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At a tender age my father introduced me to an arithmetic game: making the number 11 from five single digits using only the basic operators listed here:

+         addition
-         subtraction
*         multiplication
/         division
^         exponentiation
()        parenthesis
!         factorial
SQRT()    square root
SQ()      square (written in 1930's Netherlands using a small *square* as exponent)

The challenge is to construct an expression with value 11 ten times, each time using exactly five instances of a single digit from 0 to 9.

For example,

11 = (4 + 4 + 4) - (4 / 4)

is a correct solution for five 4's.

You are tasked with posing a solution for all 9 remaining combinations of five identical digits. One is much more challenging to solve than the others, but I will leave that to readers to identify. Solutions for all ten cases of five identical digits are known to exist.

Back when Ontario car licence plates were two letters and five digits, the challenge for a ten year old was to solve each such puzzle before another car came into view.

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11 Answers 11

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  1. $ (5^2)-\frac{5!}{5}+5+5=11$

  2. $ (3*3)+3-\frac{3}{3}=11$

  3. $(4*4)-4-\frac {4}{4}=11$

  4. $ [\frac {66}{6}]+\{66\} =11 $, where [ ] and {} represents greatest integer and fractional part function. (Does this counts)

  5. $6+\frac {6}{6}+6-[\sqrt(6)]= 11$, where [ ] represents greatest int function. (Does this counts)

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  • $\begingroup$ I understand the shorthand for squaring, in the absence of a superscript box. However concatenation to make 66 from two 6's is not allowed. Only the specified operations may be applied to single digits. (Not my down-vote; but I will up-vote if you make suitable corrections.) $\endgroup$ – Pieter Geerkens Jun 8 '16 at 20:43
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0:

SQ(0!+0!+0!)+0!+0!

1:

SQ(1+1+1)+1+1

2:

2+SQ(2)+SQ(2)+2/2

3:

3+3+3!-3/3

4:

4+4+4-4/4

5:

5+5+(5*5)/SQ(5)

6:

6+6-(6*6)/SQ(6)

7:

7+SQ(7/7+7/7)

8:

8+(8+8+8)/8

9:

9+9/9+9/9

7 without SQ():

SQRT(((7+7)/7)^7-7)

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  • $\begingroup$ Maybe I shouldn't have allowed my Dad's SQ function - seven is meant to be much harder than that. Well done! $\endgroup$ – Pieter Geerkens Jun 8 '16 at 1:59
  • $\begingroup$ Can you find a solution for 7 without using SQ()? $\endgroup$ – Pieter Geerkens Jun 8 '16 at 2:00
  • 2
    $\begingroup$ Yes, concatenation is not allowed; I know of a particularly beautiful solution that took me hours to discover. If necessary I will pose that as another problem. $\endgroup$ – Pieter Geerkens Jun 8 '16 at 2:08
  • $\begingroup$ Aren't there simpler solutions? As in, I personally can come up with simpler solutions $\endgroup$ – Xylius Jun 9 '16 at 8:50
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Working without the square/(root) functions:

0) Possibly too small

  1. Possibly too small

  2. $\frac{(2+2)!}{2} -\frac{2}{2}$

  3. $\frac{3^3 -3}{3} + 3$

  4. Given by OP

  5. $\frac{5(5+5)+5}{5}$

  6. Given by krnxprs

  7. Unknown

  8. Given in accepted answer
  9. $9 + \frac{9}{9} + \frac{9}{9}$

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  • $\begingroup$ @BrunoCosta Pardon? Do you have any improvements? $\endgroup$ – Inazuma Jun 8 '16 at 11:43
  • $\begingroup$ Yes, namely 0 and 1 $\endgroup$ – Bruno Costa Jun 8 '16 at 11:49
  • $\begingroup$ @BrunoCosta What are your suggestions? $\endgroup$ – Inazuma Jun 8 '16 at 11:54
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    $\begingroup$ @BrunoCosta Curse this internet again. I edited it once to have 'root' in brackets, but it mustn't have gone through. Not the first time this has happened on this site today, either. Arghhh. I'll correct it now, anyway. Thanks. $\endgroup$ – Inazuma Jun 8 '16 at 12:12
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    $\begingroup$ I must admit you got me on a corner though :p $\endgroup$ – Bruno Costa Jun 8 '16 at 12:13
5
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Got all ten numbers! This was very fun!

0:

SQ(0! + 0! + 0!) + 0! + 0!

