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On the TV channel SBS, in Australia, there is a TV show in which contestants have six numbers and the operations of addition, subtraction, multiplication and division with which to produce a three digit number.

My question is whether, for any 6 numbers, this is always possible, and if so, does it hold for any choice of 6 single digit numbers? What is the lower bound on how many numbers are required?

EDIT: The numbers must all be different. So (1,1,1,1,1,1) and (1,1,1,99,99,99) aren't allowed.

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  • $\begingroup$ Given 1,1,1,1,1,1, I don't think you can get higher than 9, so no. You can experiment here crosswordtools.com/numbers-game $\endgroup$
    – Granger
    Commented Jun 7, 2011 at 14:18
  • $\begingroup$ Ed Pegg Jr. wrote on the "one complexity" of a number, which is the least number of ones needed to produce the number where only addition and multiplication are allowed. The least complex number greater than 100 is qpi which takes 13 ones to produce in this way. Guy's Unsolved Problems In Number Theory has some more on this problem. Gerhard "Ask Me About System Design" Paseman, 2011.06.07 $\endgroup$
    – Gerhard Paseman
    Commented Jun 7, 2011 at 14:32
  • $\begingroup$ Sorry, I meant any six numbers subject to the restrictions on the show, so six different numbers. So the 6-tuple (1,1,1,1,1,1) is not allowed. $\endgroup$
    – Shannon
    Commented Jun 7, 2011 at 14:35
  • $\begingroup$ qpi is smartphone for 108. Gerhard "Wants A Bigger, Smarter Keyboard" Paseman, 2011.06.07 $\endgroup$
    – Gerhard Paseman
    Commented Jun 7, 2011 at 14:36
  • $\begingroup$ 6 is the minimum, as suggested by the set 0,1,2,3,4,5. Gerhard "Ask Me About System Design" Paseman, 2011.06.07 $\endgroup$
    – Gerhard Paseman
    Commented Jun 7, 2011 at 14:40

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I assume the six numbers are all distinct, and each must be used exactly once. A brute-force program found, e.g. that with the six numbers 4, 6, 8, 16, 32, 64 the possible results did not include 571, 581, 587, 619, 623, 631, 649, 657, 661, 671, 673, 679, 681, 695, 709, 713, 721, 731, 743, 793, 811, 817, 821, 823, 827, 839, 841, 845, 849, 851, 853, 855, 857, 859, 863, 865, 871, 873, 877, 878, 879, 881, 887, 905, 911, 913, 917, 919, 921, 923, 933, 935, 937, 941, 943, 979, 983, 985, 987, 991, or 993.

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    $\begingroup$ Robert: my favourite version of this question involves making 24 out of 8,3,8,3 (and you have to use all the digits). The answer is 8/(3-(8/3)), and you have to leave the world of the integers to find this. Did you allow for such possibilities with your program? $\endgroup$
    – Kevin Buzzard
    Commented Jun 7, 2011 at 17:58
  • $\begingroup$ @Robert: Each number CAN be used once, but does not have to be. $\endgroup$
    – Shannon
    Commented Jun 7, 2011 at 19:37
  • $\begingroup$ @Shannon: Robert's suggestion of 4, 6, 18, 32, 64 cannot produce 571, even if each number does not necessarily have to be used once. I didn't do the math, but rather used Granger's link above: crosswordtools.com/numbers-game $\endgroup$
    – Vince
    Commented Jun 7, 2011 at 22:57
  • $\begingroup$ Robert: see also 7,7,3,3 to make 24. It's 7*(3+3/7). $\endgroup$
    – Michael Lugo
    Commented Oct 19, 2012 at 21:13
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The answers to the MathOverflow question entitled "Optimal Countdown" might be of interest. One poster has results where certain tuples of 6 numbers can yield more than the answers from 1 to 1000. (Link may be provided by some helpful soul without a phone-sized screen.)

Gerhard "Ask Me About System Design" Paseman, 2011.06.07

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To resolv this kind of problem there is a program CEB you can find here

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