18
$\begingroup$

Can you use the digits 2, 0, 1 and 7 each only once to create the number 88?

$\endgroup$
  • 3
    $\begingroup$ I got 42 with 7! mod 102. :S $\endgroup$ – darkdemise Feb 3 '17 at 14:31
  • 3
    $\begingroup$ Just out of curiosity, what's so special about 88? $\endgroup$ – user27263 Feb 3 '17 at 20:41
  • 2
    $\begingroup$ Are other digits allowed? $\endgroup$ – Pharap Feb 4 '17 at 2:32
  • 7
    $\begingroup$ Please specify the rules. $\endgroup$ – theonlygusti Feb 4 '17 at 17:57
  • 4
    $\begingroup$ In the spirit of hexomino's and Andrew's answers... ;) $\endgroup$ – Walt Feb 5 '17 at 3:56

16 Answers 16

57
$\begingroup$

What about this

$\left(\frac{0!}{.\overline1}\right)^2 + 7 = 88$

where

$.\overline1 = 0.1111\ldots$

$\endgroup$
  • 5
    $\begingroup$ Why is this accepted? $\endgroup$ – theonlygusti Feb 4 '17 at 18:00
  • 2
    $\begingroup$ @the why shouldn't it be? $\endgroup$ – Nick T Feb 5 '17 at 22:43
  • 1
    $\begingroup$ @Nick well, the OP never actually told us what they were after. I'm just wondering why the OP considered this a better answer than all the others. $\endgroup$ – theonlygusti Feb 5 '17 at 22:48
  • 2
    $\begingroup$ @theonlygusti, then ask OP and not the answerer. :P $\quad$ $\endgroup$ – tilper Feb 6 '17 at 13:52
  • 1
    $\begingroup$ This one also seems to add the least amount of handwavium to get to the solution. $\endgroup$ – tfitzger Feb 8 '17 at 21:39
51
$\begingroup$

No rules?

enter image description here

Looks like 88 to me if I squint.

$\endgroup$
33
$\begingroup$

Because modern math is done with computers, here's some Python:

>>> int(str(0 + 1 + 7) * 2)
88
$\endgroup$
  • 3
    $\begingroup$ I disagree that modern maths is done with computers. See the IMO and other related mathematics competitions. $\endgroup$ – boboquack Feb 3 '17 at 23:52
  • 1
    $\begingroup$ Similarly in C#: new string((1 + 7).ToString()[0], 2) $\endgroup$ – Jeppe Stig Nielsen Feb 4 '17 at 11:15
  • 3
    $\begingroup$ @boboquack Long-distance running is one of the best known competitions in the Olympics; still, modern commuters drive. $\endgroup$ – Nat Feb 4 '17 at 16:19
  • $\begingroup$ If you're going to do something like this, why not use BF? You just put 2017 in the front, and then put the BF code after. $\endgroup$ – Qwerp-Derp Feb 5 '17 at 4:27
  • 2
    $\begingroup$ I'm not sure is string manipulation is a mathematical operation $\endgroup$ – user902383 Feb 6 '17 at 10:17
26
$\begingroup$

For that matter:

In base 86: $12 + 0*7$

$\endgroup$
  • 30
    $\begingroup$ This does answer the question. In base 86 the numerals 12 = 88 in base 10. Adding the 0*7 in just a cheeky way of disposing of two useless numbers. How do I create 88 from 1, 2, 0 and 7? Change the base! $\endgroup$ – EvSunWoodard Feb 3 '17 at 20:20
  • 3
    $\begingroup$ In base 86, the number OP asked for doesn't exist. So no, it doesn't answer the question. $\endgroup$ – Josh Part Feb 3 '17 at 21:23
  • 11
    $\begingroup$ @Josh Sure it does. 88 base 10 = 12 base 86, which Bob used. Or, the other way around 88 base 86 = 696 base 10. $\endgroup$ – Graipher Feb 3 '17 at 22:46
  • $\begingroup$ Here is a good example of where the question asked needs to have proper constraints. I see another answer which involved changing the base. And to take the unconstrained problem one step further, someone even suggested that superimposing 1 over 2, and 7 over zero looks like 88. In my own defense, no one would argue that 0xf is 15. I don't see how saying "12 base 86 is 88" is any different. $\endgroup$ – Bob DeMattia Feb 6 '17 at 2:34
  • $\begingroup$ @BobDeMattia Really? Because the only information I can find on 0xf is that it's the number 15. $\endgroup$ – JMac Feb 6 '17 at 13:06
21
$\begingroup$

If floor were allowed, then this works:

$\left\lfloor\sqrt{10!!}\right\rfloor + 27 $

because

$10!!$ is $10\cdot 8\cdot 6\cdot 4\cdot 2 = 3840 $
$\sqrt{3840} = 61.9677335393\cdots.$

$\endgroup$
13
$\begingroup$

The only digits used here are 2,0,1,7 to reach 88:

$(\textbf{10}+(i\times i))^\bf2\rm+\bf7 = 88$

$\endgroup$
  • 3
    $\begingroup$ With this you could literally make any number though. Start at some number and add any amount of i^4 as required. $\endgroup$ – orlp Feb 5 '17 at 19:13
12
$\begingroup$

We can do it without the $0$...

