3
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These are numbers that can also be read upside down as the same number, or different number: 0, 1, 6, 8, 9, 10, 11, 16, 18, so on. Each number can only be used once.

Using these numbers, try to get all numbers from 1-100.

EDIT: 1-100 has been completed by Marius, so keep going if you want to...

Here are 1 to 5 so you can understand.

1: 1

2: 8 - 6

3: 9 - 6

4: 10 - 6

5: 6 - 1

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  • 1
    $\begingroup$ did you delete the part saying we cannot use a number twice? if so we can do everything with 1+1+... $\endgroup$ – sousben Oct 19 '17 at 18:18
  • $\begingroup$ do you mean each number, or each digit? e.g. is 11 valid? $\endgroup$ – sousben Oct 19 '17 at 18:21
  • $\begingroup$ I'll allow that, also I added that to the original post. $\endgroup$ – user41716 Oct 19 '17 at 18:23
  • $\begingroup$ Why not 2 and 5? If we are talking traditional digital displays here, those two are definitely paired and ought to be included. Even handwritten, they can look quite similar when rotated. $\endgroup$ – feelinferrety Oct 19 '17 at 23:07
  • $\begingroup$ I was going to do that but there would be too many numbers, so I just decided to make it numbers that are upside-down in the font of the website. 1 isn't perfect but I allowed it anyways @feelinferrety $\endgroup$ – user41716 Oct 20 '17 at 20:58
3
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I love these puzzles.

$0 = 0$
$1 = 1$
$2 = 11-9$
$3 = 11 -8$
$4 = 10 - 6$
$7 = 1+6$
$8 = 8$
$9 = 9$
$10 = 10$
$11 = 11$
$12 = 11 + 1$
$13 = 6+8-1$
$14 = 6+8$
$15 = 6+8+1$
$16 = 16$
$17 = 16+1$
$18 = 18$
$19 = 19$
$20 = 10 + 9 + 1$
$21 = 10 + 11$
$22 = 10 + 11 + 1$
$23 = 16 + 8 - 1$
$24 = 16 + 8$
$25 = 16 + 8 - 1$
$26 = 10 + 16$
$27 = 11 + 16$
$28 = 11 + 16 + 1$
$29 = 19 + 10$
$30 = 10 + 19 + 1$
$31 = 11 + 19 + 1$
$32 = 18 + 8 + 6$
$33 = 18 + 8 + 6 +1$
$34 = 19 + 16 - 1$
$35 = 19 + 16$
$36 = 19 + 16 + 1$
$37 = 10 + 11 + 16$
$38 = 1 + 10 + 11 + 16$
$39 = 18 + 11 + 10$
$40 = 18 + 11 + 10 + 1$
$41 = 61 + 1 - 10 - 11$
$42 = 61 - 19$
$43 = 61 - 19 + 1$
$44 = 1 + 6 + 10 + 11 + 16$
$45 = 8 + 10 + 11 + 16$
$46 = 1 + 8 + 10 + 11 + 16$
$47 = 61 - 8 - 6$
$48 = 61 + 1 - 8 - 6$
$49 = 9 + 10 + 11 + 19$
$50 = 1 + 9 + 10 + 11 + 19$
$51 = 61 - 10$
$52 = 61 + 1 - 10$
$53 = 61 - 8$
$54 = 61 - 8 + 1$
$55 = 61 - 6$
$56 = 61 - 6 + 1$
$57 = 61 - 9 + 6 - 1$
$58 = 61 - 9 + 6$
$59 = 61 - 8 + 6$
$60 = 61 - 1$
$61 = 61$
$62 = 61 + 1$
$63 = 61 + 9 - 6 - 1$
$64 = 61 + 9 - 6$
$65 = 61 + 9 - 6 + 1$
$66 = 61 + 6 - 1$
$67 = 61 + 6$
$68 = 68$
$69 = 69$
$70 = 69 + 1$
$71 = 61 + 10$
$72 = 81 - 9$
$73 = 61 + 11 + 1$
$74 = 68 + 6$
$75 = 81 - 6$
$76 = 81 - 6 + 1$
$77 = 61 + 10 + 6$
$78 = 81 - 9 + 6$
$79 = 81 + 9 - 11$
$80 = 81 - 1$
$81 = 81$
$82 = 81 + 1$
$83 = 81 - 6 + 8$
$84 = 81 - 6 + 9 $
$85 = 81 + 10 - 4$
$86 = 86$
$87 = 88 - 1$
$88 = 88$
$89 = 89$
$90 = 91 - 1$
$91 = 91$
$92 = 91 + 1$
$93 = 81 + 11 + 1$
$94 = 96 - 8 + 6$
$97 = 96 + 1$
$98 = 98$
$99 = 99$
$100 = 99 + 1$

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  • $\begingroup$ given other answers, I believe the question was not narrowly enough defined, but as I understand it, your answer is a possible one $\endgroup$ – sousben Oct 20 '17 at 8:52
  • $\begingroup$ Good job, I edited the original post incase anyone wanted to keep going. $\endgroup$ – user41716 Oct 20 '17 at 13:39
7
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Here are answers for all numbers in the range 0 to 249.

0 to 9 are:

0 = 0
1 = 1
2 = 8-6
3 = 9-6
4 = 9-6+1
5 = 6-1
6 = 6
7 = 6+1
8 = 8
9 = 9

For the rest, do this:

To these numbers you can add up to four multiples of ten expressed as (91-81), (96-86), (98-88) and (99-89), so it is then trivial to get any number in the range 0 to 49.
By adding multiples of fifty, expressed as (61-11), (66-16), (68-18), or (69-19) to these you get all numbers up to 249.

