-2
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You can use the numbers 1, 5, 0, and 4 but you can't square and cube, you can't use any other numbers, you can't use the numbers more than once, your answer can't be rounded, and you MUST use all the digits

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  • 3
    $\begingroup$ What else can be used? Concatenation, decimal point, factorial, square root, etc etc. $\endgroup$ Mar 5 at 11:16
  • 3
    $\begingroup$ You should list all the things that can be used rather than those that can't, otherwise your question may be closed for being too vague. $\endgroup$
    – hexomino
    Mar 5 at 11:46

1 Answer 1

4
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Worked out by hand using only $ () + - / \times \, !$ and no concatenation

$1 = 1 \times 5 - 4 + 0 $
$2 = 5 - 4 + 1 + 0 $
$3 = 5 - 4 + 1 + 0! $
$4 = \frac{4!}{5 + 1 + 0} $
$5 = \frac{4!}{5 + 1} + 0! $
$6 = \frac{4!}{5 - 1 + 0} $
$7 = 5 + 4 - 1 - 0! $
$8 = 5 + 4 - 1 - 0 $
$9 = 5 + 4 + 1 - 0! $
$10 = 5 + 4 + 1 + 0 $
$11 = 5 + 4 + 1 + 0! $
$12 = 4 \times (5 - 1 - 0!) $
$13 = 4 \times (1 + 0!) + 5 $
$14 = 5 \times (1 + 0!) + 4 $
$15 = (5 - 1) \times 4 - 0! $
$16 = (5 - 1) \times 4 + 0 $
$17 = (5 - 1) \times 4 + 0! $
$18 = 5 \times 4 - 1 - 0! $
$19 = 5 \times 4 - 1 - 0 $
$20 = 5 \times 4 + 1 - 0! $
$21 = 5 \times 4 + 1 + 0 $
$22 = 5 \times 4 + 1 + 0! $
$23 = (5 + 1) \times 4 - 0! $
$24 = 5 \times (4 + 1) - 0! $
$25 = 5 \times (4 + 1) + 0 $
$26 = 5 \times (4 + 1) + 0! $
$27 = 4! + 5 - 1 - 0! $
$28 = 4! + 5 - 1 - 0 $
$29 = 4! + 5 - 1 + 0! $
$30 = 5 \times (4 + 1 + 0!) $

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  • $\begingroup$ I suggest making 4 less complicated using $(5 + 1) * 0 + 4$ You could use something similar for five and six $5 * 0 + 1 + 4 = 5$ & $6 = 4 * 0 + 1 + 5$ $\endgroup$
    – WOWOW
    Mar 5 at 17:53
  • $\begingroup$ @WOWOW you could make 5 and 6 in a similar way too, but I just wrote the first one I thought of. There are several way to make some of them. $\endgroup$ Mar 5 at 17:57
  • $\begingroup$ I would have written my own answer, with some of my ideas, but the question is already closed so... $\endgroup$
    – WOWOW
    Mar 5 at 17:58
  • $\begingroup$ @WOWOW with your suggestions, no division is needed. I was already typing a comment when you edited the first one. $\endgroup$ Mar 5 at 17:59

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