A calculator has the usual keys for arithmetic and for “memory”. It fell into
the bathtub and now several keys (like × and ÷ ) don’t work at all.
The only keys that work are:
numbers keys, parentheses keys, recall memory keys, and the following three
operation keys:
(+) Plus key
(-) Minus Key
(1/x) Inverse key
If A and B are stored in calculator memories, can the product AB be computed
using only those few working keys?
Stated in another way:
Is there a formula for AB in terms of A, B, parentheses, and those three operations?
the bathtub and now several keys (like × and ÷ ) don’t work at all.
The only keys that work are:
numbers keys, parentheses keys, recall memory keys, and the following three
operation keys:
(+) Plus key
(-) Minus Key
(1/x) Inverse key
If A and B are stored in calculator memories, can the product AB be computed
using only those few working keys?
Stated in another way:
Is there a formula for AB in terms of A, B, parentheses, and those three operations?
-
Found a solution on the web (see link below).
Here it is in a procedural format (i.e., steps to take when punching things into the broken calculator)
Note that the following assumes you can store multiple values (not just A and B).
First compute and store A^2:
1) recall A, invert, store result. [ 1/A ]
2) recall A, add 1, invert, store result. [ 1/(A+1) ]
3) recall entry 1 above, subtract entry 2 from it, store result. [ 1/(A^2+A) ]
4) recall entry 3, invert, subtract A from it, store result. [A^2]
Now do the same steps for B to compute and store the values... we'll call each result
5) [ 1/B ]
6) [ 1/(B+1) ]
7) [ 1/(B^2+B) ]
8) [ B^2 ]
Here is where it gets complicated...
9) recall A, add B, subtract 1, invert, store result. [ 1/(A+B-1) ]
10) recall A, add B, inverse and store result. [ 1/(A+B) ]
11) recall entry 9 and subtract entry 10 from it -- that is 1/(A+B-1) - 1/(A+B) ,
which is the same as 1 / ((A+B)(A+B−1)) ,
which in turn is the same as 1 / (A^2+B^2+2AB−A−B). Store this result.
Home stretch: get rid of all but one of the terms from the denominator:
12) recall result from entry 11, invert, add A, add B, subtract entry 4, subtract entry 8, invert and store result. [ 1/2AB ]
13) now you can simply add entry 12 to itself and invert the results, which will be AB
I'll leave it up to you to figure out how to put this into a formula...
Here it is in a procedural format (i.e., steps to take when punching things into the broken calculator)
Note that the following assumes you can store multiple values (not just A and B).
First compute and store A^2:
1) recall A, invert, store result. [ 1/A ]
2) recall A, add 1, invert, store result. [ 1/(A+1) ]
3) recall entry 1 above, subtract entry 2 from it, store result. [ 1/(A^2+A) ]
4) recall entry 3, invert, subtract A from it, store result. [A^2]
Now do the same steps for B to compute and store the values... we'll call each result
5) [ 1/B ]
6) [ 1/(B+1) ]
7) [ 1/(B^2+B) ]
8) [ B^2 ]
Here is where it gets complicated...
9) recall A, add B, subtract 1, invert, store result. [ 1/(A+B-1) ]
10) recall A, add B, inverse and store result. [ 1/(A+B) ]
11) recall entry 9 and subtract entry 10 from it -- that is 1/(A+B-1) - 1/(A+B) ,
which is the same as 1 / ((A+B)(A+B−1)) ,
which in turn is the same as 1 / (A^2+B^2+2AB−A−B). Store this result.
Home stretch: get rid of all but one of the terms from the denominator:
12) recall result from entry 11, invert, add A, add B, subtract entry 4, subtract entry 8, invert and store result. [ 1/2AB ]
13) now you can simply add entry 12 to itself and invert the results, which will be AB
I'll leave it up to you to figure out how to put this into a formula...