## Math Problem with 6’s from Scam School

TestTubeTM is a new digital network from DiscoveryTM. With shows like Stuff of Genius, Blow It Up!, and Distort (where “great ideas become reality”), it holds strong appeal for mathy folks.

My favorite show on TestTubeTM is Scam School, where magician Brian Blushwood takes you on a tour of bar tricks, street cons, and scams. In the episode “Six the Hard Way,” he poses a mathematical challenge that is a variation on one you may have seen before. As Brian explains, “it’s almost poetic how simple this is.”

The puzzle is this: Form an expression with three 1’s, three 2’s, three 3’s, and so on, up to three 9’s, so that the value of each expression is equal to 6. As an example, an expression using three 7’s is shown below. Can you find expressions using the other numbers?

0   0   0 = 6

1   1   1 = 6

2   2   2 = 6

3   3   3 = 6

4   4   4 = 6

5   5   5 = 6

6   6   6 = 6

7 – 7 ÷ 7 = 6

8   8   8 = 6

9   9   9 = 6

You can watch Six the Hard Way, but be forewarned: at least one solution for each number is given, so you may want to solve the puzzle before viewing.

Also note that some folks have posted solutions in the comments below, so scroll at your own risk.

Entry filed under: Uncategorized. Tags: , , , , , .

• 1. P-Dog  |  June 3, 2013 at 1:56 pm

(1 + 1 + 1)! = 6
(2 + 2 + 2) = 6
(3 x 3 – 3) = 6
(4 – 4 / 4)! = 6
(5 + 5 / 5) = 6
(6 + 6 – 6) = 6
(7 – 7 / 7) = 6
(Sqrt(8 + 8/ 8))! = 6
(Sqrt(9 + 9 – 9))! = 6

• 2. Srilaxmi  |  July 20, 2016 at 5:48 am

$9 - \frac{9}{\sqrt9} = 9 - 3 = 6$

• 3. Sandeep  |  August 31, 2016 at 9:21 pm

(Sqrt (8+8÷8)!)=6
Plz explain

• 4. Craig Correa  |  September 15, 2016 at 2:35 am

Using BODMAS it happens. Try to look at it like this.
(sqrt(8 + (8/8)))!
=(sqrt(8+1))!
=3!
=6

• 5. Math  |  October 13, 2016 at 12:57 am

Alright, here’s my best explanation.8/8 is 1(order of operations say division first), add 8 to get 9, square root 9 to get 3. And then finally factorial the 3 to get 6. I assume you had issues with the factorial, so I’ll explain that real quick. A factorial is symbolized by ! and is actually quite simple, you actually just multiply the number by all the numbers that come previously. For example, 3! Is 3x2x1 or 6 and 4! is 4x3x2x1 or 24. I know this is a little late to explain this to you, but if you still needed the explanation, your welcome! Fun fact, your average handheld scientific calculator can only process up to 69! (not a joke, that’s true) while more complex calculators such as the TI-Nspire CX (which I have and love) can do up to 449! which is a rediculously large number. Google calculator says it’s equal to infinity, lol. It’s actually equal to 3.851930518 E+997. P.S. a googol (a 100 zeroed number) is only E+100. Also, don’t ask me how I have all this stuff memorized, I’m just a HUGE nerd. If you want me to clarify anything else, I’d be happy to.

• 6. zachdcox  |  June 3, 2013 at 8:04 pm

I actually got all of these before watching the video … Then started watching the video and realized that my solution (which was trivial) was NOT the only one so I spent time working out their solutions too … VERY COOL problem (and the three zeros 0 0 0 = 6) is good to which you forgot to list.

• 7. venneblock  |  June 3, 2013 at 9:28 pm

Zoiks! Thanks, Zach! I can’t believe i forgot three 0’s. I’ve corrected the post.

• 8. Nanette Bright  |  June 9, 2013 at 3:58 pm

I gave this book to my mathy teenage son, and he loves it. His mathy family is also enjoying it–so much so that I bought two more copies to give to fellow math-loving folks.

