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  • Anoop Kumar

    Member • Sep 18, 2013

    x=1
    y=8
    z=5
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  • Abhishek Rawal

    Member • Sep 18, 2013

    ianoop
    x=1
    y=8
    z=5
    But that doesn't satisfies the equation.
    As per your answer, xyz=40, Hence LHS = 40+40+40 = 120. RHS = zzz = 5^3 = 125.
    LHS != RHS
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  • Anoop Kumar

    Member • Sep 18, 2013

    185+185+185 = 555 😛
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  • Abhishek Rawal

    Member • Sep 18, 2013

    Bloody hell!
    I thought question is : x*y*z + x*y*z + x*y*z = z*z*z [^_^]
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  • rahul69

    Member • Sep 19, 2013

    Abhishek Rawal
    Bloody hell!
    I thought question is : x*y*z + x*y*z + x*y*z = z*z*z [^_^]
    If that is the case, then :
    x=3;
    y=4;
    z=6;
    xyz+xyz+xyz=(3*4*6+3*4*6+3*4*6)=216=6*6*6=z*z*z
    😏
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  • Shailaja Tiwari

    Member • Sep 19, 2013

    Wow ,great you're absolutely right @ianoop.But how did you get this answer .I mean the logic if you could share would be of great help .I'd tried solving the question in several ways(including trial and error method too but couldn't get it .)
    Such questions are generally asked in BARC's RMO.
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  • Abhishek Rawal

    Member • Sep 19, 2013

    @#-Link-Snipped-#
    3xyz = zzz => xyz = zzz/3
    Take value of zzz = 111,222,333 upto 999.
    Only one will satisfy the condition & that is 555.
    Hence 555/3 = 185
    So, x=1, y=8 & z=5
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  • Anoop Kumar

    Member • Sep 19, 2013

    I started from 3 digit number multiplied by 3 = zzz
    start from unit place.
    3*z should have z at the unit place.
    start from possibilities 3*1=3, 3*6=18 you can come up with 3*5 = 15.
    now you got 5 at unit place and carry 1 that means now you need a number with
    3*y should have 4 at unit place, so adding a carry 1 will result 5.
    and same way you can find value of x.

    I hope this way is simple.
    May be there Could be other simpler approach.
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  • boombastik85

    Member • Oct 21, 2015

    Rahul 69 is right! xyz is x*y*z, 185 is an interesting answer but does not fit into rules of algebra
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