1:

SQ(1 + 1 + 1) + 1 + 1

2:

2(SQ(2)+2) - (2 / 2)

3:

(3 * 3) + 3 - (3 / 3)

4:

(4 * 4) - 4 - (4 / 4)

5:

SQ(5) - (5! / 5) + 5 + 5

6:

((SQ(6) / 6) + 6) - (6 / 6)

7:

SQ(7/7 + 7/7) + 7

8:

(SQRT(8+8)+8)-(8/8)

9:

9 + (9/9) + (9/9)

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  • 2
    $\begingroup$ That's an ingenious solution for five. I love it. $\endgroup$ – Pieter Geerkens Jun 8 '16 at 2:15
  • $\begingroup$ Also for 7. Now I wonder why I found 7 so difficult. $\endgroup$ – Pieter Geerkens Jun 8 '16 at 23:40
3
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Here is 6 without SQ.

6 + 6 - 6^(6-6)

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  • $\begingroup$ Good. this generalizes to 5 as well of course. $\endgroup$ – Pieter Geerkens Jun 8 '16 at 23:38
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I have the only valid solution for five zeroes which does not use the square function.

SQRT(((0! + 0! + 0!)! - 0!)! + 0!) = 11 <------- Answer

SQRT(((1 + 1 + 1)! - 1)! + 1) = 11

SQRT((3! - 1)! + 1) = 11

SQRT((6 - 1)! + 1) = 11

SQRT(5! + 1) = 11

SQRT(120 + 1) = 11

SQRT(121) = 11

11 = 11


And, the solution for five ones, without the square function, is similar:

SQRT(((1 + 1 + 1)! - 1)! + 1) = 11

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  • $\begingroup$ Nicely done - I will remember this one. $\endgroup$ – Pieter Geerkens Jun 29 '16 at 22:17
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Providing multiple solutions for some problems, either for fun or to remove the powerful SQ() function.

0

$SQ(0!+0!+0!)+0!+0!$
$=SQ(3)+2$
$=11$

Didn't find any solutions without squaring. You need three of the numbers just to reach 6, which leaves you very little to work with!

1

$1=0!$

Problem reduced to case 0 which we already have a solution for.

2

$(2\cdot2)!\div2-2\div2$
$=24\div2 - 1$
$=11$

3

$3+3+3+3!\div3$
$=9+6\div3$
$=11$

4

$\sqrt4=2$

Problem reduced to case 2 which we already have a solution for.

Added a solution where the problem isn't reduced to case 2.

$4+4+4-4\div4$
$=11$

5

$SQ(5)\div5+5\div5+5$
$=25\div5+1+5$
$=11$

Alternative without SQ()

$(5+5\div5)!\div5!+5$
$=6!\div5!+5$
$=11$

6

$SQ(6)\div6-6\div6+6$
$=36\div6-1+6$
$=11$

Without SQ()

$\sqrt{6\cdot6}+6-6\div6$
$=6+6-1$
$=11$

7

$7+SQ(7\div7+7\div7)$
$=7+SQ(2)$
$=11$

Without SQ(), by far the most difficult (and satisfying) to find. Tried to do it without exponentiating at first, started going for 121, used (7+7)/7 in lots of ways before I even thought of 128 - in hindsight it seems finding 128 is very obvious!

$\sqrt{((7+7)\div7)^{7}-7}$
$=\sqrt{2^{7}-7}$
$=\sqrt{128-7}$
$=11$

8

$(8+8+8)\div8+8$
$=24\div8+8$
$=11$

9

$\sqrt9=3$

Problem reduced to case 3 which we already have a solution for.

Added a solution where the problem isn't reduced to case 3.

$9+9\div9+9\div9$
$=11$

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  • $\begingroup$ Your solutions for 4 and 9 use the SQ() function; can you solve both or either without it? $\endgroup$ – Pieter Geerkens Jun 10 '16 at 19:32
  • $\begingroup$ I think you meant SQRT() function, but regardless! Those two were rather easy to solve with just +-*/, edited in one solution each. $\endgroup$ – Jakob Pamp Bengtsson Jun 13 '16 at 8:15
  • $\begingroup$ Your second/alternate solution for 0 does not count, because it is missing a matching parenthesis. $\endgroup$ – Olive Stemforn Jun 29 '16 at 18:04
  • $\begingroup$ For your alternate for the one for the sevens without the SQ(), you have a typo in the radicand. It needs to be "128 - 7" instead. $\endgroup$ – Olive Stemforn Jun 29 '16 at 18:33
  • $\begingroup$ Thank you for pointing out the errors - I have corrected the answers as per your corrections. $\endgroup$ – Jakob Pamp Bengtsson Jul 7 '16 at 7:02
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Since everyone got very near to the final solution of 10 different ways, here's the 10th solution:

SQ(5-(5/5))-SQRT(5*5)

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I got all answers with different operations. Plus two extra answers for 9

0

SQ(SQ(0! + 0!)) - (SQ(0! + 0!) + 0!)

