$(a_1, a_2, \cdots, a_n) \xrightarrow[]{{o(i, x)}^m \times {o(j, y)}^l} (b_1, b_2, \cdots, b_n)$ means (i) $n$ boxes $B_1, B_2, \cdots, B_n$ contains $a_1, a_2, \cdots, a_n$ coins before apply operations (ii) apply operation Type $i$) $m$ times at non-empty box $B_x$ and apply operation Type $j$) $l$ times at non-empty box $B_y$ to $n$ boxes $B_1, B_2, \cdots, B_n$ (iii) after apply above operations $n$ boxes $B_1, B_2, \cdots, B_n$ contains $b_1, b_2, \cdots, b_n$ coins. Type 3) $(n, 0, 0) \xrightarrow[]{o(1, 1)^2} (n-2, 2^2, 0) \xrightarrow[]{o(1, 2)^{2^2}} (n-2, 0, 2^3) \xrightarrow[]{o(2, 1)} (n-3, 2^3, 0)$ $\rightarrow \cdots \rightarrow (0, 2^n, 0)$ Type 4) Let $a_0 = 2^n$, $a_{k+1} = 2^{a_k}$ $(m, n, 0, 0) \xrightarrow[]{o(3, 2)} (m, 0, a_0, 0) \xrightarrow[]{o(2, 1)} (m-1, a_0, 0, 0)$ $\xrightarrow[]{o(3, 2)} (m-1, 0, a_1, 0) \xrightarrow[]{o(2, 1)} (m-2, a_1, 0, 0) \rightarrow \cdots \rightarrow (0, a_{m-1}, 0, 0)$ Let $b_0 = 2^4$, $b_{k+1} = 2^{b_k}$ $(1, 1, 1, 1, 1, 1) \xrightarrow[]{o(1, 1)} (0, 3, 1, 1, 1, 1)$ $\xrightarrow[]{o(1, 2) \times o(1, 3)^2 \times o(1, 4)^4 \times o(1, 5)^9} (0, 2, 1, 1, 0, 19) \xrightarrow[]{o(2, 4) \times o(2, 3) \times o(2, 2)} (0, 1, 19, 0, 0, 0)$ $\xrightarrow[]{o(1, 3)^2} (0, 1, 17, 4, 0, 0) \xrightarrow[]{o(4, 3)} (0, 1, 0, b_{16}, 0, 0)$ $\xrightarrow[]{o(2, 2)} (0, 0, b_{16}, 0, 0, 0) \xrightarrow[]{o(1, 3)^2} (0, 0, b_{16}-2, 4, 0, 0)$ $\xrightarrow[]{o(4, 3)} (0, 0, 0, b_{b_{16}-3}, 0, 0) \xrightarrow[]{o(2, 4)^\alpha} (0, 0, 0, \beta, 0, 0) \xrightarrow[]{o(1, 4)^\beta \times o(1, 5)^{2\beta}} (0, 0, 0, 0, 0, \gamma)$ ($\gamma = 2010^{2010^{2010}}, \beta = \gamma \div 4, \alpha = b_{b_{16}-3} - \beta$) $2010^{2010^{2010}} < {(2^{16})}^{{(2^{16})}^{(2^{16})}} = {(2^{16})}^{(2^{2^{20}})} = 2^{2^{(4+2^{20})}} < 2^{2^{2^{21}}} < 2^{2^{2^{b_1}}} < b_4$ so $\alpha$ is positive

This problem is definitely not hard (or am I misunderstand something?). Let $2010^{{2010}^{2010}}=2^5.k$ (k even). First,we will prove that we can transfer from $(1,1,1,1,1,1)$ to $(l,0,0,1,1,1)$,$l \geq 3$.(1) Apply this kind of transfer: $(1,1,1,1,1,1) \rightarrow (2,0,0,1,1,1) \rightarrow (1,2,0,1,1,1,) \rightarrow (1,1,2,1,1,1,) \rightarrow (3,0,0,1,1,1)$ If we have gotten $(x,0,0,1,1,1)$,we will get $(x+1,0,0,1,1,1)$ by this method: $(x,0,0,1,1,1) \rightarrow (x-1,2,0,1,1,1) \rightarrow (x-1,1,2,1,1,1) \rightarrow (x+1,0,0,1,1,1)$ So by using induction we had proved (1). After we get: $(k-2,0,0,1,1,1)$ we will transfer that configuration continually by this method: $(k-2,0,0,1,1,1) \rightarrow (k-3,2,0,1,1,1) \rightarrow (k-3,1,2,1,1,1) \rightarrow (k-3,1,3,0,0,1) \rightarrow (k,0,0,0,0,1)$ Apply k times "type 1" to box 1 we get: $(0,2k,0,0,0,1)$. Similarly,apply 2k times,4k times,8k times,16k times "type 1" to box 2,3,4,5 , respectively we get: $(0,0,0,0,0,32k)$ and that is the final configuration.

k.l.l4ever wrote: $(1,1,1,1,1,1) \rightarrow (2,0,0,1,1,1)$ How does that work?

It looks bit weird, but however. Answer: no. Lemma. If we have $k\geq 2$ boxes instead of $6$ and totally $s$ coins in them, then we may make at most $2^{k^2}s^{k-1}$ operations. Induction on $k$. Base $k=2$ is clear. Assume that we already know this for $k$ boxes and are proving for $k+1$ boxes. Consider the first box, let it contain $a$ coins. Then we make at most $a$ operations using the first box, and between any two such operations (also "before the first" and "after the last") we make at most $2^{(k-1)^2}(2s)^{k-1}$ operations by induction propose. So, totally we make at most $a+(a+1) 2^{(k-1)^2}(2s)^{k-1}$ operations. If $a=s$, then we make no operations before the first operation with the first box, hence we make at most $s2^{(k-1)^2}(2s)^{k-1}+s\le 2^{k^2} s^{k}$ operations at all. It follows from lemma that we make not more then $2^{36}6^5$ operations, so the total amount of coins may not be so large.

