Remembering Last Year

By putting the exponent sign between the two numbers
i.e.
$2^0=1^5$

$\huge \color{#D61F06}{2} \ \ \ \color{#3D99F6}{0} \ \ \ = \ \ \color{#20A900}{1} \ \ \ \color{magenta}{5}$ $\huge (\color{#D61F06}{2} \ + \color{#3D99F6}{0!})!\ \ = \ \color{#20A900}{1} \ \ + \ \color{magenta}{5}$

My terms is $\large 2+0 = \sqrt{-1+5}$

One for the logicians 2 > 0 = 1 < 5

Notice that to say Yes, one has to just make perform the operations desired, but to say no, one has to prove it.

Since any operations is not sufficiently defined, such a statement cannot be proven.

Hence, it must be yes

Does this work?

.20=1/5

A new one: $2\times 0=\left\lfloor\frac{1}{5}\right\rfloor$

$2+0=\phi(1)-\mu(5)$ $2!+0=\Gamma(-1+\phi(5))$ $-\mu(2)+0=\mu(1)\mu(5)$ $\phi(2)-0=\phi(1)^5$

Write a solution. 2^0 = 1^5

$2^{2} + 0 = -1 +5$

2 x 0=ln(1) x 5

Yes. Bcause of this

2^0=1^5

$2 \times 0 = \cos(\arcsin(1)) \times 5$

I did: 20%=1/5

2 0 = 1 5 ==> 2 0 % = 1 ÷ 5 ==> 20% = 1/5

$20 \neq 15?$

Doesn't this count?

Here's an obvious one - $2^0 = 1^5$

2+0=sqrt(-1+5)

Would it be allowed to insert a decimal? Then you could do:

.20=1/5

2 + 0 = \sqrt{ -(1-5) }

2×0=LN(1)×5

Since 0! = 1, 2-0!=1^5 is valid.

1 divided by 0.5 is 2 so: 2+0=1÷.5

2x0=cos-1(1)x5

$2-0!=1^5$

2 is 0 = 1 is 5 -> false = false

2^2+0=|1-5|

4=4

It says any function or operation can be used. Instead of thinking of a solution involving common operations I can just define a function that returns the same number for any arbitrary operands.

$(2 + 0!)!$ $= 1 + 5$

(2^4)-(0^0) = 16-1

=15

2 ^0 = 1 ^ 5

2^0=1^5

2 power 0=1

1 power 5=1

If we can use expopents: 2^0=1^5

2+0 = Sqrt(-1+5)

I agree with the exponent solution

$f(20)=g(15)$ where f=g=0

I was thinking 2 times 0 and 1 mod 5

Yes multiply both sides by zero both becomes equal

(2>0) = (1<5)

.20=1/5 using the decimal point before the 2 and the divide between the 1 and 5

(2 - 0) = sqrt(-(1-5))

(2^2)+0!=1*5 This is a bizarre solution, but it works.

2^2-0=-1+5

-2^2 +0 = 1-5

2+0 = 1/.5

2^0 = 1 1^5 = 1

1=1

2-0! = 1^5

$20_{(base 10)} = 15_{(base 15)}$

Raising ANY number in the power of 0 does ALWAYS give 1. So 2 in the power of 0 = 1.

Raising 1 in ANY power shall FOREVER repeatedly give 1. So, 1 in the power of 5 = 1 1 1 1 1 = 1.

And THUS, 2 in the power of 0 = 1 in the power of 5 = 1. :-))

(2 + 0!)! = 1 + 5

20-5=15 He said ANY operations☺️

2 0 = 1 / (5%)

2-0! = 1^5

$2+0=1\div .5$ $\beta_{\lceil \mid \rceil}$

2+0= square root of -1+5

I used: $2*0=1-(5^0)$

2^2 + 0 = -1 + 5

2 ^ 0 = 1 ^ 5

may be against the rules, but I was thinking...

