A Binary Operation

$a \otimes b=(a-4)(b-4)+4$ . Let $\sqrt{1}\otimes \sqrt{2} \ldots \otimes \sqrt{15}=x$ . Now $x \otimes \sqrt{16}=(x-4)(4-4)+4=4$ . Then for the remaining cases ( $\sqrt{17}, \sqrt{18}, \sqrt{19}, \sqrt{20}$ ), the value of the expression is $4$ , as $x \otimes \sqrt{16}=(x-4)(4-4)+4=4$ . So the value of the expression is $4$ .

Note that $a \otimes b = (ab - 4a - 4b + 16) + 4 = (a - 4)(b - 4) + 4$ .

So, $a \otimes b \otimes c = ((a - 4)(b - 4) + 4 - 4)(c - 4) + 4 = (a - 4)(b - 4)(c - 4) + 4$ .

Repeating the above, we get

$\sqrt{1} \otimes \sqrt{2} \otimes \sqrt{3} ... \otimes \sqrt{20}$

$= (\sqrt{1} - 4)(\sqrt{2} - 4)(\sqrt{3} - 4)...(\sqrt{20} - 4) + 4 = \boxed{4}$ , since $\sqrt{16} - 4 = 0$ is in the product.

Let $\begin{aligned} c &= a \otimes b = ab - 4a - 4b + 20 \\ &= (a-4)(b-4) + 4, \end{aligned}$ so that if $a' = a-4, b' = b-4, c' = c-4$ , we obtain $a \otimes b - 4 = c' = a'b'$ . It immediately follows that $\sqrt{1} \otimes \sqrt{2} - 4= (\sqrt{1} - 4)(\sqrt{2} - 4),$ and indeed, $\sqrt{1} \otimes \sqrt{2} \otimes \cdots \otimes \sqrt{20} - 4 = (\sqrt{1} - 4)(\sqrt{2} - 4) \cdots (\sqrt{20} - 4).$ But the right-hand side product contains the term $\sqrt{16} - 4 = 0$ , so the entire right-hand side is zero, and the desired value is $\boxed{4}$ .

Because we have an expression with a term $ab$ in it, it may be useful to factor it. This is simple to do: $ab - 4a - 4b + 20 = ab - 4a - 4b + 16 + 4 = (a - 4)(b - 4) + 4$ . When we re-write the operation as $a \otimes b = (a - 4)(b - 4) + 4$ , something nice becomes apparent: when we apply the operation repeatedly, as in the problem, the $+ 4$ at the end will cancel out with the $- 4$ in the $a - 4$ .

It becomes clear that $a \otimes b \otimes c = (((a - 4)(b - 4) + 4) - 4)(c - 4) + 4 = (a - 4)(b - 4)(c - 4) + 4$ and so, in general, $a \otimes b \otimes c \otimes ... \otimes z = (a - 4)(b - 4)(c - 4)...(z - 4) + 4$ . Applying this to the problem gives us that $\sqrt{1} \otimes \sqrt{2} \otimes ... \otimes \sqrt{20} =$ $(\sqrt{1} - 4)(\sqrt{2} - 4) ... (\sqrt{16} - 4) ... (\sqrt{20} - 4) + 4 = 0 + 4 = \fbox{4}$ .

This is a really nice problem:

We define some operation: $a\otimes b=ab-4a-4b+20$ There is a very suggestive factoring present, and that is: $a\otimes b=(a-4)(b-4)+4$ Which is fantastic.

To see why, let's apply the operation on another number: $a\otimes b\otimes c=([(a-4)(b-4)+4]-4)(c-4)+4=\\=(a-4)(b-4)(c-4)+4$ So, if we apply this an arbitrary number of times, say $k$ times, then: $a_1\otimes a_2 \otimes\cdots \otimes a_k=(a_1-4)(a_2-4)\cdots(a_k-4)+4$ This is interesting, though. Because, if any of the terms, $a_i=4$ , then the whole expression evaluates to 4, since all of the left-hand-side terms of the plus sign are multiplied by zero.