KuMing, perhaps I've misunderstood something, but in the last step don't you do $o(1, 4)^{\beta} \times o(1, 5)^{2\beta}$? Even though I like KuMing's notation, I don't understand what are his/her motivations, so I am going to show my solution (ok, maybe it won't be easy to feel any motivation, because this problem is meaningless, but I'll try...) First, if we have 5 boxes, trying to get big numbers, we can do this: (obs.: I will use constantly that operation 2 allows $(... x, y ...) \rightarrow (... 0, y + 2x, ...)$) $(1, 1, 1, 1, 1) \xrightarrow[]{o(1, 1)} (0, 3, 1, 1, 1) \xrightarrow[]{o(1, 2)} (0, 2, 3, 1, 1) \xrightarrow[]{o(1,4)} (0, 2, 3, 0, 3) \xrightarrow[]{o(2, 3)} (0, 2, 2, 3, 0) \rightarrow (0, 2, 2, 0, 6) \xrightarrow[]{o(2, 3)} (0, 2, 1, 6, 0) \rightarrow (0, 2, 1, 0, 12) \xrightarrow[]{o(1, 3)} (0, 2, 0, 24, 0) \xrightarrow[]{o(2, 2)} (0, 1, 24, 0, 0) \rightarrow (0, 1, 23, 2, 0) \rightarrow (0, 1, 23, 0, 4) \xrightarrow[]{o(2, 3)} (0, 1, 22, 4, 0) \rightarrow (0, 1, 22, 0, 8) \xrightarrow[]{o(2, 3)} (0, 1, 21, 8, 0) \rightarrow ... \rightarrow (0, 1, 1, 2^{23}, 0) \xrightarrow[]{o(2, 2)} (0, 0, 2^{23}, 1, 0)$ We turned the end $(23, 2, 0)$ into $(1, 2^{23}, 0)$, and we could have done one step more to end with $(0, 2^{24}, 0)$. Similarly we can turn the end above into $(1, 2^{2^{23} - 1}, 0)$. More generally, if from $(1, n, 0)$ we make $(n, 1, 0)$, we can make also $(1, 2^{n - 1}, 0)$. And to be able to do this, all we need is to have some coin in the fourth right position (i.e., fourth position counting backwards), so that we can apply operation 2. Now, in our problem we have $(1, 1, 1, 1, 1, 1) \rightarrow (1, 0, 3, 1, 1, 1) \rightarrow (0, 3, 0, 1, 1, 1) \rightarrow (0, 0, 6, 1, 1, 1)$. And these are our fourth position coins. $(6, 1, 1, 1) \rightarrow (5, 3, 1, 1) \rightarrow (5, 3, 0, 3) \xrightarrow[]{o(2, 2)} (5, 2, 3, 0) \rightarrow (5, 2, 0, 6) \rightarrow (5, 1, 0, 12) \xrightarrow[]{o(2, 2)} (5, 0, 24, 0) \xrightarrow[]{o(2, 1)} (4, 24, 0, 0) \rightarrow ... \rightarrow (4, 1, 2^{23}, 0) \xrightarrow[]{o(2, 1)} (3, 2^{23}, 1, 0) \rightarrow ... \rightarrow (3, 1, 2^{2^{23} - 1}) \xrightarrow[]{o(2, 1)} (2, 2^{2^{23} - 1}, 1, 0) \rightarrow ... \rightarrow (0, 2^{2^{2^{2^{23} - 1} - 1} - 1}, 1, 0)$ Finally, noting that 8 divides both $2010^{2010^{2010}}$ and the huge power of 2 above, which we call $4k$, we do $(0, 4k, 1, 0) \rightarrow (0, 4k, 0, 2) \rightarrow (0, 4k - 1, 2, 0) \rightarrow (0, 4k - 1, 0, 4) \rightarrow (0, 4k - 2, 4, 0) \rightarrow (0, 4k - 2, 0, 8)$. Of course we can make $o(2, 2)$ two times, or an even number of times, which is nothing but subtracting 2 of the second position of our array. We make this until the number $\frac{2010^{2010^{2010}}}{4} - 1$ appears, ending as did KuMing. We just need to verify that $2^{2^{2^{2^{23} - 1} - 1} - 1} \ge 2010^{2010^{2010}}$ or $2^{2^{2^{2^{23} - 1} - 1} - 1} > 2010^{2010^{2010}}$. Take the "binary" logs. $2^{2^{2^{23} - 1} - 1} - 1 > 2010^{2010}log2010$ To show that $x - 1 > y$, it suffices to show that $x > 2y$. $2^{2^{2^{23} - 1} - 1} > 2.2010^{2010}log2010$ Take the logs again. $2^{2^{23} - 1} - 1 > 2010log2010 + loglog2010 + 1$ But $2^23 > 1000000$ and the right side is $< 3.2010log2010 < 3.2010.11 < 100000$. Fedor, I think you cannot use the induction basis once you make a movement involving the first box, because this would alter the number of coins...

feliz wrote: KuMing, perhaps I've misunderstood something, but in the last step don't you do $o(1, 4)^{\beta} \times o(1, 5)^{2\beta}$? Even though I like KuMing's notation, I don't understand what are his/her motivations, so I am going to show my solution (ok, maybe it won't be easy to feel any motivation, because this problem is meaningless, but I'll try...) You are right. It's my mistake I just try to make big number anyway

feliz wrote: Fedor, I think you cannot use the induction basis once you make a movement involving the first box, because this would alter the number of coins... Oh, indeed, how could I miss it? We get some awful estimate on this way, but on first glance that worse that the needed.

There is my solution. We know that (n 0 0)=> (0 2^n 0).(*) And our number is divisible by 4 . If T=(2010^2010^2010), then we need the number (T/4)=A. There exists B >1002, such that (1 1 1 1 1 1)=>(0 1 0 B 0 0). Then to the last we use 2 type and we have(0 0 B 0 0 0). Use 1 type =>(0 0 1000 2*(B-1000) 0 0) => we use consequently (*) and (2 type) and after finit steps we'll have (0 0 C A 0 0) there is C < 1000 and A =(T/4). And after using (A 0 0)=> (0 0 4A) we'll have (0 0 C 0 0 4A). For (c 0 0) we only use 2type and we'll have (0 0 0). I think my proof is correct. Sorry, but I've typed it on my phone. If someone has latex, please correct it. THANKS.

