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2

Solutions for 25-36 25 = 2 - 0! + (1 + √9)! 26 = 2 + 0 + (1 + √9)! 27 = (2 + 0 + 1) * 9 28 = 20 - 1 + 9 29 = 20 + 1 * 9 30 = 20 + 1 + 9 31 = 2 || 0! + 1 + 9 32 = 2 ^ (0! + 1 + √9) 33 = (2 - 0!) || 1 * √9 34 = 2 || 0! + 1 || √9 35 = —not here yet— 36 = (2 + 0!)! ^ (-1 + 9)


2

1= ((2 + 0 + 1) / √9)! 2= 2 + 0 * 1 * 9 3 = 2 + (0 * 1 * 9)! 4 = 2^(0 - 1 + √9) 5 = 2 + 0 * 1 + √9 6 = -(2 + 0 + 1) + 9 7 = -2 + 0 * 1 + 9 8 = -2 + 0 + 1 + 9 9 = 2 * 0 * 1 + 9 10 = 2 * 0 + 1 + 9 11 = 2 + 0 * 1 + 9 12 = 2 + 0 + 1 + 9 13 = 2 + 0! + 1 + 9 14 = 2 * -(0! + 1 - 9) 15 = (2 + 0 + 1)! + 9 16 = -2 + (0! + 1) * 9 17 = -2 + 0 + 19 18 = 2 * (0 * 1 + 9) ...


6

11 for 1492: 10 for 1969:


12

Here is one way to do it


1

Again, it's likely not the optimal solutions (for still unanswered numbers 123456, 1234567, 12345678):


3

Largest 38, but could be better


2

123456789 with 38 as the highest integer, but I used the division operator once (since OP didn't list available operators):


3

Done with largest integer $60$


12

12345 (largest integer 13) 123456 (largest integer 16) 1234567 (largest integer 17) 12345678 (largest integer 24) 123456789 (largest integer 27) Bonus 123 (largest integer 9) 1234 (largest integer 12)


3

My solution, may not be optimal 12345:


0

Some "almost-solutions" (with one of the digits repeated twice) for 452, the only number which remains unsolved at the time of posting (I'm posting this as ideas, because somebody may rework them):


0

Answer: Idea similar JonMarkPerry, but slightly different, my guess: RULES: 1)BOLD NUMBER = BOLD NUMBER +2 2) right number is multiple of first two in a row multiply by number of simple divisors of each number(beginning one, not+2) different from it (EX for 3 is 1, 4 is 2 and 1, for 6 is 3 and 2 and 1 and so on) 3)right number is not bold only ...


0

My solution is... $l = left$ $r = right$ So the next number is: (I thought that the bold numbers doesn't matter)


0

My guess: because:


3

EDIT: My original solutions were invalid, sorry for being a complete idiot. New solution for 331:


7

I think this works for 97:


9

With a decimal point


4

( Partial answer) for 67:


1

It can be assumed that evaluation is left to right (or rather: clockwise), not by precedence rules. Then this reads and simplifies to or


3

Interpretation of the brackets as being the binomial coefficient I was able to get four more numbers 435, 458, 461, and 469: $435 = \binom{30}{2}\times1^4$ $458 = \binom{30}{2}+4!-1$ $461 = \binom{31}{2}-4+0$ $469 = \binom{31}{2}+4+0$


0

If subfactorial is allowed:


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