Given the equation BIG/JOKE = 0.HAHAHAHA... replace each letter with the numbers 1-9 such that it is true.?

Question

(BIG)/(JOKE) = 0.HAHAHAHAHA... (repeating)

BIG is a three digit number
JOKE is a four digit number
0.HAHAHA... is a repeating decimal with two repeating digits

Replace each letter using each of the numbers 1-9 only once such that the equation holds true.

I had solved this once before back in high school, but my solution was not very elegant. I cleverly eliminated a very large amount of the possibilities and began guess/checking on the remaining possibilities with a calculator. I am just curious if there is a more elegant solution.

Oh, and there are multiple solutions that will work (around 4 or 5 if I remember right).

Yes, the site is right. JOKE must be a multiple of 99. I had proven that to myself somehow when I originally did the problem. I forget how since I lost my work a while back.

It seems they then proceeded to go about the problem the same way I did... guess & check with a computer/calculator.

Answer 1

There are exactly seven possible solutions:

459/1683=0.272727...
594/3267=0.181818...
689/5247=0.131313...
792/4356=0.181818...
798/1254=0.636363...
972/5346=0.181818...
987/4653=0.212121...

Method: unfortunately, I wasn't able to use much in the way of theory here, just a lot of brute force. The one bit of theory I did use was the aforementioned 0.HAHAHA... = HA/99 identity, which follows thusly:

x=0.HAHAHAHA...
100x=HA.HAHAHAHA...
100x-x=

HA.HAHAHAHA...
- 0.HAHAHAHA...
------------------
HA

99x=HA
x=HA/99

As for the search itself, I used a free computer algebra system called maxima, which you can download from http://maxima.sourceforge.net/ . The approach I used was to create a function that takes a list of numbers of the form [B, I, G, J, O, K, E, H, A], determines whether BIG/JOKE = HA/99 (as per the previous identity), and then returns the original list if it does, and returns some dummy element if it doesn't. I then applied that function to the set of all permutations of the digits - the resulting set (excluding the dummy element) was the set of all permutations that worked. Here are the actual commands I used:

big(x) := if ((100*first(x) + 10*second(x) + third(x))/(1000*fourth(x) + 100*fifth(x) + 10*sixth(x) + seventh(x)) = (10*eighth(x) + ninth(x))/99) then x else [0];

map(big, permutations([1, 2, 3, 4, 5, 6, 7, 8, 9]));

This should give you a good idea of how to use maxima to brute-force your way through other cryptarithms. Note that executing the second command will take a couple of minutes, even on a fast system, so use the time to make yourself a sandwich or something.

- Pascal

Edit: I just noticed something amusing - the site linked above insists that JOKE must be a multiple of 99 and BIG must be a multiple of HA, but one of the solutions I found (798/1254=0.636363...) does not have that property. This makes sense, because a/b = c/d does not imply that a is a multiple of c or b is a multiple of d (consider 4/6 and 6/9), but the site assumes it does. Such careless mathematics...

Answer 2

this site (below) says (last problem)

since the quotient is
0.HAHAHA...
we know that
JOKE is a multiple of 99

i dunno how they got that but every time i divide something by 99 it repeats, so... i guess their method of using a spreadsheet revealed that.

um... i'm having trouble seperating the rest of their "proof" from sheer "making stuff up" but the answer they give is:

689/5247 = 0.1313131313

hope the reasoning on their site makes more sense to you than it did to me - good luck!