I am posting a solution from the wiki opened by Terence Tao http://michaelnielsen.org/polymath1/index.php?title=Imo_2010. Let $T = 2010^{2010^{2010}}$. We will be done if we can obtain the configuration $[0,0,0,T / 4,0,0]$, for then we can apply Type $1$ moves to get to $[0,0,0,0,0,T]$. To get this configuration, it is enough to get a configuration $[0,0,0,X,0,0]$ for some $X \geq T/4$, because we can then apply Type $2$ moves until the number of coins in the fourth box is reduced to $T / 4$. Finally, one can obtain such a configuration. First note that we can get $[1,1,1,1,1,1] \to [0,2,2,2,2,3] \to [0,2,1,1,8,3] \to [0,2,1,1,0,19] \to $ $\to [0,1,19,0,0,0] \to [0,1,1,36,0,0] \to [0,1,1,1,0,140] \to [0,0,140,0,0,0]$. From this we get $[0,0,139,2,0,0]$, and then by applying $139$ compound moves as that described below For $a \geq 1$ we have $[a,0,0] \to [0,2^a,0]$ via $[a,0,0] \to [a-1,2,0] \to [a-1,0,4] \to [a-2,4,0] \to [a-2,0,8] \to \ldots $ $\to [1, 0, 2^a] \to [0,2^a,0]$. For $a,b \geq 1$ we have $[a,b,0,0] \to [a-1, 2^b, 0, 0]$. This follows from the previous compound move together with a Type $2$ swap. We thus get $[0,0,0,X,0,0]$ for some $X$ much, much bigger than $T / 4$, and so we are done.

My roundabout procedure to make $[0,0,0,0,0,{2010}^{{2010}^{2010}}]$: Let $x_{i}$ be the number of coins in $B_{i}$. Then, $f([x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}])\eqref x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{4}+\frac{x_{4}}{8}+\frac{x_{5}}{16}+\frac{x_{6}}{32}$ does not change under the type 1 operations, and change by $\frac{x_{i+1}-x_{i}-4}{2^{i}}$ under the type 2 operations. From this, it is obvious that we can make the configuration we want ($[0,0,0,0,0,{2010}^{{2010}^{2010}}]$) by only using type 1 operations when we reach the moment of $f$ being $\frac{2010^{2010^{2010}}}{32}$. Also, we can make $[0,2^{n},0]$ from $[n,0,0]$. ($[n,0,0] \rightarrow [n-1,0,2^{2}] \rightarrow [n-2,2^{2},0] \rightarrow [n-2,0,2^{3}] \rightarrow [n-3,2^{3},0] \rightarrow$ $ \cdots \rightarrow [0,2^{n},0]$) $[1,1,1,1,1,1]\rightarrow[0,2,2,2,2,3]\rightarrow[0,2,2,2,0,7]\rightarrow[0,2,2,1,7,0]$ $\rightarrow[0,2,2,1,0,14]\rightarrow[0,2,2,0,14,0]\rightarrow[0,2,1,14,0,0]$ $\rightarrow [0,2,1,0,2^{14},0] \rightarrow [0,2,0,2^{14},0,0] \rightarrow [0,1,2^{14},0,0,0]$ $\rightarrow[0,1,2^{14}-1,2,0,0] \rightarrow [0,1,2^{14}-1,0,4,0]$ $\rightarrow [0,1,2^{14}-2,4,0,0] \rightarrow [0,1,2^{14}-2,0,16,0]$ $\rightarrow [0,1,2^{14}-3,16,0,0] \rightarrow [0,1,2^{14}-3,0,2^{16},0] \rightarrow [0,1,2^{14}-4,2^{16},0,0]$ $\rightarrow [0,1,2^{14}-4,0,2^{2^{16}},0] \rightarrow [0,1,2^{14}-5,2^{2^{16}},0,0]$ From the suggested procedure above, we can make numbers EXTREMELY large($16,2^{16},2^{2^{16}},\cdots$ this sequence will end up with very very large number, which has approximately $2^{14}$ twos as its bases. In fact, we can do this one more time, since $x_{2}=1$.) , and at some moment it will exceed $2010^{2010^{2010}}$. Actually, when we concern $\log_{2}(\log_{2}(2010^{2010^{2010}}))=\log_{2}(2010^{2010} \log_{2} 2010)$ $=2010 \log_{2} 2010 + \log_{2}(\log_{2} 2010)$, it's smaller than $2^{16}$ and bigger than $16$, so $2^{2^{2^{16}}}>2010^{2010^{2010}}>2^{2^{16}}$. From this inequality, though we have to consider some details, it seems that we have to stop swelling numbers. So, we should discard more chances to drastically increase numbers by just moving $x_{2}$ and $x_{3}$ to $x_{4}$. Let $\frac{2010^{2010^{2010}}}{32}$ be $f_{T}$. $ [0,1,2^{14}-5,2^{2^{16}},0,0] \rightarrow [0,0,0,2^{15}+2^{2^{16}}-6,0,0]$. Still, $f$ is smaller than $f_{T}$ since $\log_{2}(\log_{2}(2010^{2010^{2010}}))$ is still bigger than $17$ and it means that $2010^{2010^{2010}}>2^{2^{17}}>4(2^{15}+2^{2^{16}}-6)$. For any natural number $a$ smaller than $2^{15}+2^{2^{16}}-6$, we can make $[0,0,0,2^{15}+2^{2^{16}}-6-a,2^{a},0]$. Since $f$ increases as $a$ increases and $f$ exceeds $f_{T}$ when $a$ goes to $2^{15}+2^{2^{16}}-6$, there exists a natural number $N$ such that $f$ of $[0,0,0,2^{15}+2^{2^{16}}-6-N,2^{N},0]$ does not exceed $f_{T}$ while $f$ of $[0,0,0,2^{15}+2^{2^{16}}-6-N-1,2^{N+1},0]$ exceeds $f_{T}$. Let $f$ of $[0,0,0,2^{15}+2^{2^{16}}-6-N,2^{N},0]$ be $f_{N}$. Now what I want to do is, find $b$ and $c$ such that $f$ of $[0,0,0,2^{15}+2^{2^{16}}-6-N-c-1,2b,2^{N}+2c-b]$ equals $f_{T}$. The way to reach this configuration is $[0,0,0,2^{15}+2^{2^{16}}-6-N,2^{N},0] \rightarrow [0,0,0,2^{15}+2^{2^{16}}-6-N-c,2^{N}+2c-b,2b] $ $\rightarrow [0,0,0,2^{15}+2^{2^{16}}-6-N-c-1,2b,2^{N}+2c-b]$. Observe that $f$ only changes at the second step, and the degree of change is $\frac{3b-2^{N}-2c-4}{32}$. So, we have to find $b$ and $c$ satisfying $3b-2^{N}-2c-4=32(f_{T}-f_{N})$($32(f_{T}-f_{N})$ is a natural number). Note that $3 \cdot 2^{N}-2^{N}-2 \cdot 0 -4>32(f_{T}-f_{N})$ because of the definition of $N$. First, we can find $c$ among $ 0,1,2 $ satisfying $-2^{N}-2c-4 \equiv 32(f_{T}-f_{N})$. Then, let $b$ be $\frac{32(f_{T}-f_{N})+2^{N}+2c+4}{3}$. Then, $f$ of $[0,0,0,2^{15}+2^{2^{16}}-6-N-c-1,2b,2^{N}+2c-b]$ equals $f_{T}$, so by simply applying type 1 operations, we can finally make $[0,0,0,0,0,2010^{2010^{2010}}$. Minor calculations are omitted above. I don't think that this is a good solution, but it's meaningful since it suggests how to make extremely large numbers. If we did not discard chances before, we will then happen to make $4 \cdot 2^{2^{\cdots^{2^{2}}}}$, which has $2^{2^{\cdots^{2^{2}}}}$, a number with $2^{14}$ twos, twos.

