I'm Tired Of Those "1 1 1 1 = 5" Problems

Logic Level 1

$\huge{1 \, = \, 2016}$

Is it possible to make the above equation correct by inserting the appropriate operations (no numbers or constants) to the left hand side of the equation?

Assumption : Any operation or function with no usage of additional constants (digits, $\pi$ , $e$ , etc) is allowed. Ex. : square root (no higher roots); logarithms to base 10 ( $\log$ ), base $e$ ( $\ln$ ), or base 2 ( $\lg$ ); trigonometric functions ( $\sin, \cos, \arcsin, \ldots$ ); factorial; floor or ceiling; etc.

Yes No

7 solutions

Ivan Koswara
Feb 12, 2016

The answer is yes.

Of course, we have the classic solution $f(1) = 2016$ where we define $f$ to be the constant function evaluating to 2016, for example. But even if not, we can do the following:

It's reasonable to assume that $\arctan$ and $\sec$ are allowed. That's all we need for this.

Note that $\sec^2 \theta = 1 + \tan^2 \theta$ for all $\theta$ . Thus, we claim that $\sec \arctan x = \sqrt{x^2 + 1}$ . This can be proven easily; $\arctan x$ is $\theta$ such that $\tan \theta = x$ , so $\sec \arctan x = \sec \theta = \sqrt{\tan^2 \theta + 1} = \sqrt{x^2 + 1}$ .

Now that we have the above, we can simply repeat it any number of times to increase an integer to another integer. For example, $\sec \arctan \sqrt{1} = \sqrt{2}$ , then $\sec \arctan \sqrt{2} = \sqrt{3}$ , then $\sec \arctan \sqrt{3} = \sqrt{4} = 2$ , so we have $\sec \arctan \sec \arctan \sec \arctan 1 = 2$ . The pattern is clear: we just repeat $\sec \arctan$ for an appropriate number of times (to be specific, $2016^2 - 1$ times) to bring $\sqrt{1}$ to $\sqrt{2016^2} = 2016$ . Thus the answer is yes .

Moderator note:

As you stated, for such questions it's the restriction to specific operations / functions which makes this interesting. As such, you could edit the problem to say "Drawing only from the trigonometric functions and their inverses, can we ..."

Everytime I see such problems, I'm like 'why not?'. I still haven't gotten any question wrong by doing this :p

Arulx Z - 5 years, 3 months ago

I have a question.

Are we to bring $\sqrt { 1 }$ to $\sqrt { { 2016 }^{ 2 } }$ , or to $\sqrt { { 2016 }^{ 2 }-1 }$ ?

We know that,

if $x=\sqrt { n }$ , then $\sec { \arctan { x } } =\sqrt { n+1 }$ .

We can conclude that if $\sqrt { n+1 } =2016$ , then $n={ 2016 }^{ 2 }-1$ .

$x=\sqrt { n } =\sqrt { { 2016 }^{ 2 }-1 }$

So, we should bring $\sqrt { 1 }$ to $\sqrt { { 2016 }^{ 2 }-1 }$ , no? Which means we have to repeat $\sec { \arctan }$ for $({ 2016 }^{ 2 }-2)$ times.

Edit: Downvoting my post doesn't help anything unless you explain to me where I could be wrong at.

Kenneth Choo - 5 years, 3 months ago

First, I'm not the one downvoting your post.

Second, we want to bring to $\sqrt{2016^2}$ . Every $\sec \arctan$ increases the number in the radical by one, so to bring $\sqrt{1}$ to $\sqrt{2016^2}$ , we need $2016^2 - 1$ times. Where does $\sqrt{2016^2 - 1}$ come from?

Ivan Koswara - 5 years, 3 months ago

No, no, I'm not referring to you, just the specific person.

Well, from my calculations above, for $\sec { \arctan { x } } =2016$ , $x$ has to be $\sqrt { { 2016 }^{ 2 }-1 }$ . Repeating $\sec { \arctan }$ by $({ 2016 }^{ 2 }-2)$ times, $\sec { \arctan { \sec { \arctan { ......\sec { \arctan { 1 } } } } } } =2016$ .

