Four 5s make 19?

{5 - (5 / 5)}! - 5

((5)!-(5x5))/5

$(5 x 5) - 5 - 5^{0}$

Most likely intended answer: $\left( 5 - \frac{5}{5} \right)! - 5$

Another valid answer that I found first: $5 \times \phi(5) - \frac{5}{5}$

The usual hacky answer: Define $a \circ b = 19$ for any $a,b$ , then the answer is $((5 \circ 5) \circ 5) \circ 5$

5^2 - 5 -(5:5) =19 (5 squared since I'm writing from my phone)

(5-(5-5)!)! - 5 = 19

5-5+5+5=1+9

([Sqrt(5*5)]! / 5 ) - 5

((5^2)-5)-(5/5) = 19

If 'factorials ARE allowed' means 'double factorials are allowed' as well, then: 5!! + 5 - (5 / 5)

5!! means 5 x 3 x 1 (ie only odd numbers). Refer to wiki article for more information.

(Slightly hacky, I guess, but double factorials are almost essential to do the yearly 0-100 challenge. The challenge where you have to get every number from 0 to 100 using the digits of that particular year. eg Use 2 0 1 7 to get every number from 0 to 100)

(5-5/5)!-5

(5!-5*5)/5

$(5 \times 5)+(5 - 5) = 19$

Base 7 for the first half, Base 10 after

(5! /5) - $\sqrt{5 \cdot 5}$

(5!/5 -5) mod (5!)

(5+5) - (5-5) = 1 + 9

They never said you couldn't insert operators after the equals sign

(5/.5 – .5)/.5

5 + 5 + 5 - 5 = 1 + 9

√5•√5^3 - 5 - 5^0 = 19

(((5+5)^2 )-5)/5

(5!/5)%5+5=9

55+55=110-5-5-5=95/5=19

(5! -(5*5))/5=19.......(120 -25)/5= 19..... 95/5=19

5+5+5+(5 to the power of .861353116146786)

Exponent is probably against the rules, but I should get some credit for effort.

$5!/5 - \sqrt{5 \times 5}$

55/55 = 1 to the 9th

Be creative (5+5)-(5-5)=1+9

5+5-5+5=1+9

(5-(5-5)!)! - 5 = (5-0!)! - 5 = 4!-5 = 19

5 + 5 + 5 + 5 = 19 (In base 11)

5/5x5/5=1^9

{5!/(SQRT5*SQRT5)}-5

Use Base 11. 5+5+5+5=19

5+5+(5-5)=(1+9)

We have a wide variety of answers for this question

5-5+5+5=1+9

5*5-ceil(sqrt(5))-ceil(sqrt(5))

(5P5)/(5)-5

(5!/5) - 5 = 19

5!/5-√(5x5)

5 x (5 - 5/5) = 19

5 x [ 5 - ( 5 / 5 ) ] = 19

$5^2 - (5/5) - 5$

{5!/5}-(5^2/5)

[5-(5-5)!]!-1

$\frac{5!}{5} - \sqrt{5 \times 5} = 19$

5^2-(5+5/5)

{(5x5)-5}-5^0

For me the solution was (5! - (5×5)) ÷5

$\frac{√[(5 × 5)!]}{5}$ - 5 = √[5!] - 5 = √120 - 5 = 24 - 5 = 19

$5! /5-|\sqrt{5*5}|$

(5 (squared) -5) -5/5

$5^2-5-(5/5)$

{[5!/sqrt(5)]/sqrt(5)}-5 just to be a bit more fancy I guess

{5-(5/5)}! - 5.

(5*5)-5-(5/5)

This is probably the easiest solution, since it only uses basic arithmetic with no factorials.

5 squared minus 5 minus the expression 5 divided by 5.

$\frac{5!}{5}$ - ( $\sqrt{5}$ $\times$ $\sqrt{5}$ ) = 19

( 5! - 5*5 ) / 5

$5^0\cdot\frac{5!}{5}-5$

5!/5-5 = 120/5-5 = 24-5 = 19

(5!)-5+5-5 = 19

5555 >= 19

(5^2)-5-[5*(5^-1)]=19

$\frac {\sqrt {5 \times 5} !} {5}-5$

5²-5-5÷5=19

5!/5 - sqrt(5x5)

5^2 - (5/5) - 5 = 19

5^2 + 5^2 - 5^2 - 5 = 19 (Exponents) 😋

(5+5)*(5/5)=1+9. :P

5!/5-√5*√5

(5! ÷ 5 - 5 ) × 5^0

$/{5 - (5 \div 5)/}! - 5 = 19$

5!/5 - 5 and another 5 for the piggy bank.

$5!/5-5\cdot \operatorname{sign}(5)$

5+5+5-5=1+9

{5!-(5×5)}/5

5 x 5 - 5 - 5^0

5-5+(5/5) =1^9

I got (5 x 5 - 5) - 5^0 = 19

(5!)÷5-5=19

(5! / 5)-sqrt(5*5)=19

(5!-5*5)/5 = 19

5!/ $sqrt(5x5)$ -5

5! / 5 - Sqrt(5 x 5)

5! / 5 - (sqrt(5) * sqrt(5))

[ { (sqrt5 * sqrt5) ! } / 5 ] - 5

How about (5-(5/5))!-5 ?