Here is my solution, Let $ 2010^{2010^{2010}} = 2^3 * m $ (It is easy to see that it is divisible by $ 2^3 $). Now a configuration like $ (0, 0, x, 0, 0, 0) $ is equivalent to $ 2^3 * x $ in the sense that by moving the coins we can generate any even number less than that value in the last box. So to achieve the above we need to move $ m $ number of coins out of the $ x $ available. After moving them we are left with $ (0, 0, x-m, 0, 0, 2010^{2010^{2010}}) $ Now we can do Type 2 moves to reduce $ (x-m) $ to 0. It is not hard to see that we can get a $x$ as huge as $2^{2^{..^2}}$ with as many terms as $ 2^{14} $ which is clearly bigger than the given value.

A minimalistic approach. We have $2010^{2010} < (2^{2^4})^{2^{11}} = 2^{2^{15}}$, so $T = 2010^{2010^{2010}} < (2^{2^4})^ {2^{2^{15}}} = 2^{2^{4+2^{15}}} < 2^{2^{2^{16}}} = 2^{2^{2^{2^{2^2}}}}$, a tower of powers of six two's. First, obtain the configuration $[1,1,1,1,1,1] \stackrel{\textrm{T}1}{\to} [1,1,1,1,0,3] \stackrel{\textrm{T}2}{\to} [1,1,1,0,3,0] \stackrel{\textrm{T}2}{\to} [1,1,0,3,0,0] \stackrel{\textrm{T}2}{\to} $ $[1,0,3,0,0,0] \stackrel{\textrm{T}1}{\to} [0,2,3,0,0,0] \stackrel{\textrm{T}1}{\to} [0,0,7,0,0,0]$. For $a,b \geq 0$ we have $[a,2^b,0] \stackrel{\textrm{T}1}{\to} [a,0,2^{b+1}] \stackrel{\textrm{T}2}{\to} [a-1,2^{b+1},0]$, so by induction $[a,2^b,0] \stackrel{\textrm{T}3}{\to} [0,2^{b+a},0]$. Therefore, for $a,b \geq 1$ we have $[a,b,0,0] \stackrel{\textrm{T}1}{\to} [a,b-1,2^1,0] \stackrel{\textrm{T}3}{\to} [a,0,2^b,0] \stackrel{\textrm{T}2}{\to} [a-1, 2^b, 0, 0]$, so by induction, for $a \geq 1$ we have $[a,0,0,0] \stackrel{\textrm{T}1}{\to} [a-1,2,0,0] \stackrel{\textrm{T}4}{\to} [0,2^{2^{2^{\cdots 2^{2}}}},0,0]$, a tower of powers with $a$ two's. It follows we can get $[0,0,7,0,0,0] \stackrel{\textrm{T}1}{\to} [0,0,6,2,0,0] \stackrel{\textrm{T}4}{\to} [0,0,0,2^{2^{2^{2^{2^{2^2}}}}},0,0]$, where the tower of powers $X$ of seven two's is larger than $T/4$. Now get the configuration $[0,0,0,X,0,0] \stackrel{\textrm{T}2}{\to} [0,0,0,T/4,0,0] \stackrel{\textrm{T}1}{\to} [0,0,0,0,0,T]$.

As a physical aside, the number of atoms in the observable universe is estimated to be less than $10^{81}$, so those coins will not fit in myriads over myriads of possible universes! (not that I care). I always in fact took the "friendly" mathematical statements (cities connected by roads, rather than considering a graph, etc.) as ridiculous and patronizing. Could we not have said: given $6$ non-negative integers, initially of unit value, we may decrease one by $1$ and increase the next one by $2$, etc.?

mavropnevma wrote: A minimalistic approach. We have $2010^{2010} < (2^{2^4})^{2^{11}} = 2^{2^{15}}$, so $T = 2010^{2010^{2010}} < (2^{2^4})^ {2^{2^{15}}} = 2^{2^{4+2^{15}}} < 2^{2^{2^{16}}} = 2^{2^{2^{2^{2^2}}}}$, a tower of powers of six two's. First, obtain the configuration $[1,1,1,1,1,1] \stackrel{\textrm{T}1}{\to} [1,1,1,1,0,3] \stackrel{\textrm{T}2}{\to} [1,1,1,0,3,0] \stackrel{\textrm{T}2}{\to} [1,1,0,3,0,0] \stackrel{\textrm{T}2}{\to} $ $[1,0,3,0,0,0] \stackrel{\textrm{T}1}{\to} [0,2,3,0,0,0] \stackrel{\textrm{T}1}{\to} [0,0,7,0,0,0]$. For $a,b \geq 0$ we have $[a,2^b,0] \stackrel{\textrm{T}1}{\to} [a,0,2^{b+1}] \stackrel{\textrm{T}2}{\to} [a-1,2^{b+1},0]$, so by induction $[a,2^b,0] \stackrel{\textrm{T}3}{\to} [0,2^{b+a},0]$. Therefore, for $a,b \geq 1$ we have $[a,b,0,0] \stackrel{\textrm{T}1}{\to} [a,b-1,2^1,0] \stackrel{\textrm{T}3}{\to} [a,0,2^b,0] \stackrel{\textrm{T}2}{\to} [a-1, 2^b, 0, 0]$, so by induction, for $a \geq 1$ we have $[a,0,0,0] \stackrel{\textrm{T}1}{\to} [a-1,2,0,0] \stackrel{\textrm{T}4}{\to} [0,2^{2^{2^{\cdots 2^{2}}}},0,0]$, a tower of powers with $a$ two's. It follows we can get $[0,0,7,0,0,0] \stackrel{\textrm{T}1}{\to} [0,0,6,2,0,0] \stackrel{\textrm{T}4}{\to} [0,0,0,2^{2^{2^{2^{2^{2^2}}}}},0,0]$, where the tower of powers $X$ of seven two's is larger than $T/4$. Now get the configuration $[0,0,0,X,0,0] \stackrel{\textrm{T}2}{\to} [0,0,0,T/4,0,0] \stackrel{\textrm{T}1}{\to} [0,0,0,0,0,T]$. Nice and clear solution! I used configuration $[0,0,7]$ too but we can get $[0,0,7,0,0,7]$, from where by similar operations get even more: $[0,0,0,2^{2^{2^{2^{2^8}}}},1,0]$.

