of 2, 6, 10, 14...
    eg: what's the unit digit of 2^16? 16/4=4 no remainder, therefore, the units digit will be the last one in the series=6

    (and this same sequence applies to the powers of numbers 12, 22, 32... 32^3 will have a unit digit of 8, the same as 22^3 and 2^3)

    numbers 3, 7 and 8 also have sequences of 4 numbers:
    for nš3: 3/9/7/1
    eg:3^4 will have a unit digit of 1 (and 13^4 as well), 3^5 will have unit digit 3...
    the sequences of 7 (7, 9, 3, 1) and 8 (8, 4, 2, 6) follow the same rules

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    Good

    explanation moe, and time saving too!!
    thank you!

    i had a diff approach, let me know whther im right

    viz; a ^b * c^d * e ^f ; unit digit of this one will be

    a^1* b^1* c^1...???

    so in the problem above
    5*4*3 has a unit digit = 0

  7. #7
    moe
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    i'm sorry, but i'm not sure i get the method... if i understand well, in a shorter example, like 4^2*3^7, u would do 4*2*3*7? that wouldn't work on this new example...

  8. #8
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    hi moe
    sory but i cant understand this concept.can u explain again
    thanks in advance
    regards

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    helo mahi111, u can solv for the units digit by getting the patern...here's how

    4^1=4
    4^2=6
    4^3=4
    4^4=6

    So 27/2=13 remainder 1 therefore units digit is 4.

    5^1=5
    5^2=5
    5^3=5...therefore the units digit is 5...and so on..

  10. #10
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    4^27 * 5^27 * 3^27 = (4*5*3)^27 = 60^27
    unit of 60^1 = 0
    unit of 60^2 = 0
    unit of 60^3 = 0
    .
    .
    .
    .
    unit of 60^27= 0

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