1

SQ(SQ(1 + 1)) - (SQ(1 + 1) + 1)

2

SQ(2 + 2) - (SQ(2) + 2/2)

3

SQ(3) + SQ(3) - (3! + 3/3)

4

4! / 4 + 4 + 4/4

5

SQ(5) / 5 + 5 + 5/5

6

(6 * 6 + SQ(6) - 6) / 6

7

SQ(7+7) / SQ(7) + 7

8

SQRT(SQ(8) + SQ(8) - (8 - 8/8))

9

(9 * 9 + 9 + 9) / 9

9

9 + SQRT(9) - 9^(9-9)

9

(SQ(SQ(9) + 9) - SQ(9)) / (SQ(9) * 9) = 8019 / 729 = 11

Obviously all 9 solutions can be applied to 3 with SQ(3).

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  • $\begingroup$ Is 16 your favourite number. ;-) $\endgroup$ – Pieter Geerkens Jun 8 '16 at 20:41
  • $\begingroup$ @PieterGeerkens I got the ideia of using powers of two :p $\endgroup$ – Bruno Costa Jun 9 '16 at 4:25
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The challenge is to construct an expression with value 11 ten times, each time using exactly five instances of a single digit from 0 to 9.

Then I think it is correct to use this as a solution : (which is a little funny by the way)

55/5*(5/5)

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  • $\begingroup$ Concatenation is not allowed in making 55 from two fives. Only the specified operations are allowed. $\endgroup$ – Pieter Geerkens Jun 8 '16 at 20:40
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sq(0^0 + 0^0 + 0^0)+0^0+0^0=11
anything to power of 0 = 1 therefore 1+1+1=3 and
3 squared = 9 + 1 + 1 = 11

(1+1+1)^2 + 1 +1 =11

((2*2)!/2) - 2/2) 2*2=4 4!=4*3*2=24. 24/2=12 - 2/2 = 11

(( 3*3 +3) - 3/3 ) 3*3=9 +3 = 12 - 3/3 = 11

(4 + 4 + 4) - (4/4) = 11

sqrt (5!+5/5 - (5-5))=11
5!=5*4*3*2*1=120 + 5/5 = 121 add (5-5)=0 sqrt of 121 is 11

almost the same as 5
sqrt((6 - 6/6)!+6/6)=11
6 - 1 = 5! = 5*4*3*2*1=120 + 6/6 = 121 and sqrt 121 = 11

oops didnt get this one myself but solution is obvious when you see it
sq(7/7 +7/7)+ 7 = 11

(8^2 + 8 + 8 +8 )/8 = 11
8 squared = 64 + 8 + 8 = 8 = 88/8 = 11

9/9 + 9/9 = 9 = 11

seven without using square
7 + (7^0) + (7^0) + (7^0) + (7^0)=11

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  • $\begingroup$ Your solution for 0 does not count, because 0^0 is indeterminate. $\endgroup$ – Olive Stemforn Jun 29 '16 at 18:01
  • $\begingroup$ And, it uses 10 zeroes, instead of the required five zeroes. $\endgroup$ – Olive Stemforn Jun 29 '16 at 18:17
  • $\begingroup$ Your alternate solution for sevens does not count, because you also used zeroes. $\endgroup$ – Olive Stemforn Jun 29 '16 at 18:19
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    $\begingroup$ @OliveStemforn. $0^0$ is not necessarily undefined. If you calculate it with limits you get 1. google.com/?#q=0%5E0 $\endgroup$ – Marius Jun 29 '16 at 19:04
  • $\begingroup$ @google.com/? - Yes. It is necessarily undefined. It is an indeterminate form. But we are not calculating it with limits. For instance, we are not taking the limit as x approaches 0 from the right-hand side of x^x, which equals 1. 0^0 is equivalent to 0^(1 - 1) = (0^1)/(0^1) = 0/0, which is an attempt to divide by zero, which is undefined. $\endgroup$ – Olive Stemforn Jun 30 '16 at 17:56

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