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Math Help - the tens digit of 16^298 ....

  1. #1
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    the tens digit of 16^298 ....

    Please someone tell me how to work out the TENS digit of the answer of 16^{298} WITHOUT using calculator.

    I believe that it is somehow related to the principles such as (x^a)^b = x^{ab}

    the original question also included the units digit but it should always be 6.

    Thanks in advance.
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  2. #2
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    16^{298} = (2^4)^{298} = (2)^{4\times 298} =2^{1192}

    Now have a read of this article.

    How to Find the Last Digits of a Positive Power of Two - Exploring Binary
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  3. #3
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    Quote Originally Posted by ukorov View Post
    Please someone tell me how to work out the TENS digit of the answer of 16^{298} WITHOUT using calculator.

    I believe that it is somehow related to the principles such as (x^a)^b = x^{ab}

    the original question also included the units digit but it should always be 6.

    Thanks in advance.
    A few calculations (you should be able to multiply these easily w/o the abacus)

    016 = 16^1
    x16
    ===
    _56 = 16^2 (16 times 16 = 256, but we only need the tens & ones digits)
    x16
    ===
    _96 = 16^3
    x16
    ===
    _36 = 16^4
    x16
    ===
    _76 = 16^5
    x16
    ===
    _16 = 16^6 (same as 16^1, exponent difference is 5)

    16^6 \equiv 16^1 \mod 100

    298/5 = 59.9
    59*5 = 295

    1+5=6
    1+295=296

    16^1 \equiv 16^6 \equiv 16^{296} \mod 100

    &(add 2 to the exponents)

    16^3 \equiv 16^8 \equiv 16^{298} \mod 100

    see prior calculation above for the tens & ones digits for 16^3\mod 100 which is the same as  16^{298}\mod 100

    .
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  4. #4
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    Quote Originally Posted by pickslides View Post
    16^{298} = (2^4)^{298} = (2)^{4\times 298} =2^{1192}

    Now have a read of this article.

    How to Find the Last Digits of a Positive Power of Two - Exploring Binary
    2^{1192}
    \equiv (2^{22})^{54} \cdot 2^4 (mod100)
    \equiv 4^{54} \cdot 2^4 (mod100)
    \equiv 2^{108} \cdot 2^4 (mod100)
    \equiv 2^{112} (mod100)
    \equiv (2^{22})^5 \cdot 2^2 (mod100)
    \equiv 4^5 \cdot 2^2 (mod100)
    \equiv 2^{10} \cdot 2^2 (mod100)
    \equiv 2^{12} (mod100)
    \equiv 96 (mod100)
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  5. #5
    Super Member Bacterius's Avatar
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    16^{298} \equiv 16^{98} \pmod{100}

    16^{98} = (2^{98})^4 = 2^{98 \times 4} = 2^{392}

    2^{392} \equiv 2^{92} \pmod{100}

    Note that 92 = 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 1

    Thus 2^{92} = 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^1

    Note that 2^7 = 128 (easily done mentally)
    Thus 2^7 \equiv 28 \pmod{100}

    Thus 2^{92} \equiv 28^{13} \times 2 \pmod{100}

    Note that 28^{13} = 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^1
    You easily find out that 28^2 \equiv 84 \pmod{100}

    Thus, 28^{13} \equiv 84^{6} \times 28 \pmod{100}

    We get that 2^{92} \equiv 84^{6} \times 28 \times 2 \equiv 84^6 \times 56 \pmod{100}

    Note that 84^6 = 84^2 \times 84^2 \times 84^2.
    You easily find out that 84^2 \equiv 56 \pmod{100}.

    Thus, 2^{92} \equiv 56^{3} \times 56 \equiv 56^4 \pmod{100}.

    Note that 56^4 = 56^2 \times 56^2.
    You easily find out that 56^2 \equiv 36 \pmod{100}.

    Thus, 56^4 \equiv 36^2 \pmod{100}.
    You easily find out that 36^2 \equiv 96 \pmod{100}.

    Thus, 56^4 \equiv 2^{92} \equiv 2^{392} \equiv 16^{98} \equiv 16^{298} \equiv 96 \pmod{100}.

    Conclusion : the tens digit of 16^{298} is 9.



