Rephrased Problem wrote: Let $a,b,c$ be distinct naturals so that $a+b+c=2022$. Minimize $\text{lcm}(a,b,c)$. The answer is $1344$, achieved when $a=6, b=672, c=1344$. Assume $c>\max(a,b)$. Because $c>674$ we must have $a,b|c$ if we have $\text{lcm}(a,b,c)<1344$. Hence two divisors of $c$ sum to $2022-c$. So $d_1+d_2=2022-c$, where $d_1, d_2|c$. If $c<1344$, $d_1+d_2>678$. WLOG $d_1>d_2$. Notice that $c=d_1k$ for some $k$. If $k\ge 4$ then we have contradiction because $d_1>\frac{678}{2}$ and $d_1<\frac{1344}{4}$. We must check $k=2$ and $k=3$. If $k=2$, we have $3d_1+d_2=2022$. This means $3|d_2$. Letting $d_2=3r$, $d_1=674-r$, so $3r|1348-2r$. Manually checking divisors of $1348=2^2\cdot 337$, we find $r=2$ or $r=674$. The former gives $(6,672,1344)$ and the latter fails. If $k=3$, we have $4d_1+d_2=2022$. Thus $d_2=2r$ for some $r$. Hence $2d_1+r=2022$ and $2r|3d_1$. This means that $4044-4d_1|3d_1$. Hence $3d_1=4044t-4d_1t$ for some $t$, i.e. $\frac{4044t}{4+3t}\in\mathbb{Z}$. Hence $3t+4$ divides $5392=2^4\cdot 337$. This doesn't have any $3\pmod{4}$ divisors, so there are no possibilities here. This concludes the cases.

Definitely I agree with #2, this is not appropriate for the IMO.

Correct answer = 7 points.

The answer is $1344$, with the respective factors being $1344$, $672$, and $6$. Let $n$ be the answer, and $\frac n{d_1}, \frac n{d_2}, \frac n{d_3}$ be the respective factors, with $d_1 < d_2 < d_3$. We claim $d_1=1$. Assume not, then \[ 2022 = \frac n{d_1}+ \frac n{d_2},+\frac n{d_3} \leq \frac n2 + \frac n3 + \frac n4, \]but this yields $n\geq 1867$, which is much bigger than our answer. We claim $d_2=2$. Assume not. If $d_2>3$, then the sum is at most $n + \frac n4 + \frac n5$, but this yields a greater value than we got. If $d_2=3$ and $d_3>5$, then the sum is less than $\frac{3n}{2}$, and this yields a bigger value that we claimed. The only cases left to check are $d_2=3, d_3=4$ and $d_2=3, d_3=5$. But we can easily see that this yield invalid values of $n$, so we are done. Let $n = d_3 k$. Then $2022 = k + d_3 k + \frac{d_3 k}2$, or $4044 = k(3d_3 + 2)$. The only factors of $4044$ that are $2 \bmod 3$ are $2$ and $674$, and checking both cases, we see that $d_3k$ is minimized at $1344$, as desired.

Solved with plang2008 and Spectator. The answer is $\boxed{1344}$ and this is achievable by $1344 + 672 + 6 =2 022$. Now we show it's the smallest Norwegian number. Suppose some $n<1344$ was Norwegian. Notice that the sum of three distinct prime divisors of $n$ is less than $3n$, so $3n > 2022\implies n > 674$. We get that $\frac{n}{a} + \frac{n}{b} + \frac{n}{c} = 2022$. Therefore \[ S = \frac{1344}{a} + \frac{1344}{b} + \frac{1344}{c} > 2022\] WLOG that $a<b<c$. If $a > 1$, then $S < \frac{1344}{2} \cdot 3 = 2016 < 2022$, absurd. Therefore, $a = 1$, so $S = 1344 + \frac{1344}{b} + \frac{1344}{c}$. Claim: $b = 2$. Proof: Suppose $b > 2$. If $b > 3$, then $S \le \frac{1344}{1} + \frac{1344}{4} + \frac{1344}{5} < 2022$, absurd. So $b = 3$. If $c > 5$, then $S \le 1344 + \frac{1344}{3} + \frac{1344}{ 6} = 2016 < 2022$, absurd. Therefore, $c = 4$ or $c = 5$. Now it is easy to check $n + \frac{n}{3} + \frac{n}{4} = 2022$ and $n + \frac{n}{3} + \frac{n}{5} = 2022$ don't have integer solutions. Thus, $b = 3$ is impossible. So $b < 3$, which implies $ b = 2$. $\square$ Now we have $n + \frac{n}{2} + \frac{n}{c} = \frac{3nc + 2n}{2c} = 2022$, so $3nc + 2n = 4044c$. Thus, $2n = (4044 - 3n)c\implies 4044 - 3n \mid 2n$, so $4044 - 3n \mid 2n \cdot 3 + 2\cdot (4044 - 3n) = 8088$. Since $6\mid 3n$, we have $6\mid 4044 - 3n$. The divisors of $8088$ that are multiples of $6$ (essentially $6$ times the set of divisors of $\frac{8088}{6} = 1348$) are $6,12,24,2022,4044,8088$. Thus, $4044 - 3n\in \{6,12,24,2022,4044,8088\}$, so $n\in \{1346, 1344, 1340,674,4044,-1348\}$. However, since $674<n<1344$, we must have $n = 1340$. Thus, $2022 = 1340 + \frac{1340}{2} + \frac{1340}{c} \implies \frac{1340}{c} = 12$, which is absurd since $12\nmid 1340$. Therefore the smallest Norwegian number is $1344$.

