# Smallest impossible positive integer with 5 ones and 5 operators

Inspired by this question:

What is the smallest positive integer that you cannot make with 5 ones and five operators?

Rules:

• You must use all five ones and all five operators
• Allowed operators are: +, -, *, /, ^, !, Knuth up-arrows. Each up-arrow counts as one operation.
• Concatenation is not allowed. e.g. you cannot use two 1s to make 11.
• You may use parentheses freely
• By "smallest number", I assume you mean "smallest positive integer"? – user3294068 Oct 9 '15 at 19:34
• By small would -1 be smaller than 0? – Maxqueue Oct 9 '15 at 19:34
• Is concatenating 1s to make 11 allowed? (It's prohibited in the linked question, but not mentioned here.) – f'' Oct 9 '15 at 21:23
• Are you counting $\uparrow\uparrow$ as two operations, like in the linked question? – Julian Rosen Oct 9 '15 at 21:29
• I can't imagine this being answered any way except a brute force computer search, making it a boring puzzle. – xnor Oct 9 '15 at 22:21

I could generate 0-9 but I'm stuck at 10:

$0=1/1/1/1-1!$
$1=1/1/1/1/1!$
$2=1+1/1/1/1!$
$3=1+1+1/1/1!$
$4=1+1+1+1/1!$
$5=1+1+1+1+1!$
$6=(1+1+1)!/1/1$
$7=(1+1+1)!+1/1$
$8=(1+1+1)!+1+1$
$9=(1+1+1)\uparrow(1+1!)$
$10=...$
$11=...$
$12=(1+1+1)!*(1+1)$

• Can you make 5 with just threes 1s and 3 operators? – warspyking Oct 10 '15 at 4:18
• Even disregarding the number of operators, since 5 is not a factorial, the last operation was a sum, product, difference, etc. But the two operands can only be 0, 1 or 2 (since there are up to 2 ones in each side), so this never yields 5. * Assuming you meant three ones – MathET Oct 10 '15 at 16:13

Sleafar got the right answer - 10 - by forcing it. I got same by the following observation: in order to get numbers above 6 one have to make a group of (1+1+1) to get 3 to further operate with. as the remaining we have two "1" and 3 operations available. Among these 3 operations one should be reserved for connection to the first group. And to get number over 7 one must use + operator to get 2. That leads us to solve the problem having 3 and 2 and two remaining operators. Having these assumptions, 8 and 9 are possible to produce while 10 is not. BTW : the max positive integer is ((1+1+1)↑(1+1))! = 9! = 362880

• Because you are limited by 5 operators – Yavpolnom Shoke Oct 10 '15 at 5:00