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  • $\begingroup$ Can we really flip '90' to '06'? (moot point, since you could still use e.g. 86-66 to get 20), and 61 would still be valid to open up everything above 60. $\endgroup$ – LogicianWithAHat Oct 20 '17 at 12:01
  • $\begingroup$ That is a good point. The OP lists 10 as a reversible number, so I just assumed that 60, 80, 90 would be fine too. I'll edit my answer slightly. $\endgroup$ – Jaap Scherphuis Oct 20 '17 at 13:16
  • $\begingroup$ Any number that can be read upside-down can be used. Either the original one or the upside-down one $\endgroup$ – user41716 Oct 20 '17 at 13:35
5
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There only 5 numbers which can be used upside down. 0,1,6,8 and 9.

You can use any function

Ok then,

$$0=\log_ \frac 18 (\log_ 6 (((\sqrt9)!))$$

$$1=\log_ \frac 18 (\log_ 6 ((\sqrt{{\sqrt{\sqrt{(\sqrt9)!}}}})))$$

$$2=\log_ \frac 18 (\log_ 6 (\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{{\sqrt{(\sqrt9)!)}}}}}}})$$

Keep adding sqrts over the sqrt(9)! to make all the other numbers. You also can add + 0 to use all the numbers.

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  • $\begingroup$ Good job, nice loophole. I checked it because it's correct, though it kind of defeats the purpose (I don't really care though) $\endgroup$ – user41716 Oct 19 '17 at 19:02
  • $\begingroup$ How do you get 1/2 from 0, 1, 8? $\endgroup$ – jose_castro_arnaud Oct 19 '17 at 19:04
  • $\begingroup$ Sorry, I will change the typo $\endgroup$ – user35295 Oct 19 '17 at 19:16
  • 1
    $\begingroup$ (You needn't accept the first "correct" answer given - you're welcome to accept whatever answer best exemplifies the solution you envisioned. And of course you're free to change your Accepted answer at any time. Not to take anything away from Allan's solution, but it seems that @Jaap Scherphuis' answer is more in line with what you were looking for, and there's no reason you couldn't accept that answer instead.) $\endgroup$ – Rubio Oct 19 '17 at 21:48
2
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0, 1, 6, 8, 9 are given.

  • 2 = 8 - 6
  • 3 = 9 - 6
  • 4 = 10 - 6
  • 5 = 6 - 1
  • 7 = 6 + 1
  • 11 = 9 + 8 - 6
  • 12 = 9 + 8 - 6 + 1
  • 13 = 10 + 9 - 6
  • 14 = 8 + 6
  • 15 = 8 + 6 + 1
  • 16
  • 17 = 9 + 8
  • 18
  • 19
  • 20 = 160 / 8
  • 21 = 19 + 8 - 6
  • 22 = 90 - 68
  • 23 = 91 - 68
  • 24 = 16 + 8
  • 25 = 16 + 9
  • 27 = 18 + 9
  • 28 = 89 - 61
  • 29 = 90 - 61
  • 30 = 180 / 6
  • 36 = 16 R * 9 (R is the square root key, common in most simple calculators)
  • 37 = 98 - 61
  • 38 = 98 - 60

Not found (yet): 26, 31 to 35, 39 and above

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  • $\begingroup$ Oh, I seem to have misinterpreted the question. $\endgroup$ – Siddhartha Oct 19 '17 at 18:08
  • $\begingroup$ @Siddhartha No problem, I edited it as it might not have been clear. $\endgroup$ – user41716 Oct 19 '17 at 18:11
1
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Adding to Jose's answer:

  • 20 = 80 - 60
  • 22 = 88 - 66
  • 23 = 89 - 66
  • 24 = 96 / (9 - 6 + 1)
  • 25 = 100 / (10 - 6)
  • 26 = 86 - 60
  • 27 = 88 - 61
  • 28 = 88 - 60
  • 29 = 89 - 60
  • 30 = 100 / (11 - 6) + 9 + 1
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  • 1
    $\begingroup$ I think the idea is to generate all numbers from 1 to 100, not only the upside down ones, with expressions that only use upside down numbers. For example, 5 = 6 − 1. (But maybe I'm wrong.) $\endgroup$ – M Oehm Oct 19 '17 at 18:05
  • $\begingroup$ I expected to use each digit only once. Is that right? Otherwise, there is a trivial solution: 1 + 1 + 1 + ... + 1. $\endgroup$ – jose_castro_arnaud Oct 19 '17 at 18:18
  • $\begingroup$ I've just asked OP. I don't know if we can use each digit once, or each number once? $\endgroup$ – sousben Oct 19 '17 at 18:19
  • $\begingroup$ Sorry, when I edited it I forgot to add that back. Yes, each number can only be used once. $\endgroup$ – user41716 Oct 19 '17 at 18:21
  • $\begingroup$ Can we use calculator key shorthands? Keying = repeats the last operation. For instance, 3+4= gives 7, 3+4== gives 11, 3+4=== gives 15, and so on. If so, parenthesis are forbidden, since common calculators don't have them. $\endgroup$ – jose_castro_arnaud Oct 19 '17 at 19:03
1
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OP Edit: Understood the question wrong.

Numbers which are also numbers upside down: $0, 1, 6, 8, 9,$

Hence, there are $5 + 4*5 = 25$ such numbers, all of which can be generated using a simple program.

Here are the numbers: 1 6 8 9 10 11 16 18 19 60 61 66 68 69 80 81 86 88 89 90 91 96 98 99 100

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  • $\begingroup$ I may have explained it badly. I edited it, read it again. $\endgroup$ – user41716 Oct 19 '17 at 18:11

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