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.
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.
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.
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]