• 9. Garry Ruiz  |  June 21, 2013 at 4:03 pm

I gave this book to my mathy teenage son, and he loves it. His mathy family is also enjoying it–so much so that I bought two more copies to give to fellow math-loving folks.

• 10. Bronson  |  August 28, 2013 at 1:40 pm

The zeros are not answered in these comments. The way you do the zeros is:
(0!+0!+0!)=6

I also did the fours different:

sqrt4+sqrt4+sqrt4=6 so its 2+2+2=6

• 11. venneblock  |  August 29, 2013 at 10:47 pm

Good work, Bronson, but you forgot an exclamation point:
(0! + 0! + 0!)! = 6

• 12. Danika  |  June 17, 2014 at 7:05 pm

actually it would be (0!+0!+0!)!=6 because (0!+0!+0!) = 3 and it needs another to get to six so you would then do 3! which equals 6… so your work is invalid

• 13. Jack  |  May 16, 2016 at 8:25 pm

(0!+0!+0!)!
Thats the right way

• 14. Pravin shah  |  November 7, 2013 at 7:34 pm

You can also have 10 10 10 = 6

Log{ ( 10x10x10 ) squared }. = 6

Log { ( 1000 ) squared }. =. Log { 1,000,000 } = 6

( Log, to the base 10.)

• 15. venneblock  |  November 8, 2013 at 8:38 pm

I feel like adding “squared” is a bit of cheating. That’s introducing a digit of 2.

But I like the idea of extending this. I got the following for three 10’s:

$( \lfloor \sqrt{10} \rfloor )! + 10 - 10$

How far can we go? Who’s gonna post for 11’s?

• 16. Tiffany  |  December 15, 2015 at 3:12 pm

Part of the deal is no numbers added, including powers. Roots are acceptable. Typing “squared” is the same thing as adding the 2.

• 17. zachdcox  |  November 9, 2013 at 3:46 pm

$\lfloor \log 11 \rfloor + \lfloor \log 11 \rfloor + \lfloor \log 11 \rfloor$

• 18. venneblock  |  November 9, 2013 at 5:13 pm

Awesome! Who’s got 12’s?

Zach, I updated your comment to use LaTex.

• 19. Nurd  |  October 1, 2015 at 7:58 am

Three [11 log]’s only gets you about 3.1

• 20. Krigs  |  February 8, 2016 at 6:03 am

He didn’t write [] he wrote ⌊⌋ = floor (i.e round down to closest integer).

• 21. wavecollapse  |  December 2, 2013 at 6:39 pm

10 can be: (sqrt(10-10/10))!
Also, if you allow the floor operator, all numbers up to infinity can be made to equal 6. Its not easy or obvious but its possible. Try it if you like: x x x = 6 (need floor operator)

• 22. venneblock  |  January 25, 2015 at 3:32 pm

Okay, wavecollapse… I give up. What were you thinking?

• 23. Tevuz  |  December 14, 2013 at 12:19 pm

√(¬(11+11)&11)=6

• 24. Nurd  |  October 1, 2015 at 8:02 am

Using a negation symbol doesn’t make sense

• 25. will  |  October 11, 2017 at 4:35 am

can you explain further

• 26. Tevuz  |  December 14, 2013 at 12:29 pm

√(12+12+12)=6

• 27. Tevuz  |  December 14, 2013 at 12:42 pm

√(¬(-13)+¬(-13)+¬(-13))=6

• 28. Koen Verschueren  |  April 6, 2014 at 2:29 pm

can somebody explain this one?