We check if this happens in our operation at one point, and, of course $\sqrt{16}=4$ , so, we have: $\sqrt{1}\otimes \sqrt{2} \otimes\cdots\otimes\sqrt{16} \otimes\cdots\otimes \sqrt{20}=\\(\sqrt{1}-4)(\sqrt{2}-4)\cdots(\sqrt{16}-4)\cdots(\sqrt{20}-4)+4$ This is, then: $(\sqrt{1}-4)(\sqrt{2}-4)\cdots(0)\cdots(\sqrt{20}-4)+4=4$ Then the whole expression evaluates simply to: $\sqrt{1}\otimes \sqrt{2} \otimes\cdots \otimes \sqrt{20}=4$ And we are done.

The operation is commutative since interchanging a and b does not alter the value of the product. Note that ab - 4a - 4b + 20 is (a - 4)(b-4) + 4 which equals 4 when a or b is sqrt 16 (= 4). So we may write the required value as the product of a = sqrt 16 and b = the rest of the product . Hence required value is 4.

We note that

$a \otimes b=ab-4a-4b+20=(a-4)(b-4)+4$ .

Hence, When either $a$ or $b$ is $4$ , the value of binary operation $a \otimes b$ would be $4$ . Now take

$a=\sqrt{1}\otimes \sqrt{2}\otimes ...\otimes \sqrt{15}$ and

$b=\sqrt{16}=4$ . Then clearly we would have,

$a=\sqrt{1}\otimes \sqrt{2}\otimes ...\otimes \sqrt{15}\otimes \sqrt{16}=4$

After this, it does not matter if we operate to right by $\sqrt{17}$ because in this case, first term is $4$ and hence resulting value would again be $4$ . Hence for all next entries, we will again get $4$ and hence the final answer is $\boxed{4}$ .

Let $x_i = \sqrt{1}\otimes\sqrt{2}\otimes\ldots\otimes\sqrt{i}$ . We wish to find $x_{20}$ . Note that $a\otimes b = (a-4)(b-4) + 4$ , so $a\otimes b - 4 = (a-4)(b-4)$ . This means that $x_{20} - 4 = (x_{19} - 4)(\sqrt{20} - 4)$ . Similarly, $x_{19} - 4 = (x_{18} - 4)(\sqrt{19} - 4)$ , so $x_{20} - 4 = (\sqrt{1} - 4)(\sqrt{2} - 4)\cdots(\sqrt{20} - 4)$ . But this is just $0$ , since the term $(\sqrt{16} - 4)$ is in the product! Thus, $x_{20} = \boxed{4}$ .

Rewrite the definition of the binary operation as follows.

$a \otimes b = ab-4a-4b+20$

$a \otimes b = ab-4a-4b+16+4$

$a \otimes b = a(b-4)-4(b-4)+4$

$a \otimes b = (a-4)(b-4)+4$

Note that as soon as either a = 4 or b = 4, the binary operation is equal to 4. Thus, $(\sqrt{1} \otimes \sqrt{2} \otimes \cdots \otimes \sqrt{15}) \otimes \sqrt{16} = 4$ and $4 \otimes \sqrt{n} = 4$ for all n > 16.

Hey! This question was from the Brilliant Live Challenge for about a year ago!

Anyways, from the binary operation, we see that, $a\bigotimes b=ab-4a-4b+20=ab-4a-4b-16+4=(a-4)(b-4)+4$

Now if we tried applying it to 3 terms we will get;

$a\bigotimes b \bigotimes c= ((a-4)(b-4)+4)\bigotimes c$ $= ((a-4)(b-4)+4)c-4((a-4)(b-4)+4)-4c+16+4$ $=([(a-4)(b-4)+4]-4)(c-4)+4=(a-4)(b-4)(c-4)+4$

So from we we claim that; $a_1 \bigotimes a_2 \bigotimes ... \bigotimes a_{n-1} \bigotimes a_n = (a_1-4)(a_2 -4)...(a_{n-1}-4)(a_n-4)+4$

We shall prove this by induction;

Assume $a_1 \bigotimes a_2 \bigotimes ... \bigotimes a_{n-1} \bigotimes a_n = (a_1-4)(a_2 -4)...(a_{n-1}-4)(a_n-4)+4$ is true

$a_1 \bigotimes a_2 \bigotimes ... \bigotimes a_{n-1} \bigotimes a_n \bigotimes a_{n+1} = ((a_1-4)(a_2 -4)...(a_{n-1}-4)(a_n-4)+4)(a_{n+1})-4(((a_1-4)(a_2 -4)...(a_{n-1}-4)(a_n-4)+4))-4(a_{n+1}) +20$

$=(([(a_1-4)(a_2 -4)...(a_{n-1}-4)(a_n-4)+4]-4)(a_{n+1}-4))+4$

$=(a_1-4)(a_2 -4)...(a_{n-1}-4)(a_n-4)(a_{n+1}-4)+4$

So the claim is true by mathematical induction.

