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ha!

- How much is 23 ha 77?
- What is ha?
- ha is like multiplication or addition.
- In what way?
- If a and b are numbers so is a ha b.
- I see. What more should I know about ha?
- (a ha b) + c = (b + c) ha (a + c) and 0 ha (a + b) = (0 ha a) + (0 ha b).
- And you want to know how much is 23 ha 77?
- That would be nice.

Problem source: Wisconsin Mathematics Science & Engineering Talent Search.

This entry was posted on Monday, February 1st, 2010 at 12:02 am and is filed under Problem. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “ha!”

  1. mathmom Says:
    February 1st, 2010 at 11:52 am

    Let’s name the rules
    R1: (a ha b) + c = (b + c) ha (a + c)
    R2: 0 ha (a + b) = (0 ha a) + (0 ha b)

    looking at R1 with c=0 we see that a ha b = b ha a
    let’s call that R3

    23 ha 77
    = (22+1) ha (76+1) (rewrite)
    = 22 ha 76 + 1 (R1 backwards)

    if we apply R1 backwards a total of 23 times we have
    23 ha 77 = 0 ha 54 + 23

    For any x:
    2x ha x
    = x ha 0 + x (x applications of R1 backwards)
    = 0 ha (-x) + 2x (2x applications of R1 backwards)

    equating the last 2 lines and re-arranging, we have
    x ha 0 = 0 ha (-x) + x
    use R3 on LHS:
    0 ha x = 0 ha (-x) + x
    rewrite x as (2x + (-x)):
    0 ha (2x + (-x)) = 0 ha (-x) + x

    using R2:
    0 ha (2x + (-x)) = 0 ha 2x + 0 ha (-x)

    equating those two expressions for 0 ha (2x + (-x)):
    0 ha (-x) + x = 0 ha 2x + 0 ha (-x)
    therefore, for all x: 0 ha 2x = x

    So,
    23 ha 77 = 0 ha 54 + 23
    = 27 + 23
    = 50

    double checking: use 77 applications of R1 backwards to get
    23 ha 77 = -54 ha 0 + 77
    = 0 ha (-54) + 77 (R3)
    = -27 + 77
    = 50

    In my explorations, I also proved that x ha x = x for all x (use x applications of R1 backwards, and recall that 0 ha 0 = 0)

    But I couldn’t come up with a formula for 0 ha y with y odd. If ha can operate on non-integers (”number” above is vague, so I guess it can) then obviously the same formula applies.

  2. Richard Sabey Says:
    February 3rd, 2010 at 3:37 am

    Building on Mathmom’s work:

    R2 implies that 0 ha a = xa for some constant x __ [1]

    (a ha b) - b = 0 ha (a-b) [by R1 with c=-b]
    thus a ha b = x(a-b) + b __ [2]

    Thus
    x(a-b) + b = a ha b
    = b ha a (by R3)
    = x(b-a) + a (by [2])
    thus x(2a-2b) = a-b
    thus x = 1/2
    thus a ha b = (a-b)/2 + b (by [2])
    = (a+b)/2.

  3. mathmom Says:
    February 3rd, 2010 at 9:58 am

    I’m not following this part of your argument:

    R2 implies that 0 ha a = xa for some constant x __ [1]

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