10
$\begingroup$

Can you make $2471$ using first seven powers of two: $1,2,4,8,16,32,64$, each exactly once?

Only the basic operations are allowed: $+,-,\times,\div$, and parentheses $(\space)$.

Concatenation, exponentiation,... or any other trickery is not allowed.

Hint:

If parentheses were not allowed, it would be impossible to make it.


Similar Example:

Similarly, $595$ can be made with first six powers of two and same rules. Example Solution:

$595 = (1 + 4/8 + 16)\times(32 + 2)$ is unique solution.

$\endgroup$
10
$\begingroup$

Here's what I found (after way too much fiddling around):

$\mathbf{(8+32)\times(64-2-\frac{4}{16})+1 = 2471} $

Most of my useless fiddling was caused by this super annoying red herring:

$2471$ in binary is $100110100111$. That's very promisingly $100110$ repeated, plus 1, which translates into decimal as $38\times65+1$.

Alternately, it could be $100111$ repeated, minus 64, (that is $39\times65-64$), but even though there are dozens of ways to construct both these approaches (while using only six or seven powers of two), none of them will actually work because of one duplicated number or another.

Finally, I reached the solution by starting over with my initial approach, which was to

greedily go as near the goal as possible, first with two numbers: $64\times32 = 2048$, then with three: $64\times(32+8) = (64+16)\times32 = 2560 $, and then work out a way to make the remaining difference disappear. On the first time around, I had convinced myself that this wasn't going to work while sticking to integers, which was why I took the detour to binary at that point. (Bye-bye, most-of-an-hour.)

$\endgroup$
  • $\begingroup$ Congrats! This should be the unique solution If I haven't missed anything. $\endgroup$ – Vepir Aug 23 at 19:56
  • $\begingroup$ I used the "greedy method" to try to solve it too, but that got me absolutely nowhere. (Er, somewhere, but I quickly noticed what I was doing wasn't going to work. When I realized the amount of possibilities I gave up.) $\endgroup$ – Cloudy7 Aug 26 at 4:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.