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Exponent problem - what is the units digit of n?

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Exponent problem - what is the units digit of n? Post Fri Dec 05, 2008 9:16 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If [(243)^x]*[(463)^y] = n, where x and y are positive integers, what is the units digit of n?

    (1) x + y = 7

    (2) x = 4

    OA A

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    logitech GMAT Titan
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    Post Fri Dec 05, 2008 10:02 pm
    tritrantran wrote:
    If [(243)^x]*[(463)^y] = n, where x and y are positive integers, what is the units digit of n?

    (1) x + y = 7

    (2) x = 4

    OA A
    243 = 3^5

    463 ends with a 3. So we have to know how many times we will multiply 3's at the end of each numbers.

    1) 7 times - SUF
    2) we dont know Y - INSUF

    Choose (a)

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    Post Fri Dec 05, 2008 10:25 pm
    I'm still not seeing it. Is there a rule for numbers ending in 3?

    logitech GMAT Titan
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    Post Fri Dec 05, 2008 10:39 pm
    tritrantran wrote:
    I'm still not seeing it. Is there a rule for numbers ending in 3?
    3^1 = 3
    3^2 = 9
    3^3 = 7
    3^4 = 1
    3^5 = 3
    3^6 = 9
    3^7 = 7
    3^8 = 1

    So the unit digits can be 1, 3, 7 or 9 depending on the power of 3.

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    cramya GMAT Titan Default Avatar
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    Post Fri Dec 05, 2008 10:47 pm
    Most exponent problems of this type deal with what's called cyclicity of units digit.

    As Logitech mentioned in his post above it cylces as follows for 3
    3,9,7,1,3,9,7,1,3,9,7,1.......



    Stmt I

    For 3 it will be 3,9,7,1,3,9,7,1,3,9,7,1.......

    Since x+y=7

    the units digit of the product will be 7 no matter how u distribute 7 between x and y(x=4 and y=3 or x=6 and y=1 etc...)

    SUFFICIENT

    Stmt II

    x=4

    If y=1 then the units digit of the product will be 3

    3,9,7,1,3,9,7,1,3,9,7,1.......

    If y =2 then the units digit of the product will be 9

    3,9,7,1,3,9,7,1,3,9,7,1.......

    Any of the 4 values from 3,9,7,1(abive I have given 2 examples) possible

    Cannot get to one defnite answer for the units digit

    INSUFFICIENT


    A)


    Hope this helps also!
    [/b]

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    Post Fri Sep 03, 2010 11:45 pm
    Took me ages to understand this one...
    (please correct me if my logic is wrong here)...

    Since we are concerned only abt the unit's digits, lets focus on those only...
    243 & 463 both have 3 in the units place...
    to solve for the units digit's value, we can boil the equation down to :
    (3^x)(3^y) = last few digits of n ....
    so we have x+y to solve for... since we know the cyclic form of 3 to the power smths = 1 3 9 7 1 3 9 7 ...

    Stmnt 1) x + y = 7
    gives us exactly what we are looking for (x+y)
    SUFF

    Stmnt 2) x =4
    y could be ANY other +ve integer on earth...
    INSUFF

    So it should be A....

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    uwhusky GMAT Titan
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    Post Fri Sep 03, 2010 11:52 pm
    You are correct!

    Pavan M Just gettin' started! Default Avatar
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    Post Tue Mar 06, 2012 4:45 am
    I finally understood the logic. Thanks guys for your detailed explaination. since x+y=7 and both the digits are ending with 3, we don't care what x and what y will be. Smile

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