A Year Problem

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


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.

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12 Responses to A Year Problem

  1. Tjl says:


    (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’

  2. Tjl says:

    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.

  3. 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.

  4. Tjl says:

    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…

  5. someone says:


  6. someone says:

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

  7. someone says:

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

  8. someone says:

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

  9. someone says:

    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.

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