4
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Can you find a way to make:

6 5 4 3 = 1

by concatenation and/or adding any of (and only) these mathematical operators:

  • +
  • -
  • ×
  • !
  • ÷
  • ^
  • standard parentheses ()

You cannot add other numbers to the equation.
The result must be a mathematical equality.


Inspired by Make 5 5 5 5 = 19

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4
  • 1
    $\begingroup$ Yep, I totally failed to see the obvious! Hopefully 6 5 4 3 = 81 will be more fun. $\endgroup$ Aug 27 '18 at 6:25
  • 3
    $\begingroup$ Chill Everyone! Stop Downvoting Just Becasue it's obvious +1 $\endgroup$ Aug 28 '18 at 6:11
  • $\begingroup$ I can do $6\; 5 \;4\; 3\; 2 = 1$ by how $4!\div (6\times (5-3)\times 2) = 1$ :D $\endgroup$
    – Mr Pie
    Aug 28 '18 at 14:03
  • $\begingroup$ I can also complete it without the $6$ by how $(4!-3!)\div 5!! = 1$ :D $\endgroup$
    – Mr Pie
    Aug 29 '18 at 22:10
11
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One possible answer is:

(6 - 5) * (4 - 3)

By request another:

-(6 + 5) + (4 * 3)

Some tongue in cheek answers:

(6 - 5)! * (4 - 3)!
(6 - 5) ^ (4 - 3)
(6 - 5) / (4 - 3)
(6 - 5)! / (4 -3)!
plus other combinations of factorials of 1

Another solution:

6 + 5 - 4 - 3!

More complicated:

(6!) / (5 * 4! * 3!)

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5
  • $\begingroup$ Nice and simple one... any more? :) $\endgroup$ Aug 27 '18 at 5:38
  • $\begingroup$ Ho hum. I epic fail :D guess I'll try again... $\endgroup$ Aug 27 '18 at 5:41
  • $\begingroup$ Should I go on? $\endgroup$
    – wolfram42
    Aug 27 '18 at 5:47
  • $\begingroup$ LOL, see my previous comment $\endgroup$ Aug 27 '18 at 6:04
  • $\begingroup$ @wolfram42 Yes, please go on there ;-) $\endgroup$
    – xhienne
    Aug 28 '18 at 14:23
6
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I'm going to try:

$$\dfrac{-6+5+4}{3}=1$$

and:

$$\dfrac{6\times5}{4!+3!}=1$$

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4
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The simplest:

$65 - 4^3$

(I wrote a program.)

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1
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Some exponential solutions

$\frac{6^{5-4}}{3!} $
$(6 - 5^{4-3})$

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1
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A simple solution

${3 - \dfrac{6 + 4}{5}}$

In code form

3 - (6 + 4)/5

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1
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Very simple.

(6-5)*4-3
6-5=1 => 1*4=4 => 4-3=1

Cannot get simpler than this.

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2
  • $\begingroup$ Hello and welcome to PSE! Please hide your answers in spoilers using the >! prefix on the lines needing hiding. $\endgroup$
    – rhsquared
    Aug 29 '18 at 13:12
  • 1
    $\begingroup$ sure will do from now.. Thanks for letting me know! $\endgroup$ Aug 30 '18 at 3:54
1
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Simple and in order:

(6 - 5 - 4 + 3)!

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1
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Here are some solutions I found:

  • (6-5)^(4-3)
  • (6-5)*(4-3)
  • (6-5)/(4-3)
  • (6+5)-(4+3!)
  • (6+5)+(4*3)
... way too many!

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1
  • $\begingroup$ Hello, and welcome to PSE! Unfortunately your answer doesn't add anything new to the existng ones (wolfram42 already gave the same answers like you did). Anyways, feel free to stick around, check out the Tour and enjoy your stay at the site! $\endgroup$
    – Christoph
    Sep 11 '18 at 20:54
0
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Assuming the following is possible:

(6+5)->(11)->(1+1)

then: 1)

((6+5)-4)+3)=1

2)

((11)-(4)+3)=1

3)

((1+1)-(4)+3)=1

4)

(2-4)+3=1

Here is another one:

((6*5)-4!)/3!

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