Just asking, what is the largest value we can go? I manage to get $2^{2^{2...}}$ with the number of 2's is $2^{2^{2...}}$ with the number of 2's is $2^{14}$. \[(1,1,1,1,1,1)\to(2,2,2,0,7)\to (2,2,1,7,0)\to (2,2,0,14,0)\to (2,1,14,0,0)\to (2,1,0,2^{14},0)\to (1,2^{14},0,0,0)\to (1,0,f(2^{14}),0,0)\to (f(2^{14}),0,0,0)\to (f(f(2^{14})),0,0)\] where $f(n)=2^{2^{2...}}$ with $n$ number of 2's. Note: we can get $(n,0,0)\to (0,2^n,0)$ and $(n,0,0,0)\to (0,f(n),0,0)$.

Anyone knows the marking scheme of this problems? How many pts would getting the (k, 0, 0) --> (0, 2^k, 0) get?

This problem is very tough and requires multiple insights...

How would one come up with a solution to this problem?

This problem is hard because it's not obvious initially if the answer is Yes or No. And you could waste a lot of time trying to prove the answer is No. It's not even immediatley obvious that you can get really large numbers with the given operations. So the first step is to see if you can even create really large numbers like $2010^{2010}$ with the given operations. Type 1 moves increase the number of coins by 1, so we want to use this operation repeatedly to create larger numbers. Its quite easy to see how to get $2k$ from $k$. Its also not hard to realize that you can get $2^k$ from $k$ as Konisgberg suggested above, and this observation is an important first step. But you need to make much larger numbers than $2^k$ starting from $k$. The next observation which is somewhat hard to come across until you play around a lot, is that you can build towers of $2$'s like $2^{2^k}$ and $2^{2^{2^k}}$ and so on. Now the problem is essentially solved because you now know how to create numbers larger (much larger actually) than $2010^{2010}$. The problem now is that we need EXACTLY $2010^{2010}$ in the 6th box. Knowing that we can get numbers larger than that, the task now is to find how to get rid of coins in a systematic manner. Type 2 moves allow us to decrease the total number of coins by 1 whilst swapping the contents of two other boxes. Thus if we have two consecutive empty boxes we can systematically decrease the total number of coins and nothing else is affected. So you want a very large number of box $B_4$ (in the form of a tower of $2$'s) and boxes $B_5$ and $B_5$ empty. That way you can decrease the number of coins in $B_4$ until you get $2010^{2010}/4$ coins and then you use the $k$ to $2k$ procedure twice to get exactly $2010^{2010}$ coins in box 6.

The idea of trying to generate large numbers by shifting back and forth, becomes clear after some experimentation, and thought. So it points to the fact that it is possible, however in my opinion, finding the exact algorithm to generate the end state, is lengthy and uses a lot of trial and error which is downright painful and against the spirit of pleasurable problem solving.

Yes, it is possible. We claim that it is is possible to reach $(0)(0)(34)(0)(0)(0)$ by $(1)(1)(1)(1)(1)(1) \to (0)(3)(0)(3)(0)(3) \to (0)(3)(0)(2)(3)(0) \to (0)(2)(2)(2)(2)(2) \to (0)(1)(4)(2)(2)(2) \to (0)(1)(1)(8)(2)(2) \to (0)(1)(1)(1)(16)(2) \to (0)(1)(1)(1)(0)(34) \to (0)(0)(34)(0)(0)(0)$. Now, consider a position $(0)(0)(2n)(0)(0)(0)$. We can reach $$(0)(0)(n)(n)(2n)(0) \to (0)(0)(n)(n)(0)(4n) \to (0)(0)(n)(n-1)(4n)(0) \to (0)(0)(n)(n-1)(0)(8n) \to$$$$ \to ... \to (0)(0)(n)(1)(0)(2^{n+1}n) \to (0)(0)(n-1)(2^{n+1}n)(0)(0) \to (0)(0)(n-1)(2^nn)(2^{n+1}n)(0) \to (0)(0)(n-1)(2^nn)(0)(2^{n+2}n)$$$\implies$ repeating $(2)$ on the fourth box followed by $(1)$ on the fifth box, until the fourth box contains $1$ coin, we will have $(0)(0)(n-1)(1)(0)(2^{n+2^nn+1}n \to (0)(0)(n-2)(2^{n+2^nn+1}n)(0)(0)$. Doing the same thing, we can reach $(0)(0)(n-3)(2^{n+2^nn+1+2^{n+2^nn}}n(0)(0)$. In general, from $(0)(0)(n-k)(2^{a_k}n)(0)(0)$, we can reach $(0)(0)(n-k-1)(2^{a_k+2^{a_k-1}}(0)(0) \implies$ at each step, the coefficient of the fourth box is powered to the two, so at the end of the process, we will have $(0)(0)(0)(T)(0)(0)$, with $T \geq 2^{2^{2^{...2}}}$, where $2$ appears $17$ times, which is clearly greater than $M=2010^{2010^{2010}}$. Now, we perform $(2)$ on the fourth box until it has $\frac{M}{4}$. Then, we just make $(0)(0)(0)(\frac{M}{4})(0)(0) \to (0)(0)(0)(0)(\frac{M}{2})(0) \to (0)(0)(0)(0)(0)(M)$, as desired. $\blacksquare$