Just like for $\sec { \arctan { x } } =\sqrt { 4 } =2$ , $x$ has to be $\sqrt { 3 }$ . Repeating $\sec { \arctan }$ by $2$ times, $\sec { \arctan { \sec { \arctan { \sec { \arctan { \sqrt { 1 } } } } } } } =\sqrt { 4 } =2$ .

Kenneth Choo - 5 years, 3 months ago

@Kenneth Choo – Oh, you have one $\sec \arctan$ counted first. Yes, there are thus $2016^2 - 2$ more remaining. Just like because $\sec \arctan \sqrt{3} = \sqrt{4}$ , we need two $\sec \arctan$ to bring 1 to $\sqrt{3}$ ; this, together with the one mentioned earlier, makes three $\sec \arctan$ to bring 1 to 2.

Ivan Koswara - 5 years, 3 months ago

@Ivan Koswara – Yeah, I probably used the wrong choice of words by saying "repeating". For my case, I add $(n-1)$ number of $\sec { \arctan }$ to $\sec { \arctan { x } }$ while yours is including the original $\sec { \arctan { x } }$ .

I misunderstood your saying of bringing $\sqrt { 1 }$ to $\sqrt { { 2016 }^{ 2 } }$ . I thought you're referring both the numbers to $x$ . I was actually saying that $x$ has to be increased from $\sqrt { 1 }$ to $\sqrt { { 2016 }^{ 2 }-1 }$ to get $2016$ .

Alright, everything is sorted out now.

Kenneth Choo - 5 years, 3 months ago

this solution doesn't meet all conditions as it specifically says that powers other than square roots aren't allowed because they use digits.

Jess Nudalo - 5 years, 3 months ago

There is no power involved. The only operations used are $\sec$ and $\arctan$ . I proved that $\sec \arctan \sqrt{x} = \sqrt{x+1}$ , but that doesn't mean I'm using $\sqrt{\cdot}$ 's and $+1$ 's.

Ivan Koswara - 5 years, 3 months ago

Epic solution, you're a genius.

Christopher Lim - 5 years, 3 months ago

Rishik Jain
Feb 18, 2016

@Calvin Lin The answer to these types of questions is always yes unless you can prove beyond reasonable doubt that the statement is not possible which is sometimes not even possible. So the questions should be on how to make this true rather than asking if it is true or false.

That is the point; I'm mocking all those similar problems because the answer is always yes. Yes, I did say "always", because without any limitation on the operations you can always define your own function as in the first line of my solution.

Ivan Koswara - 5 years, 3 months ago

David Serero
Feb 17, 2016

((2+0+1)!-6)! = 1 is that correct?

But you can only add operators to the left hand side (the 1).

Deejesh Subramanian - 5 years, 3 months ago

Yes it is correct

Divyansh T - 4 years, 10 months ago

that's cheating.

Luke Gabriel Balgan - 2 years, 6 months ago

Farman Saifi
Feb 18, 2016

1 != 2016

1 not equal to 2016

You may only modify the left hand side of the equation. $!=$ doesn't exist in math, and if you use $\neq$ that modifies something other than the left hand side.

Ivan Koswara - 5 years, 3 months ago

r u a beginner??????? != does this not exist in maths????? check your knowleadge kid

Farman Saifi - 5 years, 3 months ago

$!=$ doesn't exist in math; inequality is called $\neq$ . != exists in certain programming languages, but that doesn't mean it exists in math; it's just because the modern keyboard doesn't have a symbol for $\neq$ .

Ivan Koswara - 5 years, 3 months ago

It is true that we don't use $!=$ in math, hence it doesn't exist in math. We only use $\neq$ . No need to be so rude.

Kenneth Choo - 5 years, 3 months ago

Fahim Abid
Feb 17, 2016

Use base 1 log

First, you're not allowed to use extra digits. Second, there is no such thing as base-1 log.

Ivan Koswara - 5 years, 3 months ago

Reidel Nabut
Feb 17, 2016

1 = 2^(0*16)

Only the left-hand side of the equation.

Kenneth Choo - 5 years, 3 months ago

Takis Psaltis
Feb 17, 2016

(2^0)*(1^6)

Only the left-hand side of the equation.

Kenneth Choo - 5 years, 3 months ago

I'm Tired Of Those "1 1 1 1 = 5" Problems

7 solutions

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