terrible problem for oly(not as bad as 2010 c1 but still why) The answer is $\boxed{1344}$, which works because $1344+672+6=2022$. Assume FTSOC that there is a smaller solution, that is, let $n<1344$ be a Norwegian number. Therefore, $\frac na+\frac nb+\frac nc=2022$, where $a<b<c$ are positive divisors of $n$. Note that $(a,b,c)=(1,2,224)$ for $1344$. Claim. $a=1$. Proof. Suppose not. Then if $a\ge 3$, we would need $n>2022$ which is bad, and if $a=2$, we would need $b=3$(since if $b=4$ we have $n>2022$), then $c=4$ or $c=5$, neither of which work. Therefore, our task turns to maximizing $\frac1b+\frac1c$ since \[n\cdot\frac{bc+b+c}{bc}=2022.\] Claim. $b=2$. Proof. Suppose not. If $b\ge 4$ then $\frac1b+\frac1c$ is smaller than $\frac12+\frac1{224}$, so $b=3$. Then we still need $c\in\{4,5\}$ for $\frac1b+\frac1c$ to be big enough, and neither of these work. So we just want to minimize $c$. But \[n\cdot\frac{3c+2}{2c}=2022,\]and the gcd between the numerator and denominator is a divisor of $4$ by Euclid. So \[3c+2\mid 8088=2^3\cdot 3\cdot 337.\]If $3c+2<337$, then it must be $2$ or $8$, impossible. Therefore, $3c+2$ is at least $337\cdot 2=674$, so $c$ is at least $224$. $\blacksquare$

tfw you typo "ISL" instead of "AMC"

With regards to programming this, giving ChatGPT a small nudge yields the following valid program.

ISL 2022 N1. A number is called Norwegian if it has three distinct positive divisors whose sum is equal to $2022$. Determine the smallest Norwegian number. (Note: The total number of positive divisors of a Norwegian number is allowed to be larger than $3$.) Solution. The answer is $1344$. Note that $1344$ satisfies the problem condition as $1344+672+6=2022$. To show that $1344$ is minimal, assume that $n<1344$, $n\left(\frac1a+\frac1b+\frac1c\right)=2022$, $a<b<c$. Use usual size arguments to show that if $a\ge 2$, then $n>1866>1344$. Hence it suffices to consider the case when $a=1$. We then get the bound $b\le 3$ in the same way. Then it is casework and dealing with simple divisibilities.

This can be even done by simply working backwards and seeing that the closest number that has many divisors to $2022$ is $2016$, which upon closer inspection yields that $1344+672+6=2022$, where $1344,672,6$ are divisors of $1344$. They should add N0 for such problems if they insist putting it on the ISL.