Now we go to the question where we need to calculate:

$\sqrt{1} \bigotimes \sqrt{2} \bigotimes ... \bigotimes \sqrt{19} \bigotimes \sqrt{20}$

Hence;

$\sqrt{1} \bigotimes \sqrt{2} \bigotimes ... \bigotimes \sqrt{19} \bigotimes \sqrt{20}=$

$=(\sqrt{1}-4)(\sqrt{2}-4)...(\sqrt{16}-4)...(\sqrt{19}-4)(\sqrt{20}-4)+4$

$=(\sqrt{1}-4)(\sqrt{2}-4)...(4-4)...(\sqrt{19}-4)(\sqrt{20}-4)+4$

$=(\sqrt{1}-4)(\sqrt{2}-4)...(0)...(\sqrt{19}-4)(\sqrt{20}-4)+4$

$=0+4=\boxed{4}$

(a * b)=20 - 4a - 4b + ab = 4 + (4 - a)(4 - b)

therefore,(a * b) * c = 4 - (4 - a)(4 - b)(4 - c)

proceeding so we have , $a_{1}$ * $a_{2}$ * $a_{3}$ ........... $a_{n}$ = 4+ $(-1)^n$ (4 - $a_{1}$ )(4 - $a_{2}$ ).........(4 - $a_{n}$ )

=> $\sqrt{1}$ * $\sqrt{2}$ ......... . $\sqrt{20}$ = 4 + (4 - $\sqrt{1}$ )(4 - $\sqrt{2}$ ).......(4 - $\sqrt{20}$ )

since, 4 - $\sqrt{16}$ = 0

the answer is, 4+0 = $\boxed{4}$

Let us consider a number n . n*4=4n-4n-16+20=4. Therefore as the operation crosses 4 i.e $\sqrt{16}$ the result becomes 4 irrespective of the other operand . Hence final answer becomes 4

In advance, I apologise for any mistakes in the format. After the latest update, my broswer doesn't display mathmode anymore.

With all these kind of special operations, one starts by calculating some of the first terms. Very quickly you realise that the amount of work increases rapidly, so this is not a good idea.

Secondly one might try to see if you can rearrange terms to retry your calculations. Expanding $a\otimes b \otimes c$ and $a \otimes c \otimes b$ you realise this is not going to work either.

If we are to find a clever solution, it must come from the definition of the operation. Seeing a term $ab$ , a term $-4a$ and a term $-4b$ should spark the idea to try $4-a)(4-b)$ .

Expanding gives us a nice formula: $a \otimes b = (4-a)(4-b)+4$ .

The $+4$ might cause one to think this is leading nowhere, but this is where one should stick to the plan. This MUST be the right idea.

Indeed:

$\sqrt{1} \otimes \sqrt{2} \otimes \ldots \otimes \sqrt{20} =$

$(4-\sqrt{1} \otimes \sqrt{2} \otimes \ldots \otimes \sqrt{19})(4-sqrt{20})+4 =$

$(4-(4-\sqrt{1} \otimes \sqrt{2} \otimes \ldots \otimes \sqrt{18})(4- \sqrt{19})+4)(4-sqrt{20})+4 =$

$\ldots$

$(4-sqrt{1})(4-sqrt{2})*\ldots*(4-sqrt{20})+4$ .

Now all that is left is for the realisation to hit you that $4-sqrt{16}$ is one of the terms in the product. So the entire product results in $0$ .

We are left with $0+4=4$ .

The operation is commutative since interchanging a and b does not alter the value of the product. Note that ab - 4a - 4b + 20 is (a - 4)(b-4) + 4 which equals 4 when a or b is sqrt 16 (= 4). So we may write the required value as the product of a = sqrt 16 and b = the rest of the product . Hence required value is 4.