There are more year problems than years. But since I have been pondering on this particular one, I will present it here. You are allowed to use +, -, / and * (plus, minus, division and multiplication) signs and bracketing. These signs you can put between the numbers

1963

to form mathematical expressions. You must put at least one sign between two numbers and – cannot be used as “negative”, thus -1+9+6+3 is not allowed, but 1-9+6+3 is allowed.

The question is: *what is the smallest natural number that cannot be expressed in this way?*

Hinty example: 1*9-6-3=0 and (1/9)*(6+3)=1, so the answer is greater than 1.

Level 3.5/5

6

(got lazy and used Perl; all the below is in one line, and it’s not my fault if it formats your hard drive)

perl -e ‘@s=qw|+ – / *|;for $a (@s) {for $b (@s) {for $c (@s) {push @e, “((1$a 6)$b 9)$c 3”, “(1$a 6)$b(9 $c 3)”, “1 $a (6 $b (9 $c 3))”; }}}; print map{eval($_) . ” : $_\n”} sort {eval($a)eval($b)} @e’

didn’t compile :(

syntax error at -e line 1, near “)eval”

Execution of -e aborted due to compilation errors.

Ahh, it seems the blog system erased the “” (less-than equals greater-than) operator between eval($a) and eval($b) because it looks like a HTML tag.

Now it says

bash: syntax error near unexpected token `;’

sorry i’m not good at perl, but it looks like you did it for 1693, not 1963.

Ah, that’s another thing caused by the blog software: it seems it has changed the apostrophe after perl -e to an inverse one (i.e., ascii character 39 to ascii character 60.

Whoops, you are correct about the year. The correct answer is then 14, which is quite astonishingly high…

2=1+9/(6+3)

3=1*9/(6-3)

4=1+9/(6/2)

5=(1+9)/(6/3)

6=1*9-6+3

7=1+9-6+3

8=1+9-6/3

9=1*(9-6)*3

10=1+(9-6)*3

11=1*9+6/3

12=1+9+6/3

13=1+9+6-3

Good job, someone. The 4 is mistaken though.

What do you mean by mistaken? 1+9/(6/2)=1+9/3=1+3=4.

Sorry, my plunder. There is 2, not 3.

It should be 4=1+9/(6-3).

Yes. Any elegant proof why did you stop at 13? : )

I’m not sure about the elegant method. If we work in Z/3Z then we have integers 1, 0, 0, 0 and 2. But 1+0=1, 1*0=0, 1-0=0 so we have to use division at least once. If we form a fraction then we see that denominator is divisible by 3, so numerator should also be divisible by three. Therefore the expression is of the form 1*(something)=14, where something contains numbers {x/3,y,z} for some {x,y,z}={3,6,9}. Now to get 2 mod 3, we see that x must be 6 so the original expression is 1*(9A6/3) where A is one of +, -, *, /. I checked every case and found no solution.