wlog $c>b>a$ the problem can be written as $a+b+c=2022$ minimize $lcm(a,b,c)$ to minimize the lcm, $a$ divides $b$, which divides $c$, so $lcm(a,b,c)=c$ then $a+b+c$ divides $a$ and $2022$ divides $a$ so $a$ can be $2,3,6,337,$or higher than $337$ $a \ge 337$ does not work because min val of $b$ is $337*2=674$ and min val of $c$ is $337*4=1348$ and they add to more than $2022$ $a=3$, then let $b=3x$, $c=3x*y$ $3+3x+3xy=2022$ $x+xy=673$ $x(1+y)=673$ But $673$ is prime, so it cannot be the product of two numbers more than $1$ $a=2$, then let $b=2x$, $c=2x*y$ $2+2x+2xy=2022$ $x+xy=1010$ $x(1+y)=1010$ $x=2, y=504$ $x=5, y=201$ $x=10, y=102$ $x=101, y=9$ $x=202, y=4$ Trying each of them results in $c=1616, b=404, a=2$ --> $1616$ as min $a=6$, then let $b=6x$, $c=6x*y$ $6+6x+6xy=2022$ $1+x+xy=337$ $x(1+y)=336$ $x=2,y=167$ $x=3,y=111$ $x=4,y=83$ $x=6,y=55$ $x=7,y=47$ $x=8,y=41$ $x=12,y=27$ $x=14,y=23$ $x=16,y=20$ $x=21,y=15$ $x=24,y=13$ $x=28,y=11$ $x=42,y=7$ $x=48,y=6$ $x=56,y=5$ $x=84,y=3$ $x=112,y=2$ Trying each of them results in $c=6*224=\boxed{1344}, b=672, a=6$ I the lost game 1344

We claim the answer $6\cdot 224=1344,$ which works since $6+3\cdot 224+6\cdot 224=2022.$ Take an arbitrary Norwegian number $n\leq 1344$ such that $\frac na+\frac nb+\frac nc=2022$ for its positive divisors $a<b<c.$ Note that $r=\frac 1a+\frac 1b+\frac 1c\geq \frac{2022}{1344}=\frac 32 +\frac{1}{224}.$ Note that $a=1$ since otherwise $r\leq \frac 12 +\frac 13 +\frac 14=\frac{13}{12}<\frac 32.$ Next, note that $b=2.$ Indeed, otherwise either $b\geq 3,$ $c\geq 6$ and so $r\leq \frac 32,$ or $b\geq 4$ and so $r\leq \frac {29}{20}<\frac 32,$ or $b=3,$ $c=4$ and so $n\notin \mathbb Z,$ or $b=3,$ $c=5$ and so $n\notin \mathbb Z$ by a straightforward check. Finally, $n=\frac{4044c}{3c+2}\in \mathbb Z\implies (3c+2)\mid 8\cdot 337.$ Since $c>2$ we thus get $3c+2\in \left \{ 337, 2\cdot 337, 4\cdot 337, 8\cdot 337\right \}.$ Considering the reminders mod $3$ it follows that $c\in \left \{ 224,\frac{8\cdot 337-2}{3}\right \}.$ On the other side $\frac 1c\geq \frac{1}{224},$ implying $c=224$ and so $n=1344,$ we are done.

This seems kind of like an AIME problem?

This works since $1344+672+6 = 2022$. Now, we will show that there is no possible value greater than $1344$. We need to relate the divisors to the number, so let's say the number is $x>1344$ and the divisors are $\frac{x}{p}, \frac{x}{q}, \frac{x}{r}$. Then we are given that $\frac{x}{p}+\frac{x}{q} + \frac{x}{r} = 2022$. Now clearly we need some sort of bound on the factors; WLOG $p<q<r$ and notice that if $p>1$ then we have $\frac{x}{p}+\frac{x}{q} + \frac{x}{r} \geq \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{13}{12}x$. For this sum to be equal to $2022$ we have $1866 < x < 1867$ which is clearly much bigger than $1344$, so we must have $p = 1$. Using similar logic we get $q = 2$, so now we have $x\left(\frac{3}{2} + \frac{1}{r}\right) = 2022$ or equivalently $x\left(\frac{3r+2}{2r}\right) = 2022 \implies x(3r+2) = 4044r$. But this is clearly not possible since there will always be an integer term on the LHS, so thus the answer is $\boxed{1344}$.

Truly the ISL of all time The answer is $1344$. Suppose some $n<1344$ also works, then let $\tfrac {n}{a}>\tfrac{n}{b}>\tfrac{n}{c}$ sum to $2022$. $~$ If $a\ge 2$ then \[2022 \le \frac n2 + \frac n3 + \frac n4 < 1456 \]which is absurd. Thus, $a=1$. $~$ If $b>3$, then \[2022 \le n + \frac n4 + \frac n5 < 1950\]Thus, we have reduced to finite case check and we're done.