$(2+0)^2=|(1-5)|$

2+0 = 1^0 + 5^0

2^2 + 0 = -1 + 5

2- cos(0) = 1^5

2 + 0 = $\sqrt {|1-5|}$

-(2**2)=1-5

-2^2 + 1 - 5 Make two negative, square it.

2 to the 0 power = 1

1 modulo 5 = 1

$2 \times 0 = \ln{1} \times 5$

Bit rusty on what constitutes an operation or function, but this is my guess. .20 = 1/5

$2^2+0=-1+5$

2 $2$ +0=-1+5

2^2 + 0 = -1 + 5

2 squared = -1+5

-5 + 20 = 15

2 x 0 = \arccos(1) x 5

(2 - 0)^2 = 1 - 5

2 0 = 1 5

solution .20 = 1÷5

No one is finding what is arguably the simplest way to solve this problem.

2*0 = /frac {1}{5}

I don’t know how to insert the greatest integer symbol around the fraction, but since the greatest integer to 1/5 is 0, this is correct.

d/dx(2)+d/dx(0)=d/dx(1)+d/dx(5)

2^0 = 1^5

since x^0 is always 1 so as 1^x is always 1 too.

20 >= 15 should be a valid solution to this puzzle. And if spaces cannot be removed then this should do: 2 - 0 >= 1 - 5

2 + 0 = sqrt(-1 + 5)

2^2 + 0 = -1 + 5

2^2+0^0=1*5

-(2^2)+ 0 =1-5

2^0 = 1 || 5

20 - 5 = 15

Meh.

2 + 0 = sqrt(|1-5|)

You could use the inequality operator to return true;

20 != 15

2^2-0 = |1-5|

This works too

$\lfloor\sqrt{2+0}\rfloor = \left\lceil\frac15\right\rceil$

My solution was

2^2 + 0 = -1 + 5

It may have some flaws though since adding the 2 may be cheating

If percentage counts as an operation -- and I don't see why it wouldn't -- then this can be a fraction conversion sentence.

20% = 1/5

2 - 0! = 1 ^ 5

If an arithmetical operation consists of any function mapping a set of natural numbers to a natural number, then the function f(20) = 15, f(x) = x otherwise, is a solution. For another solution, define a function called 'predecessor' 'p' such that p(0) = 0 and p(n + 1) = n for any natural number n. Similarly, define f(0) = 0, f(1) = 0, f(2) = 0, f(3) = 0, f(4) = 0, f(n + 5) = n. So, f(20) = 15 and thus works as such a function.

2^2+0=-1+5

putting a decimal pint before 2 and a dividing 1 by 5

.20 = 1/5

2 - 0! = 1^5

-(2²) + 0 = 1 - 5

2^2 +0= -1 + 5

ANY operations and functions, eh?

Well, then let say $f(2) = 0$ , $g(0) = 0$ , $h(1) = 0$ , and $j(0) = 1$ .

Then $f(2) \times g(2) = h(1) \times j(5)$

I know this isn't the intent, but either say what you mean or mean what you say.

-(2^2)+0=1-5

Simplest: .20=1/5

20% = 1/5 I know % isn’t a function. But as given, the numbers aren’t assigned values. Also works as: .20 =1/5

-(2^2)-0=1-5

2(0)=d/dx 5

2^0=1^5 any number to the zero power is 1; 1 to the fifth power is 1

u(15)=u(20) where u is the unit step function

2^0 * 0^0=1^0 * 5^0

1* 1=1* 1

1 = 1

-(2pow2)+0=1-5

2^2+0 =-1+5

2^2+0=1-5 makes them both 4

.20=1/5 This shows a fraction and a percentage.

2square + 0 = -(1-5)

Make left side of equation to base ten and right side of equation to base fifteen

.20=1/5 by putting decimal on lhs and by diving 1 by 5 on rhs

(2+0)^2 = - 1+5

2^2+0^2= -(1)^2 + 5

Thought basically this just means:

2^2 + 0 = -1 + 5

2^0 = 1^5 Both results simplify down to the number one.