so weird If an ordered tuple is written, it means the configuration with that number of coins in the boxes in order. Lemma 1: $(a,0,0)$ can move to $(0,2^a,0)$. This can be done by using a Type 1 move on the left box, using Type 1 moves on the second box until it runs out, and performing swaps using the left box and repeat this for as long as possible. For example, $$(3,0,0)\rightarrow (2,2,0)\rightarrow(2,0,4)\rightarrow(1,4,0)\rightarrow(1,0,8)\rightarrow(0,8,0). $$ We perform the following sequence of moves. Indicate a move using Lemma 1 by (L1). $$(1,1,1,1,1,1)\rightarrow (1,1,1,1,0,3)\rightarrow(1,1,1,0,3,0)\rightarrow(1,1,0,3,0,0)\rightarrow(1,0,3,0,0,0)$$$$\rightarrow(3,0,0,0,0)\rightarrow(2,2,0,0,0)\rightarrow(2,1,2,0,0)\rightarrow(2,1,0,4,0) (L1)\rightarrow(2,0,4,0,0)$$$$\rightarrow(1,4,0,0,0)\rightarrow(1,3,2,0,0)\rightarrow(1,3,0,4,0)(L1)\rightarrow(1,2,4,0,0)\rightarrow(1,2,0,16,0)(L1)$$$$\rightarrow(1,1,16,0,0)\rightarrow(1,1,0,65536,0)(L1)\rightarrow(1,0,65536,0,0)\rightarrow(65536,0,0,0).$$From here, we apply Lemma 1 65536 times, making a Type 2 move on the left box after each time, to reach $$(2\uparrow\uparrow65536,0,0)$$(where $\uparrow\uparrow$ denotes tetration). We then "waste" $$2\uparrow\uparrow65536-\frac{2010^{2010^{2010}}}{4}$$coins by making a Type 2 move on the left box that many times, to reach $$(\frac{2010^{2010^{2010}}}{4},0,0).$$Using Type 1 moves for the rest of the moves finishes. Remark: what I did here is pretty suboptimal (as I could have went straight to (3,1,1,1,1) at the start instead of going to (3,0,0,0,0)), but I did this because i knew it would still be large enough and it makes the steps after that less complicated

Define a type $3$ operation to be doing operation $2$ on box $B_k$, then doing operation $1$ on box $B_{k+1}$, and this operation can only be done when box $B_{k+1}$ is empty and $B_{k+2}$ is not empty. This is equivalent to doubling the number of coins in box $B_{k+2}$ and removing $1$ coin from box $B_k$. First, do operation $1$ on box $B_1$ to get $0,3,1,1,1,1$ coins in the boxes. Now, do operation $1$ on box $B_3$ to get $0,3,0,3,1,1$ coins. Now, we can do operation $3$ on box $B_2$ to get $0,2,0,6,1,1$. Doing operation $2$ on box $B_2$, we get $0,1,6,0,1,1$. Now, we can do operation $3$ $4$ times on box $B_3$ to get $0,1,2,0,16,1$. Doing operation $2$ on box $B_3$, we get $0,1,1,16,0,1$. We can do operation $3$ $15$ times on box $B_4$ to get $0,1,1,1,0,2^{15}$. Now, we do operation $2$ on boxes $B_4$, $B_3$, and $B_2$ in that order to get $0,0,2^{15},0,0,0$. Doing operation $1$ on box $B_3$ and doing operation $1$ on box $B_4$ twice, we get $0,0,2^{15}-1,0,4,0$. Now, we can do operation $3$ $2^{15}-3$ times to get $0,0,2,0,2^{2^{15}-1},0$. Now, we do operation $2$ on box $B_3$ to get $0,0,1,2^{2^{15}-1},0,0$. Applying operation $1$ to $B_4$ once, then $B_5$ twice gives $0,0,1,2^{2^{15}-1}-1,0,4$. Now, doing operation $3$ $2^{2^{15}-1}-2$ times on box $B_4$ gives $0,0,1,1,0,2^{2^{2^{15}-1}}$. Finally, applying operation $2$ on $B_4$ and $B_3$ gives $0,0,0,2^{2^{2^{15}-1}},0,0$. We claim $2^{2^{2^{15}-1}}>2010^{2010^{2010}}$. We have \begin{align*} 2010^{2010^{2010}}&<(2^{11})^{(2^{11})^{2^{11}}}\\ &=2^{11\cdot2^{11\cdot2^{11}}}\\ &<2^{2^4\cdot2^{11\cdot2^{11}}}\\ &=2^{2^{11\cdot2^{11}+4}}\\ &<2^{2^{2^{15}-1}}. \end{align*} Therefore, we can apply operation $2$ on box $B_4$ until we get $0,0,0,\frac14\cdot2010^{2010^{2010}},0,0$, then apply operation $1$ to get $0,0,0,0,0,2010^{2010^{2010}}$.

First, we get rid of the preciseness condition. Let $N = 2010^{2010^{2010}}$. Claim. If we can get to $(0, 0, 0, M, 0, 0)$ where $M > \frac N4$, then we are okay. Perform the second operation with $k=4$ until $M = \frac N4$. Now perform the first operation on $j=4$ and $j=5$. $\blacksquare$ Thus, our goal is to blow up the fourth box using all means necessary. Claim. [Initiating Gravity] We can get from $(a, 0, 0)$ to $(0, 2^a, 0)$. Simply go from $(a, 0, 0) \to (a-1, 2, 0) \to (a-1, 0, 4) \to (a-2, 4, 0)$. Now we can go from $(a-2, 4, 0) \to (a-2, 0, 8) \to (a-3, 8, 0) \to (a-3, 0, 16) \to \cdots$ as long as necessary until the first box is totally emptied. $\blacksquare$ With our new powerful force, we can now collapse the core of the boxes. Notice that we can go from $(a, b, 0, 0) \to (a, 0, 2^b, 0) \to (a-1, 2^b, 0, 0)$, and repeat until we have $2^b$ as the exponent upon $a$ tetrations of $2$'s, which we will call $T(a, b)$. To set this up, we can perform $$(1, 1, 1, 1, 1, 1) \to (0, 1, 1, 1, 17, 1) \to (0, 1, 1, 1, 0, 35) \to (0, 0, 35, 0, 0, 0).$$Now the boxes are ready to go supernova, and indeed after a great gravitational collapse we can obtain the mass of $T(35, 2)$ coins within a rapidly expanding black hole that once was the fourth box. However, $T(35, 2) > 2010^{2010^{2010}}$ obviously (say, by taking $\log_2$), so by the first claim we are done. Remark. Actually, the first claim is not strictly needed anyways because box $B_6$ will have been sucked into the same singularity as the $T(35, 2)$ coins, so the coins must also lie inside $B_6$.

