Find all 3-digit numbers that can be formed from their own digits.
arithmetic/digits/equations/own.digits New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
Find all 3-digit numbers that can be formed from their own digits. Rules: each digit is to be used exactly once. Allowed operators are addition, subtraction, multiplication, division, exponentiation, square-root and factorial, and also concatenation of 2 digits. Example: 660 = 6! - 60 -- Jacques Haubrich <jhau@wsdw10.win.tue.nl>, December 1995
Show solution
Well, my brute-force search has finished, and provided the following 41 numbers. Four of the answers use a somewhat dubious function, which could arguably be within the stated rules. I have noted these. Many of the numbers have more than one solution, but most secondary solutions are trivial (e.g. based on 1 = 1!). 120 = ( (2+1)! - 0!)! 121 = 11^2 125 = 5^(1+2) 126 = 21 * 6 127 = 2^7-1 128 = 2^(8-1) 143 = 3! * 4! -1 144 = (4-1)! * 4! 145 = 1! + 4! + 5! (particularly nice) 147 = 21 * 7 (dubious - where 2 = SQR(4)) 153 = 51 * 3 184 = (4!-1) * 8 216 = 6^(2+1) 240 = (4+0!)!*2 289 = (8+9)^2 305 = (61) * 5 (dubious - where 6=3!, 1=0!) 324 = (4! - 3!) ^2 337 = 7^3 - 3! 343 = (3+4)^3 347 = 7^3+4 354 = (5!-SQR(4))*3 355 = 3 * 5! - 5 360 = 60 * 3! 384 = SQR(8^4)*3! 441 = 21^(SQR(4)) (dubious - where 2 = sqrt(4)) 456 = (5!-6) * 4 464 = 6! - 4^4 624 = 26 * 4! 625 = SQR(5^(6+2)) 648 = SQR(6^8/4) 660 = 6! - 60 688 = 86 * 8 690 = 6! - 30 (dubious - where 30 = SQR(9)0) 693 = 6! - 3*9 713 = (1*3!)!-7 715 = (7-1)!-5 720 = (7-2+0!)! 729 = (7+2)^(SQR(9)) 736 = (3^6)+7 744 = (7+4!)*4! 936 = 3!^SQR(9) + 6! -- James Tinsley <jtinsley@iol.ie>