Or maybe (((5+5) -(.5))/(.5)) ?

How about with three fives?! (5!)/5 -5

5exp2 -5 -5/5

(5!-5x5)/5

(5^3)-5!+5+5=19

• example A. shows the symbols needed.
• example B. shows the first (and last) step of order of operations.
• example C. shows the answer of both equations.
• A. 5 x 5 - 5 / 5 = ?
• B. 20 - 1 = ?
• C. ? = 19

(5^0)*5!/5-5

(5! - 5*5) / 5

The answer to these style questions will always be yes, because it would be near impossible to test all possibilities to say NOTHING will make the equation true.

((5!) - (5 5) / 5) I'm not sure if I wrote it right, but here's a step by step: (120 #the factorial of 5# - (5 5)) / 5 = 95 / 5 = 19

$(5-5/5)! - 5$

(( \sqrt{5} * \sqrt{5} )! / 5) - 5

Many have already shown solutions using single factorial, like:

{5 - (5 / 5)}! - 5

How about double factorial (!!) An operation defined as n*(n-2)!!. So in example 5!! = 5 * 3 * 1 = 15

5!! + 5 - ( 5/5 ) = 19

5^2-5-(5/5)=19

(5 * 5) / (5 * 5) = 1 ^ 9

5!-5^3+5^2-5^0 =120-125+25-1=19

(5! - 5x5) / 5

$((5! / 5) - 5) \times 5^{0}$

((5!)-5*5)/5

({5! / (5^2)} x 5) - 5

(5*5 -5)-5^0

(5!/5)-5/5

5555 = 19 in base 5546

5*5/5+5=1+9 ;-P

S(S(S(S(5*5-5-5)))) where S is the successor function.

Outside of (5!-5x5)/5, I initially thought that we could insert anything anywhere, so I did 5+5-5/5=(1)9

Assuming exponents are operators, not numerical inclusions 5^2-5-5/5

I noticed 5! = 5 * 4 * 3 * 2, then divided by 5 gave 4 * 3 * 2 = 24
24 - 5 gives 19 but I have not one 5 but 2 5's to get the solution

Stuck for a while at 5! / 5 -5 (with that pesky extra 5). Solution sqrt(5) * sqrt(5) gives 5 Putting it together

5! / 5 - (sqrt{5} *sqrt{5})

(5!/5)-5=19 . 5 Factorial divided by 5 =120. 120/5=24. 24-5=19

{(5^3 -5^2)/5}-5^0

5! - 5 - (5/5)

[5! -(5x5)]%5

(5!-(5*5))/5

$\lceil \sqrt{555} \rceil - 5 = 19$

Because: $\sqrt{555} \approx 23.558$ ; so rounded up (ceiling function) that is $24$ ; and $24-5=19$ .

Using the ceiling or floor functions opens up many possibilities with non-integer intermediate terms …

Another very logical and funny one (IMHO) that I haven't seen in the comments yet (I'm proud of it):

$\neg ( 5 5 5 5 = 19 )$

Because: " $5555 = 19$ " is a $false$ statement; and the logical NOT is an (unary) operation that literally makes the equation true: $\neg ( 5 5 5 5 = 19 ) = true$ ◕‿◕

You probably can't get closer to the problem's wording.

I went through pretty much all of the comments and collected the approaches to solving the problem:

• $(5! - 5 × 5) / 5 = 19$
• $5! / 5 - 5 = 19$
  • As this uses $5$ only three times, we've got to somehow get another $5$ in. Three ways were used:
    1. replacing one of the 5s by: $\sqrt{5×5}$
    2. replacing one of the 5s by: $\sqrt{5} × \sqrt{5}$
    3. appending the modulo function in a way that does not change the value, e.g. … $\mod 5!$ . Instead of $5!$ any value bigger than 19 would do, of course.
  • i.e. $\sqrt{5×5}! / 5 - 5 = 19$ , or $5! / \sqrt{5×5} - 5 = 19$ , or $5! / 5 - \sqrt{5×5} = 19$
  • i.e. $(\sqrt{5}×\sqrt{5})! / 5 - 5 = 19$ , or $5! / (\sqrt{5}×\sqrt{5}) - 5 = 19$ , or $5! / 5 - (\sqrt{5}×\sqrt{5}) = 19$
  • i.e. $(5! / 5 - 5) \mod 5! = 19$
• $(5-1)! - 5 = 19$
  • Here the $1$ somehow has to be replaced by a term with two fives. I've seen these:
    1. $1 = 5/5$
    2. $1 = (5-5)!$ (mind that $0! = 1$ )
  • i.e. $(5-5/5)! - 5 = 19$ , or $(5-(5-5)!)! - 5 = 19$
• $(5 / .5 - .5) / .5 = 19$ (that's a nice one, using the decimal point as sort of a notational operation)
• As the problem does not define the allowed operations and notations, we are free to use any in existance in the mathematical world, e.g. ceiling and floor functions, double factorials, logarithms, modulo, notations used in combinatorics, Euler's phi function ( $\varphi$ ), increment (++) or decrement (--) functions as in programming languages … whatever.
  • i.e. $5×5 - 5 - \lceil \log{5} \rceil = 19$ , or $5×5 - 5 - \lfloor \ln{5} \rfloor = 19$
  • i.e. $5×5 - \lceil \sqrt{5} \rceil - \lceil \sqrt{5} \rceil = 19$
  • i.e. $5 × \varphi(5) - 5/5 = 19$
  • i.e. $5 × (\textrm{--}5) - 5/5 = 19$ , or $(\textrm{--}5)! - 5 + 5 - 5 = 19$ , or $5+5+5+(\textrm{--}5) = 19$ …
  • i.e. $5!! + 5 - 5/5 = 19$
  • i.e. $(5\textrm{P}5) / 5 - 5 = 19$ (with notation $n\textrm{P}r = \binom{n}{k}$ )
• A nice out-of-the-box approach is to squeeze operations into the right side $19$ . Why not, as there is no given rule about spacing.
  • i.e. $5+5 +5-5 = 1+9$
  • i.e. 5/5 +5-5 = 1\textrm{^}9 , or 55/55 = 1\textrm{^}9