• 29. MD  |  December 22, 2013 at 2:58 pm

My method is a cheat method…
$(0!+0!+0!)! =6$
You will start seeing why…
$\vspace{2mm} \big((\frac{1}{\infty})!+(\frac{1}{\infty})!+(\frac{1}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{2}{\infty})!+(\frac{2}{\infty})!+(\frac{2}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{3}{\infty})!+(\frac{3}{\infty})!+(\frac{3}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{4}{\infty})!+(\frac{4}{\infty})!+(\frac{4}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{5}{\infty})!+(\frac{5}{\infty})!+(\frac{5}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{6}{\infty})!+(\frac{6}{\infty})!+(\frac{6}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{7}{\infty})!+(\frac{7}{\infty})!+(\frac{7}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{8}{\infty})!+(\frac{8}{\infty})!+(\frac{8}{\infty})!\big)!=6\\ \vspace{2mm} \big((\frac{9}{\infty})!+(\frac{9}{\infty})!+(\frac{9}{\infty})!\big)!=6$

• 30. dravig  |  March 16, 2014 at 4:51 pm

$(\cos 0 + \cos 0 + \cos 0)! = 6$

• 31. venneblock  |  March 17, 2014 at 11:56 am

Nice! I hadn’t thought to use trig functions. Love the creativity!

• 32. Daniel  |  October 7, 2015 at 11:05 pm

Yeah… true. Since infinity isn’t TECHNICALLY a number, you haven’t technically added any numbers. Clever girl….

• 33. Dardan  |  October 25, 2016 at 8:22 am

Ett is this

• 34. Dylan Dang  |  March 14, 2014 at 11:32 pm

My solution for 8 didn’t involve factorial.

$8 - \sqrt{\sqrt{8 + 8}} = 6$

• 35. Stephen Xu  |  September 21, 2016 at 2:41 pm

brilliant!

• 36. Bob  |  October 28, 2014 at 6:31 pm

General solution for n n n: (Count(n, n, n))!

• 37. Raghava  |  January 15, 2015 at 3:47 am

$(n\cdot3)- 3\cdot(n-2) \\ (0\cdot3)-3\cdot(0-2)=0-(-6)=6 \\ (1\cdot3)-3\cdot(1-2)=3-(-3)=6 \\ (2\cdot3)-3\cdot(2-2)=6-0=6 \\ (3\cdot3)-3\cdot(3-2)=9-3=6 \\ (4\cdot3)-3\cdot(4-2)=12-6=6 \\ (5\cdot3)-3\cdot(5-2)=15-9=6 \\ (6\cdot3)-3\cdot(6-2)=18-12=6 \\ (7\cdot3)-3\cdot(5-2)=21-15=6 \\ (8\cdot3)-3\cdot(8-2)=24-18=6 \\ (9\cdot3)-3\cdot(9-2)=27-21=6$

• 38. venneblock  |  January 15, 2015 at 8:55 am

Your function always results in 6, Raghava, but I think it violates the rules of the game: it introduces two 3’s and a 2 in each expression, and it only uses the required digit twice.

• 39. JOS  |  February 15, 2015 at 2:09 am

my solution is

$2 + 2 + 2 = 6 \\ 3 \times 3 - 3 = 6 \\ \sqrt{4} + \sqrt{4} + \sqrt{4} = 6 \\ 5 \div 5 + 5 = 6 \\ 6 \times 6 \div 6 = 6 \\ 7 - 7 \div 7 = 6 \\ \sqrt[3]{8} + \sqrt[3]{8} + \sqrt[3]{8} = 6 \\ \sqrt{9} \times \sqrt{9} - \sqrt{9} = 6 \\$

• 40. venneblock  |  February 18, 2015 at 10:05 am

Jos — I updated your reply to use $\LaTeX$. Your solution looks good except for 8s. I think by using the cube root, you’ve added three 3s, which seems like a violation. Is there another symbol that means cube root? I’m not aware of one.

• 41. yo  |  October 14, 2015 at 4:23 pm

great job man it helped me a lot

• 42. Hunter  |  March 1, 2015 at 7:02 pm

if you think about it the rule was ANYTHING other than a number so technically this works:
0 + 0 + 0 ≠ 6
1 + 1 + 1 ≠ 6
2 + 2 + 2 ≠ 6
3 * 3 * 3 ≠ 6
4 + 4 + 4 ≠ 6
5 + 5 + 5 ≠ 6
6 + 6 + 6 ≠ 6
7 + 7 + 7 ≠ 6
8 + 8 + 8 ≠ 6
9 + 9 + 9 ≠ 6

• 43. venneblock  |  March 2, 2015 at 12:09 pm

Not sure I agree with this solution, Hunter. The problem statement asks for an equation (“the value of each expression is equal to 6″) — yours would be considered an inequality — and an equal sign was included with each group of three numbers.