$S=\{1,2,7\}$

$(\sum{S}-|S|)\times\prod{S}-\sum{S}$

(using the sum, $\sum$, cardinality, $||$, and product,$\prod$, of the set $S$.)

Evaluated:
$=(10-3)\times 14-10$
$=7\times 14-10$
$=98-10$
$=88$

So, obviously we could just add zero afterwards.

Mind you, I suppose that we could also do it with just one of the numbers in that case too.

$S = \{x\}$

$(|S|+|S|+|S|+|S|+|S|+|S|+|S|+|S|+|S|+|S|+|S|)\times(|S|+|S|)\times(|S|+|S|)\times(|S|+|S|)$
$=(1+1+1+1+1+1+1+1+1+1+1)\times(1+1)\times(1+1)\times(1+1)$
$=11\times2 \times2 \times2$
$=88$

...so
For $x = $...$7$, $2$, or $1$ just multiply the rest together and add them on.
For $x=0$ one can add $(7\times 2)\pmod{1}=0$, or $(7+1)\pmod{2}=0$.


The only question is: Does doing what I have done here count as using the given numbers more than once?


Here is an alternative, sneaky way...

Subtract the one from the seven, turn the resulting six upside-down, append the zero, then subtract the two.

$7-1=6$
$\text{turn}(6)=9$
$\text{append}(9,0)=90$
$90-2=88$

$\endgroup$
  • 1
    $\begingroup$ Just add a $+0$ onto the end. $\endgroup$ – Deusovi Feb 3 '17 at 16:36
  • 4
    $\begingroup$ I'm a big fan of the sneaky way. $\endgroup$ – EvSunWoodard Feb 3 '17 at 20:20
  • 1
    $\begingroup$ Why not x = 2 * 0 * 7 + 1, and then x + x + x + ... + x = 88? Seems like once you're allowed to define variables, the challenge is lost. $\endgroup$ – Dietrich Epp Feb 4 '17 at 8:26
  • 1
    $\begingroup$ if you are allowed to use sets you don't need any digits. $\endgroup$ – Jasen Feb 4 '17 at 9:01
  • 2
    $\begingroup$ define sneakyfunc(2,0,1,7) = 88. $\endgroup$ – Neil W Feb 4 '17 at 9:20
6
$\begingroup$

If ceiling or nearest integer function is allowed,

$\lceil{\tan^{-1}(27+0!)}\rceil = 88^{\circ}$

$\endgroup$
  • $\begingroup$ That's the cleverest use of the 1, I think... to turn tan into arctan. $\endgroup$ – NH. Nov 21 '17 at 18:42
6
$\begingroup$
  • Use "2" and "0" as digital numbers to combine them to form "8"
  • Add "1" and "7" mathematically and the result is "8"

So "88"

enter image description here

$\endgroup$
5
$\begingroup$

As a perl one-liner you could write:

perl -le 'for ($_=-1-2,$i = 0; $i<7; $i++) {$_+= $i*$i }; print'

or without a zero:

perl -le 'print ((7+1)x2)'

or without a zero OR a two in the bash shell:

x=$((7+1)) && echo $x$x

or without a zero, one, or two in bash:

false || x=$((7+$?)) && echo $x$x

or without any numbers at all:

false || x=$(($?+$?+$?+$?+$?+$?+$?+$?)) && echo $x$x

$\endgroup$
  • 1
    $\begingroup$ The puzzle is tagged [mathematics], not [programming]. $\endgroup$ – Glorfindel Feb 3 '17 at 21:43
  • 4
    $\begingroup$ @Glorfindel the puzzle is also tagged [calculation-puzzle]. my answer involves both mathematics and calculations. $\endgroup$ – gogators Feb 3 '17 at 21:49
  • 1
    $\begingroup$ You could build any number you want. this way. No fair! :D $\endgroup$ – GameDeveloper Feb 4 '17 at 4:08
  • 1
    $\begingroup$ ...or in English eighty-eight. $\endgroup$ – Jonathan Allan Feb 4 '17 at 11:32
3
$\begingroup$