I also like Ραμών Αδάλια's generic solution: "Use a quaternary operand which sends 5 5 5 5 to 19, easy just define it and that's it". Define $▼(a,b,c,d)≔19$ for any $a,b,c,d$ . Then $▼(5,5,5,5) = 19$ .

Or Ivan Koswara's version of this idea: "The usual hacky answer: Define $a \circ b ≔ 19$ for any $a,b$ , then the answer is $((5 \circ 5) \circ 5) \circ 5 = 19$ "

That is mathematics! :-)

In a way, such a loosly given (under-defined) problem gives much more space to creativity as a precise unambiguous one does. ◕‿◕

the simple solution is (5)^2-(5+5/5)

5^2–5-(5/5) = 19

you can do it with only 3

(5!/5)-5

but for the question's sake,

((5!/5)-5)x5^0

5^2-5-(5/5)

5^(2) - (5/5) - 5

Do 5 squared. Divide 5 by 5 to make 1. You have a 5 left over. The sum of 5 squared is 25. Minus the extra five from this. You have 1 minus the one and you get 19.

I’ve already seen the basic arithmetic solutions, including the one I have thought of. By the way, if you really want you can use logarithms (Since those are allowed). For example, ${5}^{log5(14)}$ +5 - 5 + 5, or even ${5}^{log5(x)}$ - 5 + 5 - ${5}^{log5(19-x)}$ etc. thus an infinite amount of solutions

(5-(5:5))!-5= 4!-5=24-5=19

5! / 5 - ( sqrt( 5 x 5 ))

5(squared) - 5 - (5/5) = 19 Writing on my phone so couldn’t properly show 5 squared

5^2 - 5 - (5 / 5)

(((5^3)-5)/5)+5

NO ONE wrote this one: 5×(5-5^0)-5^0; And yet it is one of the simplest answers!

Yes, $(5 - \frac{5}{5})! - 5$ .

Solutions using powers, including zero, aren't really correct as the power index is another number. Unless of course the index is 5 and is counted (good luck with that one!) This includes square roots (where the index is $\frac {1}{2}$ ) such as $5! / 5 - \sqrt{5 \times 5}$ . Also, square roots have two solutions, so this could equally give 29.

((5! / 5)-5)*5^0

5 * 5 - 5 - tanh5 = 19

5! ÷ 5 - (5÷5)

|5/5-5x5|=19

(5!-5×5)÷5

$5!/5-\sqrt{5*5}$

((5! / 5) - 5) x 5^0

5!/5-(5)(5^0)

$5!:5-(5^2:5)$

5²-5-5÷5 = 25-5-1 = 19.

5!/5-[5x $5^0$ ] = 19.

(5! - 5*5)/5

[5!-(5*5)]/5

5^{2}-5-(5/5)

(5! - (5 * 5)) /5

((5!)/5)-5=19

(5!(5+5))÷5

-(5-5)! + 5^2 - 5

5^2 -5 -5/5 =19

5^2-5-(5/5)

(5!)/5-5*1 is also possible

(5!)5÷(5^2)-5

5!+5 -(5/5)

5!/5 - SQRT(5*5)

5^2 - (5 + (5/5))

$\frac{5!}{5}-\sqrt{5 \times 5}$

(5!/5 x 5^0)- 5

5+5+5+5-(5/5)

Use a quaternary operand which sends 5 5 5 5 to 19, easy just define it and that's it

(5*5)-(5)-5°

(5!)/5-5*5^0

{(5/5)*5}!-5

(5! / 5) - (5 * 5^0)

(5^2 / 5)! / 5 - 5

(5!-(5X5))/5

5^2 - 5^0 - 5*5^0

5!/5 - sqrt(5x5)

(5!-(5*5))/5

( 5! - 5 × 5 ) ÷ 5

( 5! / (5+5) ) + 5 = 19

5 / 5 = 1

5 - 1 = 4

4! = 24

24 - 5 = 19

((5!) / 5) - 5