Express 6

I was given this puzzle at the last Gathering for Gardner.

Use arithmetic operations to express 6 using three identical digits. For each digit from 0 to 9 find at least one way to express 6.

For example, I can express 6 using three twos in many ways: 2 + 2 + 2, or 2 · 2 + 2, or 22 + 2. But the problem doesn’t ask for many ways. One way is enough, but you need to do it for every digit. So nine more cases to go: 0, 1, 3, 4, 5, 6, 7, 8, and 9.

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12 Comments

  1. George R.:

    Well, I’ll try the easy ones first 🙂
    6*(6/6)
    5+5/5
    7-7/7
    sqrt4 +sqrt4 +sqrt4
    3*3 -3
    sqrt (9*9)-sqrt9
    (0! +0!+0!)!
    (1+1+1)!
    I don’t have a solution for 8’s yet.

  2. Tanya Khovanova:

    George, that was fast. One more to go.

  3. Konstantin Knop:

    8 – sqrt(sqrt(8+8))

  4. Tanya Khovanova:

    Konstantin,

    Nice, I know a different one.

  5. Leo:

    log(cbrt(8), 8*8)

  6. Leo:

    OK, now seriously:

    log(sqrt(8)*8)/log(sqrt(sqrt(8)))

  7. Leo:

    This works for all digits except 0 and 1, too!

  8. tanyakh:

    The one I knew for 8: (sqrt(8+8/8))!

  9. tanyakh:

    Also, 8!!/(8*8).

  10. tanyakh:

    Also for 8: ceiling(8*.8-.8).

  11. tanyakh:

    Also for 9: floor(.9) + sqrt(9) + sqrt(9).

  12. tanyakh:

    For 8: ceiling((8+8)/ceiling(sqrt(8)));
    8!!/sqrt(8*8);
    ceiling(sqrt(8))! + 8 – 8;
    ceiling(sqrt(8))*ceiling(sqrt(8))-ceiling(sqrt(8)),
    For 9: sqrt(sqrt(9))sqrt(sqrt(9)) + sqrt(9);
    9 – 9/aqrt(9);
    sqrt(9)! + 9 – 9.

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