June 2017

[science_man_88]

Originally Posted by LaurV

grrr... another silly problem... the formulation is unclear, where and whenever you can/may/must place signs, and what's happening if you don't...

"while other seven-digit numbers simply cannot be solved, such as 0314157"

0+3-1+41+57=100

skip.

if the 7 weren't there I could get 100. as to finding unsolvable ones other than the one given under only multiplication 100=2*2*5*5 =4*5*5 = etc. which translate to any number with three 1's , two 2's, and two 5's is solvable under straight multiplication as is any with four 1's, one 4, and two 5's. etc. edit: and to uau nope but -1^2 = i^4 so we technically are working in powers of them etc.

[Batalov]

There are a few interesting technical questions hidden in the very first half an hour of thinking about coding the problem (and no, unlike last month's problem, for this one there is no way to avoid coding!):

1. Given two digits i and j, how many operations can you do? Well, 4, duh. ...and with the unary minus? You could think of 16, but doing unary minus on j will give redundant results, so maybe it is 8? ... ...and so will unary minus on i, in fact! So it is still 4 - with one huge unary minus operation reserved for the last operation on the whole train of digits. It is equivalent to all the ways of using unary minus operation anywhere else.

2. How many non-redundant ways of applying pairwise operations can you enumerate? Brute-force it would be 720 which is 6! first you can combine any tow digits, 6 ways; then repeat on what's resulting - 5 ways, and so on... but in fact it is much lower, because brute force will count [I](a op b) c (d op e) f g[/I] twice

[Xyzzy]

https://www.research.ibm.com/haifa/p.../June2017.html

[Batalov]

Machine-generated solutions are simultaneously ugly and riveting! ;-)

((0+3)*6)*(5+5/3/3)