19
$\begingroup$

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

17 Answers 17

58
$\begingroup$

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
7
$\begingroup$
  • 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

$\endgroup$
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
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 $088=88$.

$\endgroup$
0
$\begingroup$

If subfactorial is allowed:

$!(7-2)\times(1+0!)=44\times2=88$

$\endgroup$

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