    I know my method is long, boring and repetitive, but eh ? it gets the job done
    Last edited by Bacterius; December 23rd 2009 at 11:09 PM.
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  6. #6
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    Quote Originally Posted by Bacterius View Post
    16^{298} \equiv 16^{98} \pmod{100}

    16^{98} = (2^{98})^4 = 2^{98 \times 4} = 2^{392}

    2^{392} \equiv 2^{92} \pmod{100}

    Note that 92 = 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 1

    Thus 2^{92} = 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^1

    Note that 2^7 = 128 (easily done mentally)
    Thus 2^7 \equiv 28 \pmod{100}

    Thus 2^{92} \equiv 28^{13} \times 2 \pmod{100}

    Note that 28^{13} = 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^1
    You easily find out that 28^2 \equiv 84 \pmod{100}

    Thus, 28^{13} \equiv 84^{6} \times 28 \pmod{100}

    We get that 2^{92} \equiv 84^{6} \times 28 \times 2 \equiv 84^6 \times 56 \pmod{100}

    Note that 84^6 = 84^2 \times 84^2 \times 84^2.
    You easily find out that 84^2 \equiv 56 \pmod{100}.

    Thus, 2^{92} \equiv 56^{3} \times 56 \equiv 56^4 \pmod{100}.

    Note that 56^4 = 56^2 \times 56^2.
    You easily find out that 56^2 \equiv 36 \pmod{100}.

    Thus, 56^4 \equiv 36^2 \pmod{100}.
    You easily find out that 36^2 \equiv 96 \pmod{100}.

    Thus, 56^4 \equiv 2^{92} \equiv 2^{392} \equiv 16^{98} \equiv 16^{298} \equiv 96 \pmod{100}.

    Conclusion : the tenths digit of 16^{298} is 9.



    I know my method is long, boring and repetitive, but eh ? it gets the job done
    according to a computer the tenth digit is 7.

    http://www.wolframalpha.com/input/?i=10th+digit+of+16^298
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  7. #7
    Super Member Bacterius's Avatar
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    Quote Originally Posted by BabyMilo View Post
    according to a computer the tenthS digit is 7.

    http://www.wolframalpha.com/input/?i=10th+digit+of+16^298
    I mean the before-last digit, of course ... not the 10th digit from the left

    16^{298} = 67259685376506838351825694125205607357734081036908 94500677045631406539332516306882577142397357830536 43998864024685999522991598722297525821288734761044 09052127576185436156214378455780845781465327759636 97038693955559119408307906472347614286627358580697 16218116239821639793538226085261772687392217674908 12009984323096665122511297763735177027009515287152 894672896

    I was talking about the red digit, (I believe it is the question) ...
    Last edited by Bacterius; December 23rd 2009 at 11:10 PM.
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  8. #8
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    Hello, ukorov!

    My method is similar to Aiden's . . .


    Find the tens digit of the answer of 16^{298} without using calculator.
    Note the last two digits of consecutive powers of 16 . . .

    . . \begin{array}{c|c}16^n & \text{ends in} \\ \hline 16^1 & 16 \\ 16^2 & 56 \\ 16^3 & 96 \\ 16^4 & 36 \\ 16^5 & 76 \\ 16^6 & 16 \\ \vdots & \vdots \end{array}

    We find that the last two digits have a 5-digit cycle: . 16, 56, 96, 36, 76

    . . Hence: . 16^{5n+1} \equiv 16 \text{ (mod 100)}


    Since 298 \:=\:[5(59) + 1] + 2

    . . 16^{298} \;\;\equiv\;\; 16^{[5(59) + 1] + 2} \;\;\equiv\;\;16^{5(59) + 1}\cdot16^2  \;\;\equiv\;\; 16\cdot16^2 \;\;\equiv\;\;16^3 \;\;\equiv\;\;96\text{ (mod 100)}


    Therefore, the tens digit of 16^{298} is {\color{red}9}.



    Edit: corrected typo . . . Thanks, Galactus!
    Last edited by Soroban; December 24th 2009 at 07:01 AM.
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  9. #9
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    workin mod 100 noting 16^3=4^6=-4

    16^298= 16(16)^297
    =16(-4)^99
    =16(-4)^3.(-4)^96
    =-24(4)^16
    =-24(4^4)((-4)^2)
    =-24.-4=96
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