If we use base 36

We now have access to the digits 2, 0, 1, A, N, D, 7. So:
$= (N \times D) - 7 + \left(\frac{A}{2}\right) - 1 + 0$
$= 8B - 7 + 5 - 1 + 0$
$= 8B - 3$
$= 88$
Using base 10 math gives us 2, 0, 1, 10, 13, 23, 7
$= (23 \times 13) - 7 + \left(\frac{10}{2}\right) - 1 + 0$
$= 299 - 7 + 5 - 1 + 0$
$= 299 - 3$
$= 296$
296 is 88 in base 36

$\endgroup$
  • 2
    $\begingroup$ But that's not what the question is asking for. $\endgroup$ – Rand al'Thor Feb 3 '17 at 17:19
  • 1
    $\begingroup$ The question says using those four digits once only. Your answer is the equivalent of saying "well, if I also use these other digits, I can get the answer!" which is completely besides the point. $\endgroup$ – Nij Feb 3 '17 at 22:49
  • $\begingroup$ Using the digits "2 0 1 and 7", if extract_digits=lambda x:set(" ".join(x).split()). However, the usual definition is extract_digits=lambda x:set(parse_english(x)). $\endgroup$ – wizzwizz4 Feb 4 '17 at 15:26
  • $\begingroup$ Why are we allowed to use $D$? I can agree with using $A$ in base 36 as a creative way of using $1$ and $0$, especially since the question never specified base 10, but why is $D_{36} = 13_{10}$ ok? $\endgroup$ – tilper Feb 6 '17 at 13:59
2
$\begingroup$

In base 9: $$88 = 102 - \lceil\sqrt 7\rceil$$

$\endgroup$
1
$\begingroup$

Here is my first answer after about 5 minutes of brute-force checks!

$\lceil\log{\sqrt{102!}}\rceil+7=88$

where log means logarithm in base 10.

By the way, as a wild guess, I think that 88 is very likely to be the OP's birth year.

$\endgroup$
  • $\begingroup$ Or, perhaps, age. You can never tell, unless you're a data miner or have access to said data miner's data. $\endgroup$ – wizzwizz4 Feb 4 '17 at 15:29
  • $\begingroup$ @wizzwizz4 This could have been another puzzle. And we can never be sure about a puzzle's solution until the puzzle owner reveals the answer... $\endgroup$ – user27263 Feb 4 '17 at 15:33
  • 1
    $\begingroup$ It's the visual representation of a 7-bar numerical display. Interpreting it as such gives 1111111 1111111. ASCII has seven bits; interpreting this number as ASCII gives DEL DEL. The OP wants to delete all puzzles. $\endgroup$ – wizzwizz4 Feb 4 '17 at 15:41
  • $\begingroup$ @wizzwizz4 Add some evil laughs to the background $\endgroup$ – user27263 Feb 4 '17 at 17:28
  • $\begingroup$ I thought $88$ was a reference. $\endgroup$ – tilper Feb 6 '17 at 14:01
1
$\begingroup$

Here is yet another one.

$$\lceil{\sqrt{\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left( 2 + 0! + 1\right)!}\rceil}$$

Breaking it down:

$$7! = 5040$$ $$\sqrt{7!} = \sqrt{5040} = 70.992957$$ $$\sqrt{\sqrt{7!}} = \sqrt{70.992957} = 8.42573$$ $$\left(\sqrt{\sqrt{7!}}\right)! = 8.42573! = 101358.44566$$ $$\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} = \sqrt{101358.44566} = 318.368411$$ $$ \sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left(2 + 0! + 1\right)! = 318.368411 \times 24 = 7640.84$$ $$\sqrt{\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left(2 + 0! + 1\right)!} = \sqrt{7640.84} = 87.412$$

$$\lceil{\sqrt{\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left( 2 + 0! + 1\right)!}\rceil} = \lceil{ 87.412 \rceil} = 88$$

$\endgroup$
1
$\begingroup$

Another use of mathematical functions and flooring...

$\lfloor\ln\Gamma(\frac{7\times10}{2})\rfloor$

$=\lfloor\ln\Gamma(35)\rfloor$

$=\lfloor88.58082754219768\rfloor$

$=88$

Reference: $\ln\Gamma(x)$

$\endgroup$
1
$\begingroup$

$0!-(.7-.1)\times.2$

$= 1 - (0.6)(0.2)$

$= 1 - 0.12 = 0.88$

remove the decimal point to get $088=88$.

$\endgroup$

protected by Aza Feb 5 '17 at 11:15

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.