Can you use the digits 2, 0, 1 and 7 each only once to create the number 88?
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3$\begingroup$ I got 42 with 7! mod 102. :S $\endgroup$– darkdemiseCommented Feb 3, 2017 at 14:31
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4$\begingroup$ Just out of curiosity, what's so special about 88? $\endgroup$– user27263Commented Feb 3, 2017 at 20:41
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2$\begingroup$ Are other digits allowed? $\endgroup$– PharapCommented Feb 4, 2017 at 2:32
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7$\begingroup$ Please specify the rules. $\endgroup$– user21469Commented Feb 4, 2017 at 17:57
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4$\begingroup$ In the spirit of hexomino's and Andrew's answers... ;) $\endgroup$– WaltCommented Feb 5, 2017 at 3:56
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17 Answers
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8
What about this
$\left(\frac{0!}{.\overline1}\right)^2 + 7 = 88$
where
$.\overline1 = 0.1111\ldots$
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5$\begingroup$ Why is this accepted? $\endgroup$– user21469Commented Feb 4, 2017 at 18:00
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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. $\endgroup$– user21469Commented Feb 5, 2017 at 22:48
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2$\begingroup$ @theonlygusti, then ask OP and not the answerer. :P $\quad$ $\endgroup$– user33097Commented Feb 6, 2017 at 13:52
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1$\begingroup$ This one also seems to add the least amount of handwavium to get to the solution. $\endgroup$– tfitzgerCommented Feb 8, 2017 at 21:39
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No rules?
Looks like 88 to me if I squint.
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7
Because modern math is done with computers, here's some Python:
>>> int(str(0 + 1 + 7) * 2)
88
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1$\begingroup$ Similarly in C#:
new string((1 + 7).ToString()[0], 2)$\endgroup$ Commented Feb 4, 2017 at 11:15 -
3$\begingroup$ @boboquack Long-distance running is one of the best known competitions in the Olympics; still, modern commuters drive. $\endgroup$– NatCommented Feb 4, 2017 at 16:19
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$\begingroup$ If you're going to do something like this, why not use BF? You just put
2017in the front, and then put the BF code after. $\endgroup$ Commented Feb 5, 2017 at 4:27 -
2$\begingroup$ I'm not sure is string manipulation is a mathematical operation $\endgroup$ Commented Feb 6, 2017 at 10:17
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7
For that matter:
In base 86: $12 + 0*7$
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31$\begingroup$ 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! $\endgroup$ Commented Feb 3, 2017 at 20:20
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4$\begingroup$ In base 86, the number OP asked for doesn't exist. So no, it doesn't answer the question. $\endgroup$ Commented Feb 3, 2017 at 21:23
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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. $\endgroup$– GraipherCommented Feb 3, 2017 at 22:46
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$\begingroup$ 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. $\endgroup$ Commented Feb 6, 2017 at 2:34
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$\begingroup$ @BobDeMattia Really? Because the only information I can find on 0xf is that it's the number 15. $\endgroup$– JMacCommented Feb 6, 2017 at 13:06
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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 $
$\sqrt{3840} = 61.9677335393\cdots.$
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The only digits used here are 2,0,1,7 to reach 88:
$(\textbf{10}+(i\times i))^\bf2\rm+\bf7 = 88$
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4$\begingroup$ With this you could literally make any number though. Start at some number and add any amount of i^4 as required. $\endgroup$– orlpCommented Feb 5, 2017 at 19:13
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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.
$7-1=6$
$\text{turn}(6)=9$
$\text{append}(9,0)=90$
$90-2=88$
answered Feb 3, 2017 at 16:32
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4$\begingroup$ I'm a big fan of the sneaky way. $\endgroup$ Commented Feb 3, 2017 at 20:20
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2$\begingroup$ 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. $\endgroup$ Commented Feb 4, 2017 at 8:26
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2$\begingroup$ if you are allowed to use sets you don't need any digits. $\endgroup$– JasenCommented Feb 4, 2017 at 9:01
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2
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If ceiling or nearest integer function is allowed,
$\lceil{\tan^{-1}(27+0!)}\rceil = 88^{\circ}$
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$\begingroup$ That's the cleverest use of the 1, I think... to turn tan into arctan. $\endgroup$– NH.Commented Nov 21, 2017 at 18:42
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- 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"
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4
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
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1$\begingroup$ The puzzle is tagged [mathematics], not [programming]. $\endgroup$ Commented Feb 3, 2017 at 21:43
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4$\begingroup$ @Glorfindel the puzzle is also tagged [calculation-puzzle]. my answer involves both mathematics and calculations. $\endgroup$– gogatorsCommented Feb 3, 2017 at 21:49
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1$\begingroup$ You could build any number you want. this way. No fair! :D $\endgroup$ Commented Feb 4, 2017 at 4:08
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1$\begingroup$ ...or in English
eighty-eight. $\endgroup$ Commented Feb 4, 2017 at 11:32
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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
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2$\begingroup$ But that's not what the question is asking for. $\endgroup$ Commented Feb 3, 2017 at 17:19
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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. $\endgroup$– NijCommented Feb 3, 2017 at 22:49
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$\begingroup$ Using the digits "2 0 1 and 7", if
extract_digits=lambda x:set(" ".join(x).split()). However, the usual definition isextract_digits=lambda x:set(parse_english(x)). $\endgroup$ Commented Feb 4, 2017 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? $\endgroup$– user33097Commented Feb 6, 2017 at 13:59
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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.
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$\begingroup$ Or, perhaps, age. You can never tell, unless you're a data miner or have access to said data miner's data. $\endgroup$ Commented Feb 4, 2017 at 15:29
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$\begingroup$ @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... $\endgroup$– user27263Commented Feb 4, 2017 at 15:33
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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 givesDELDEL. The OP wants to delete all puzzles. $\endgroup$ Commented Feb 4, 2017 at 15:41 -
$\begingroup$ @wizzwizz4 Add some evil laughs to the background $\endgroup$– user27263Commented Feb 4, 2017 at 17:28
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$\begingroup$ I thought $88$ was a reference. $\endgroup$– user33097Commented Feb 6, 2017 at 14:01
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In base 9: $$88 = 102 - \lceil\sqrt 7\rceil$$
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Here is yet another one.
$$\lceil{\sqrt{\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left( 2 + 0! + 1\right)!}\rceil}$$
Breaking it down:
$$7! = 5040$$ $$\sqrt{7!} = \sqrt{5040} = 70.992957$$ $$\sqrt{\sqrt{7!}} = \sqrt{70.992957} = 8.42573$$ $$\left(\sqrt{\sqrt{7!}}\right)! = 8.42573! = 101358.44566$$ $$\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} = \sqrt{101358.44566} = 318.368411$$ $$ \sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left(2 + 0! + 1\right)! = 318.368411 \times 24 = 7640.84$$ $$\sqrt{\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left(2 + 0! + 1\right)!} = \sqrt{7640.84} = 87.412$$
$$\lceil{\sqrt{\sqrt{\left(\sqrt{\sqrt{7!}}\right)!} \times \left( 2 + 0! + 1\right)!}\rceil} = \lceil{ 87.412 \rceil} = 88$$
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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: $\ln\Gamma(x)$
answered Feb 7, 2017 at 2:10
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$0!-(.7-.1)\times.2$
$= 1 - (0.6)(0.2)$
$= 1 - 0.12 = 0.88$
remove the decimal point to get $088=88$.
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If subfactorial is allowed:
$!(7-2)\times(1+0!)=44\times2=88$