We use the notation $(b_1,b_2,\dots ,b_6)$ to denote the number of coins in each box. We start with $(1,1,1,1,1,1)$. We prove that we can turn $(x,y,0)$ into $(x-1,2y,0)$. Indeed, \[(x,y,0)\to_1 (x,0,2y)\to_2 (x-1,2y,0)\]We will call this operation of type 3. Thus, we can turn $(x,0,0)$ to $(0,2^x,0)$ by induction: \[(x,0,0)\to (x-1,2,0)\to (0,2^x,0)\]We will call this operation of type 4. Now, we have \[(1,1,1,1,1,1)\to_1 (1,1,0,3,0,3)\to_2 (1,1,0,2,3,0)\to_3 (1,1,0,0,12,0)\to_2 (1,0,1,12,0,0)\]\[\to_4 (1,0,1,0,2^{12},0)\to_2 (1,0,0,2^{12},0,0)\to_1 (0,2,0,2^{12},0,0)\to_2 (0,1,2^{12},0,0,0)\]\[\to_4 (0,1,0,2^{2^{12}},0,0)\to_1 (0,0,2,2^{2^{12}},0,0)\to_4 (0,0,2,0,2^{2^{2^{12}}},0)\to_2 (0,0,1,2^{2^{2^{12}}},0,0)\]\[\to_4 (0,0,1,0,2^{2^{2^{2^{12}}}},0)\to_2 (0,0,0,2^{2^{2^{2^{12}}}},0,0)\]We now note that \[2010^{2010^{2010}}<2^{10\cdot 2010^{2010}}<2^{2010^{2011}}<2^{2^{20110}}<2^{2^{2^{15}}}<2^{2^{2^{2^{12}}}}\]so \[(0,0,0,2^{2^{2^{2^{12}}}},0,0)\to_2 (0,0,0,\frac14 \left(2010^{2010^{2010}}\right),0,0)\to_1 (0,0,0,0,0,2010^{2010^{2010}})\]

I think this is actually not that bad for a 2010 p5 (exactly what i said for the china problem too lol) Let the number we need in $B_6$ be $x$. There's two main ideas here: 1. If we get $B_4$ to at least $x/4$, we're done. This is clear because we would then just apply operation 1 repeatedly to get $x/2$ in $B_5$->$x$ in $B_6.$ 2. We have two algorithms: a) $(a,0,0)->(0,2^a,0).$ This follows immediately by induction from $(a-k,2^k,0)$->$(a-k,0,2^{k+1})$->$(a-(k+1),2^{k+1},0)$ with base case $(a,0,0)$->$(a-1,2,0).$ b. Let $P_n$ denote $2^{2^{\cdots}}$ where n represents the number of 2s. Then $(n,0,0)$->$(0,P_n,0).$ This follows immediately by induction from a) and then applying it again for the next number, whence $(0,P_n,0,0)$->$(0,0,2^{P_a}=P_{n+1},0)$ with base case a). Now we just do it. $(1,1,1,1,1,1)$->$(1,1,1,1,0,3)$->$(1,1,1,0,3,0)$->$...$->$(0,3,0,0,0,0)$->$(0,0,P_3=2^{2^2}=16,0,0,0)$->$(0,0,0,P_{16},0,0).$ Now note $$2010^{2010^{2010}}<(2^{11})^{(2^{11})^{2010}}=(2^{11})^{2^{22110}}=2^{11*2^{22110}}<2^{2^{65536}}=2^{2^{2^{16}}}=2^{2^{2^{2^{2^2}}}}$$, which is clearly less than $P_{16}. \blacksquare$

The answer is yes. Claim: We can go from $(a,0,1)$ to $(0,2^{a-1}-1,1)$. Proof: $(a,0,1) \to (a-1,1,0) \to (a-1,0,2) \to (a-2,2,0) \to \cdots \to (1,0,2^{a-1}) \to (0,2^{a-1},0) \to (0,2^{a-1}-1,1)$. $\blacksquare$ Now given our starting position $(1,1,1,1,1,1)$, perform the operations $$(1,1,1,1,1,1) \to (0,3,1,1,1,1) \to (0,2,3,1,1,1) \to (0,2,0,7,1,1) \to (0,1,7,0,1,1) \to (0,1,0,14,1,1) \to (0,0,14,0,1,1) \to (0,0,13,2,1,1) \to (0,0,13,0,5,1).$$Henceforth ignore $B_1$ and $B_2$ and represent the remaining boxes with a $4$-tuple. We now repeat our claim $11$ times as follows: $$(13,0,5,1) \to (12,5,0,1) \to (12,0,15,1) \to (11,15,0,1) \to (11,0,16383,1) \to \cdots.$$To lower bound the final configuration, which will be of the form $(2,0,x,1)$ for some $x$, we can "lower bound" each operation by going from $(a,0,1)$ to $(0,\sqrt{2}^a,1)$ and prove that $x \gg 2010^{2010^{2010}}$. I am too lazy to write this part out. Now apply the claim one more time, except don't go from $(0,2^{a-1},0)$ to $(0,2^{a-1}-1,1)$, so we get some configuration $(1,0,M,0)$, and from here we can go to $(0,M,0,0)$; note that $M$ is a massive power of $2$. Henceforth ignore $B_3$ and represent the remaining boxes with a $3$-tuple. Now perform the operations $$(M,0,0) \to (M-2010^{2010^{2010}}/2,2010^{2010^{2010}}/2,0) \to (M-2010^{2010^{2010}}/2,0,2010^{2010^{2010}}),$$and note that $M-2010^{2010^{2010}}/2$ is even, hence we can "burn through" all the coins in $B_4$ by swapping $B_5$ and $B_6$ an even number of times to finish the problem. $\blacksquare$

First, we begin with the following operations: 111111 031111 031103 031030 030300 023000 007000 Key Claim: After operations, 007000 can become $0,0,0,2^{2^{2^{2^{2^{2^2}}}}},0,0$ From here, we first make 006200 and then apply the below. We prove that from $00ab00$, we can get $0, 0, a, 0, 2^b, 0$ and then $0, 0, a - 1, 2^b, 0, 0$. To prove this, we first make it $0, 0, a, b - 1, 2, 0$ and see that $00abc0$ can become $00ab02c$ and then $0, 0, a, b - 1, 2c, 0$ $\square$ Now, we see that $2010^{2010^{2010}} < 2^{11 \cdot 2^{11 \times 2010}} < 2^{2^{32,000}} < 2^{2^{2^{16}}} = 2^{2^{2^{2^{2^2}}}}$ which is clearly less than the tower of 7 2's found earlier. Therefore, from here we simply remove $\frac{2010^{2010^{2010}}}{4}$ coins from box 4 and add $2010^{2010^{2010}}$ to box 6. Now, we swap boxes 5 and 6 repeatedly to move all the remaining coins from box 4.