I’d also point out that 2 + 2 + 2 is, in fact, equal to 6, so that’s a problem, too.

But nice try. I appreciate out-of-the-box thinking!

• 44. Hunter  |  March 4, 2015 at 11:04 pm

It would seem I have missed the fine print and also failed even in my outside the box thinking:
0 + 0 + 0 ≠ 6
1 + 1 + 1 ≠ 6
2 * 2 * 2 ≠ 6
3 + 3 + 3 ≠ 6
4 + 4 + 4 ≠ 6
5 + 5 + 5 ≠ 6
6 + 6 + 6 ≠ 6
7 + 7 + 7 ≠ 6
8 + 8 + 8 ≠ 6
9 + 9 + 9 ≠ 6
was my intended post however it would seem to fail according to “the value of each expression is equal to 6″

• 45. venneblock  |  April 13, 2015 at 11:30 pm

Here’s my thought about great problems: they get people thinking! I don’t care so much that people think about different things, as long as they’re thinking!

• 46. onepatas  |  March 31, 2015 at 4:47 pm

This is really nice problem! Does anyone of you know about something similar to this one? … I would like to mention, that my friend discovered this method for number eight:
$\bigl(\sqrt{\frac{8}{8}+8}\bigr)!$

😀

• 47. Thijs  |  April 25, 2015 at 3:58 pm

My solution for the first 13:

(Cos0+cos0+cos0)!
(1+1+1)!
2+2+2
3*3-3
√4+√4+√4
5/5+5
6+6-6
-7/7+7
√(8/8+8)!
√9*√9-√9
√(-10/10+10)!
√(12+12+12)

And 11…:

11-(Arctan(tan(-Π√(11e)))/Π)-√(11e) = 6

No cheaty methods as Math.Floor..

• 48. venneblock  |  April 26, 2015 at 10:17 pm

Love it, Thijs! Not sure about 11, though. Seems like you added π and e, which I’d classify as numbers, not functions. We’ll have to ask the judges for an official ruling…

• 49. Old Math Teacher  |  June 2, 2015 at 10:49 am

I have an objection to using SQRT or a radical sign. This is technically inserting numerals into the problem. I watched the first few minutes of the Scam School video and remember the explicit direction of not inserting numerals into the problem. The example given was not using the cube root of something because that would insert a 3 into the problem.

If you can’t cube root you can’t square root. The index of the radical maybe unwritten but it is understood to be 2. If the puzzle is not to include extra digits I don’t see how it’s proper to use roots. The only root that I would permit would be to use a left-hand number as the index of the root of its right-hand neighbor.

• 50. venneblock  |  June 4, 2015 at 8:47 am

I like the suggestion of using a number as the index of the root for the number to its right. I don’t know if that actually helps get a sum of 6 in any case, but it’s creative.

As for not including SQRT, perhaps it’s a valid point. But the Scam School guys used them, and I wonder how many of these are possible without SQRT. Do you have solutions for all of them that don’t use roots?

• 51. Old Math Teacher  |  June 5, 2015 at 11:27 am

I’ve rewatched the Scam School video several times to listen to the language of the setup. I realize that — although I think it woud be more interesting to exclude the insertion of all specific or implied nurmerals — the original setup was that “any symbol could be used … you just can’t draw another digit.” The not “drawing” a digit does grant permission to use log, ln, sqrt, etc. that might have implied indices or bases.