By inserting a decimal on the left and a division sign on the right:

.20 = 1÷5

I used scientific notation to make 1= 100. then I divided it by 5 to make 20. dabs So satisfying!

(2 + 0!)! = 1 + 5

2^2 + 0! = 1 × 5

2^0 = 1 and 1^5 =1

$2 \times 0 = \frac{\log1}{\log5}$

2 = 0-1+5. I moved the = !

2^0 = 1^5

2x5=1+9 Played with multiples of 5and10

20× .75=15

Or you could go unconventially and make it 20≠15

20 - 5 = 15

2 0=ln(1) 5

20%=1:5..... I think this will be the answer

20=1/(5%) =1/5/100

Counting in a base 5 system.

One can use exponent and modulus operations. Like this: $2^0 = 1$ mod $5$

Put a decimal in front of the 2, make the right side into a fraction.

.20 = 1/5

2^2 + 0 = -1 + 5

Assuming we can use any base logarithm...

$log_{20}20 = log_{15}15$

$1 = 1$

2+0=sqrt(-1+5)=2

I was thinking 2+0=√(-1+5)

2 x 0 = (log1) x 5

2 - 0 = sqrt(-1+5)

2+0=√(-1 + 5)

$2 + 0 = \sqrt{-(1-5)}$

(2)^2 +0=-1+5

2(-2)-0=1-5

2*0 = 5 ln(1)

2^2 + 0 = |1-5|

(2 - 0!) = 1 ^ 5

X Y = X + (0.2 * Y)

"Any operations and functions can be used." I define the function f(x)=x-5

f(20) = 15

I'm not sure if this is used in mathamathics, but the mod function from CS is possible 2%0 = 1%5

20% = $\frac{1}{5}$

Decimal 2 is 2/10. 1/5 = 2/10. Assumes that inserting a decimal is permissible .2+0=1/5

I think this one could be a solution: $(2+0)^2=|1-5|$

2x + 0 = x + 5

2x - x = 5 - 0

x = 5

then 2 . 5 + 0 = 5 + 5

10 = 10

-(2)^2 + 0 = 1 - 5

20^0 = 15^0

2:0=infinity

2 ^ 0 = 1 mod 5

2 * 0 = arccos(1) * 5

By multiplying 2 and 0 in LHS and taking the step function of the fraction obtained by dividing 1 with 5 (0.2) in RHS, we get 0=0 i.e.,

2 * 0 = [ 1 / 5]

Using modular arithmetic i.e. 20 mod5 = 15 mod5 =0

2^2 +0=-1+5

$2 - 0! = 1^5$

[2+cos(0)]! =1+5

$2^2$ + 0 = -1 + 5

2 (addition modulo 2) 0=1 (addition modulo 2) 5

Also works with multiplication modulo 5

2² - 0 = |1-5|

20 ≠ 15 .... took the easy root just drew a line through the equal sign ... TADA

2 xor 0 = 1 xor 5

2 + 0 = the square root of (-1+5)

2*0=1 * (((\frac{d}{dx})) 5)

$2^{2} = -1+5$

2 x 0 = {1} x {5} where {.} is fractional part

2 x 0 = arcsin(1) x 5

2 x 0 = arccos(1) x 5

2^2 + 0 = |1-5|

2^0 = 1 mod(5)

$2 + 0 = | \sqrt{|1-5|}|$

2 - 0! = 1^5

2 * 0 = 1%5

2^0 = 1^5 1 = 1

Subtract 5 from 1, take absolute value, then square root. Subtract 2 from 0. 2=2 i.e. $2+0=\sqrt{\left | 1-5 \right |}$

$2+0 = \sqrt{-1+5}$

20=15 Add an operation so that 20 becomes 15 Eg 20 - 5 = 15 Or 3/4 of 20= 15 Etc

2 * 0 = log(1) * 5

0.20 = 1/5

2^4-0!=15 is my answer

2^2 + 0 = -1 + 5

-(2)²-0 = 1-5

2 + 0 = 1 + 5^0

2 - 0! = 1^5

.20 = 1 /5