The answer is yes. Claim: If we have consecutive boxes with coins $(k,n,0,0)$, then we can operate on them to get $(k-1,2^n,0,0)$. Proof: \[(k,n,0,0) \to (k,n-1,0,4) \to (k,n-2,4,0) \to (k,n-3,8,0) \to \cdots \to (k,0,2^n,0)\]which concludes the proof. Now do the following \[(1,1,1,1,1,1) \to (0,3,0,3,1,1) \to (0,2,3,0,1,1) \to (0,2,2,0,5,1) \to (0,2,1,5,0,1)\]\[\to (0,2,1,0,10,1) \to (0,1,10,0,0,1)\]Now, our claim yields we can get to \[(0,0,2^{10},0,0,1) \to (0,0,2^{10}-1,2,0,1) \to (0,0,2^{10}-1,1,1,0) \to (0,0,2^{10}-2,3,0,0)\]Now our claim yields we can get to \[(0,0,0,\text{Some really big number $\ge 2010^{2010^{2010}}$},0,0)\]Now, if we just repeat a ton of the second operation (switching the $0$s in box $5$ and $6$), we can get to \[\left(0,0,0,\frac{2010^{2010^{2010}}}{4},0,0\right)\]Now, two more of the first operation finishes.

$ $ This one is funny.

Start by noticing that from $(x,0,0)$ you can get $(0,2^x,0)$ by taking all the 1's from $x$ and dupping them twice then using another one to swap back and repeating, so in fact you can go from $(a,b,0,0)$ to $(a-1,2^b,0,0)$ which is a huge improvement, now it remians to be able to use this a couple times for the win!. \[(1,1,1,1,1,1) \to (0,3,1,1,1,1) \to (0,2,2,3,1,1) \to (0,2,2,2,2,3) \to (0,2,1,1,0,19) \to (0,1,19,0,0,0)\]\[(0,1,19,0,0,0) \to (0,0,2^{19},0,0,0) \to (0,0,0,2^{2^{19}},0,0) \to (0,0,0,0,2^{2^{2^{19}}},0) \to (0,0,0,0,0,2^{2^{2^{19}}+1})\]And we can check that $2010^{2010^{2010}}<(2^{11})^{{2^{11}}^{2^{11}}}=2^{11 \cdot 2^{11 \cdot 2^{11}}}<2^{2^{11 \cdot 2^{11}+4}}<2^{2^{2^{19}}}$ therefore we are done .

Just to ask: have anyone found and proved the possible largest value we can reach?

The answer is yes. Note that we can execute \begin{align*} &(1, 1, 1, 1, 1, 1) \\ \implies &(0, 3, 1, 1, 1, 1) \quad (1)\\ \implies &(0, 1, 5, 1, 1, 1) \quad (1) \\ \implies &(0, 1, 1, 9, 1, 1) \quad (1) \\ \implies &(0, 1, 1, 1, 17, 1) \quad (1) \\ \implies &(0, 1, 1, 1, 0, 35) \quad (1) \\ \implies &(0, 1, 1, 0, 35, 0) \quad (2) \\ \implies &(0, 1, 0, 35, 0, 0) \quad (2) \\ \implies &(0, 0, 35, 0, 0, 0) \quad (2) \end{align*}and therefore ignore the first two boxes hereon. Observe that we can also execute \begin{align*} &(0, r, 0, 0) \\ \implies &(0, r - 1, 2^1, 0) \quad (1) \\ \implies &(0, r - 1, 0, 2^2) \quad (1) \\ \implies &(0, r - 2, 2^2, 0) \quad (2) \\ \implies &(0, r - 2, 0, 2^3) \quad (1) \\ \implies &(0, r - 3, 2^3, 0) \quad (2) \\ \implies &(0, r - 3, 0, 2^4) \quad (1) \\ \implies &(0, r - 4, 2^4, 0) \quad (2) \\ \implies &\dots \\ \implies &(0, 0, 2^r, 0) \quad (2) \end{align*}(call this sequence $\dagger$) and hence we can execute \begin{align*} &(35, 0, 0, 0) \\ \implies &(34, 2, 0, 0) \quad (1) \\ \implies &(34, 0, 2^2, 0) \quad (\dagger) \\ \implies &(33, 2^2, 0, 0) \quad (2) \\ \implies &(33, 0, 2^{2^2}, 0) \quad (\dagger) \\ \implies &(32, 2^{2^2}, 0, 0) \quad (2) \\ \implies &(32, 0, 2^{2^{2^2}}, 0) \quad (\dagger) \\ \implies &(31, 2^{2^{2^2}}, 0, 0) \quad (2) \\ \implies &\dots \\ \implies &(0, {^{35}2}, 0, 0) \quad (\dagger) \end{align*}where ${^a}b$ denotes tetration. Indeed, let $M = {^{35}2}$ and let $N = 2010^{2010^{2010}}$. Note that \[\log_2 \log_2 N = 2010 \log_2 2010 + \log_2 \log_2 2010 < 2010 \cdot 11 + \log_2 11 < 22114.\]Hence $N / 4 < N < 2^{2^{22114}} < 2^{2^{65536}} = {{^6}2} < M$, so we can execute \begin{align*} &(0, M, 0, 0) \\ \implies &(0, M - 1, 0, 0) \quad (2) \\ \implies &(0, M - 2, 0, 0) \quad (2) \\ \implies &(0, M - 3, 0, 0) \quad (2) \\ \implies &\dots \\ \implies &(0, N / 4, 0, 0) \quad (2) \\ \implies &(0, 0, N / 2, 0) \quad (1) \\ \implies &(0, 0, 0, N) \quad (1) \end{align*}as desired. (We can indeed make $N / 4$ since $4 \mid N$ obviously.)