I also realized that when I first posted, that the using a left or right hand digit as an index or exponent for the right or left hand neighbor wasn’t useful above the 2nd or 3rd root or power. It was just a “no extra digits of any kind” way to use radicals and exponents 😉

• 52. Ciarán Lawler  |  September 2, 2015 at 12:10 pm

here’s my solution to solve for absolutely every real or complex n up to (but not including) infinity

6=[∇n. !+∇n. !+∇n. !]! ∀n∈{ℝ|ℂ}

• 53. Daniel  |  October 7, 2015 at 11:01 pm

The solution is the same for every number. For any number y, perform the following function:

[(d/dx y)! + (d/dx y)! + (d/dx y)!]! = 6

This works for any real number. Sorry to ruin your efforts to find answers for as many numbers as you can… I guess I found the answer to all of them.

• 54. Ziki  |  November 11, 2015 at 5:14 am

Also there is
{[(√8+8)!] / 8}!

Since:
8 + 8 = 16
√16 is 4
4! is 24
24 / 8 is 3
and finally 3! is 6

• 55. Ziki  |  November 11, 2015 at 5:18 am

(√(8 + 8)! ÷ 8)! is my method of solving 8 8 8 = 6

• 56. john bert  |  December 21, 2015 at 12:05 am

i have a question to all.
how can u add a number to number 9 and the result is 6 only using additon???

sorry for my grammar!!!

• 57. venneblock  |  January 3, 2016 at 4:57 pm

If I understand your question correctly, John, I think the answer is -3. That is, 9 + (-3) = 6, and this uses only addition. I don’t know your age, so perhaps you’ve not yet learned about negative numbers.

• 58. Ciarán Lawler  |  January 8, 2016 at 5:02 pm

({ℵ}Σ•range±∇[x,y,z,…]!)!∀(x,y,z,…)∈{{}|ℵ|ℝ|ℂ}

General solution for all numbers, real or complex, or sets of numbers, and any number of these and any combination

• 59. Bob  |  January 9, 2016 at 8:40 am

If looking for a universal solution, a much simpler approach is:

(Count(n,n,n)) !

• 60. venneblock  |  January 9, 2016 at 12:44 pm

Beautiful, Bob! Probably the most elegant general solution we’ve seen to this problem.

• 61. Matias  |  January 20, 2016 at 5:51 pm

(0! 0! 0!) = 6

1 + 1 + 1 = 6

2 + 2 + 2 = 6

3 x 3 – 3 = 6

4 – 4 / 4 = 6

5 / 5 + 5 = 6

6 + 6 – 6 = 6

7 – 7 / 7 = 6

√8 – √8 / √8 = 6

√9 x√9 – √9 = 6

• 62. Ali Akm  |  February 15, 2016 at 1:43 pm

Hi guys, I have a solution for 11 11 11 = 6
You can:
$\log\,(11 - \frac{11}{11})^6 = \log\,(11 - 1)^6 = \log\,10^6 = 6$

• 63. Chris McKenna  |  February 29, 2016 at 6:58 pm

Log base 10 is assumed. It’s the default for many calculators so it seems fair to use it. The use of a ( . ) doesn’t seem to violate the rules either.

( sqrt ( 11 – log (11 / .11) ) )! = 6

( 13 – ( 13 / 1.3) )! = 6

• 64. Henry Ngo  |  August 12, 2016 at 1:28 am

(13 – antilog(13/13))! = 6
(13 – antilog(1))! = 6
(13 – 10)! = 6
3! = 6

• 65. Kelvin  |  April 6, 2016 at 8:30 pm

Puzzle: 7 + 7 + 7 + 7 is equal to 28… How can you switch two sticks to make an expression equal to 6?

• 66. venneblock  |  October 25, 2017 at 1:29 pm

Kelvin, I assume this is a matchstick problem where each of the 7’s and the +’s are made with two sticks. Is that correct?

• 67. Yusuf Hidayat  |  September 8, 2016 at 9:05 am

1. 2+2+2=6
2. 3×3-3=6
3. 4-4/4=6
4. 5+5/5=6
5. 6*6/6=6
7. 7-7/7=6
8. √8+8/8=6
9. √9√9:√9=6

• 68. Yusuf Hidayat  |  September 8, 2016 at 9:06 am

1. 2+2+2=6
2. 3×3-3=6
3. 4-4/4=6
4. 5+5/5=6
5. 6*6/6=6
7. 7-7/7=6
8. √8+8/8=6
9. √9√9:√9=6

• 69. Zem Andrei Zamora  |  October 9, 2016 at 3:05 am

My solution:
(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 × 3 – 3 = 6
√4 + √4 + √4 = 6
5 + 5 ÷ 5 = 6
6 + 6 – 6 = 6
7 – 7 ÷ 7 = 6
8 – √[√(8+8)] = 6
√9 × √9 – √9 = 6

• 70. Vaibhav  |  December 15, 2016 at 11:50 am

= 0! 0! 0!
= (1+1+1)!
= 6

• 71. Math Magic  |  January 9, 2017 at 8:52 pm

This is the best answer… ❤

(Count(n,n,n))! a universal solution. (Bob)

• 72. Leo  |  January 20, 2017 at 1:06 am

|{n,n,n}|!=6 make more sense 🙂

• 73. Newbie-Hello  |  February 8, 2017 at 8:01 am

(0+0+0)! = 3! = 3x2x1 = 6
(1+1+1)! = 3! = 3x2x1 = 6
2+2+2 = 6
(3×3)-3 = 6
4+4-(√4) = 8-2 = 6
5+(5÷5) = 5+1 = 6
(6×6)÷6 = 6
7-(7÷7) = 7-1 = 6
³√8 + ³√8 + ³√8 = 2+2+2 = 6
(9+9)÷√9 = 18÷3 = 6

• 74. Ahlad  |  November 2, 2017 at 10:19 am

How (0+0+0)!= 3!

• 75. venneblock  |  November 2, 2017 at 12:59 pm

It’s not. As Bob noted below, (0 + 0 + 0)! = 1, not 3. That should read: (0! + 0! + 0!)! = 3.

• 76. Bob  |  March 3, 2017 at 10:31 pm

The zero line is incorrect since parentheses are performed first:
(0+0+0)! = 0! = 1

• 77. Apon  |  June 6, 2017 at 12:39 am

six=9 how.?

• 78. will  |  October 11, 2017 at 4:32 am

do you know 11 13 and 14

• 79. venneblock  |  October 13, 2017 at 3:07 pm

Will, I’m not sure I understand the question. Do you mean, do I know how to make 11, 13, or 14 with three 6’s? I don’t, but if you confirm that that’s your question, I can give it some thought.

• 80. Ahlad  |  November 2, 2017 at 10:18 am

( Cos 0 + Cos 0 +Cos 0 )! = 6

• 81. Zackack  |  January 26, 2018 at 4:43 pm

(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 * 3 – 3 = 6
√4 + √4 + √4 = 6
5 ÷ 5 + 5 = 6
6 + 6 – 6 = 6
7 – 7 ÷ 7 = 6
√(8 ÷ 8 + 8)! = 6
√9 * √9 – √9 = 6
√(10 – 10 ÷ 10)! = 6
√√(11!!! + 11) ÷ 11)! = 6
√(12 + 12 + 12) = 6
(13 – (13 ÷ 1.3))! = 6

I’ve got that so far, anybody got 14?

• 82. venneblock  |  February 4, 2018 at 2:11 pm

I don’t think inserting decimal points would be allowed, as that fundamentally changes the number. So I think your solution for 13 would be disqualified. And I suspect there’s a typo in your solution for 11 — I love the use of four factorials and two square roots, but 11!!! is a number much too large to get back to 6 even with multiple square roots.

• 83. venneblock  |  February 4, 2018 at 2:22 pm

My inelegant solution for 13 is:
$\left (\sqrt{\left [\,\sqrt{13}\,\right ] + \left [\,\sqrt{13}\,\right ] + \left [\,\sqrt{13}\,\right ]} \right )!$

That same approach would work for 14 and 15, too.

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