We claim, that the answer is $\boxed{1344}$, which obviously works, since, $1344+672+6=2022.$ FTSOC, we assume that there exists a $n<1344$, which satisfies, these conditions. Let $p_1, p_2, p_3$, be the $3$, factors, then, $2022=\sum_{k=1}^{3}\frac{n}{p_k}$, so WLOG, assume that $p_1>p_2>p_3$, now, if $p_3$, is at least $2$, we have contradiction. Because, then the minimum value of $n$, is $\frac{n}{2}+\frac{n}{3}+\frac{n}{4}=\frac{13n}{12}$, but then, $n>1344$, hence, proved. Now, if $p_2 \ge 4$, then we have that, the minimum, is $\frac{n}{1}+\frac{n}{4}+\frac{n}{5}=\frac{29n}{20}$, but then $n>1344$, so contradiction. Then, quickly, checking the cases, when $p_2=3$, and $p_2=2$, no $n$, works. Hence, the smallest $n$, is $\boxed{1344}.$

The answer is $1344$, which has divisors $1344+672+6=2022$. We prove that this is optimal. Assume FTSOC that there exist positive integers $n \le 1343$ and $a \le b \le c$ such that $\tfrac{n}{a}$, $\tfrac{n}{b}$, and $\tfrac{n}{c}$ are distinct integers summing to $2022$. This means $\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}=\tfrac{2022}{n}>\tfrac{2022}{1348}=\tfrac{3}{2}$, so we have $\tfrac{1}{a}>\tfrac{1}{2} \implies a=1$ and $\tfrac{1}{b}>\tfrac{1}{4} \implies b \le 3$. Notice that $\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c} \le \tfrac{11}{6}$, so $n \ge 2022 \cdot \tfrac{6}{11}>1011$. If $b=3$, then we must have $b=4$ or $b=5$. However, neither $1+\tfrac{1}{3}+\tfrac{1}{4}=\tfrac{19}{12}$ and $1+\tfrac{1}{3}+\tfrac{1}{5}=\tfrac{23}{15}$ have a numerator dividing $2022$, so $2022$ does not become an integer when divided by either, a contradiction. If $b=2$, then we have $\tfrac{3}{2}n+\tfrac{n}{c}=2022 \implies 3n+\tfrac{2n}{c}=4044$, so $\tfrac{n}{c} \mid 4044$. Since $1011<n \le 1343$, we have $337 \nmid n \implies 337 \nmid \tfrac{n}{c}$, which means $\tfrac{n}{c} \mid 12$. If $\tfrac{n}{c}=12$, then $n=1340$, which is a contradiction as $12 \nmid n$. Otherwise $\tfrac{n}{c} \le 6$, which gives $n \ge 1344$, a contradiction. $\square$

Hello, what's the motivation for $1344$ ?

The answer is $\boxed{1344}$ with divisors $1344$, $672$, and $6$. Let $N$ be a Norwegian number with three divisors $\frac{N}{a}$, $\frac{N}{b}$, and $\frac{N}{c}$. WLOG assume $1\leq a<b<c$. The condition implies, $$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})N=2022$$To minimize $N$ is equivalent to maximize $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ Case 1: $a=1$ and $b=2$ This gives $N=\frac{4044c}{3c+2}\Rightarrow 3c+2|4044c$. If $c$ is odd then $3c+2|4044$, which has no odd solutions. If $c$ is even then $c|8088$, the smallest such $c$ is $224$. We have one working triplet of $(1,2,224)$ which gives $N=1344$. Case 2: All other triplets We only need to check triplets $(a,b,c)$ with $\frac{3}{2}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ and $(a,b)\neq (1,2)$. These triplets are just $(1,3,4)$ and $(1,3,5)$ which both don't work.

i got a 0 because i miswrote my answer as 1434 instead of 1344

asdf334 wrote: i got a 0 because i miswrote my answer as 1434 instead of 1344 The moment you lost both the game and the points LOL

asdf334 wrote: i got a 0 because i miswrote my answer as 1434 instead of 1344 They had no mercy.

I can't believe I spent half an hour of my life doing this The answer is $n = 1344$, where the divisors $1344, 672, 6$ clearly work. Assume for the sake of contradiction that there is a smaller $n$. Now let $\frac na+\frac nb+\frac nc = 2022$ with $a<b<c$. If $a \geq 2$, then \[\frac n2 + \frac n3 + \frac n4 \geq \frac na+\frac nb+\frac nc = 2022\]which implies $n > 1866 > 1344$. So $a = 1$, and it follows that $n = \frac{2022bc}{bc+b+c}$. If $b = 2$, then $n = \frac{4044c}{3c+2} = 1348 - \frac{2696}{3c+2}$. As $c > 2$, we have $3c+ 2 > 8$, which yields that $c=224$ yields the maximum value of $\frac{2696}{2c+2}$. If $b \geq 3$, we just have to destroy the cases $(a, b, c) = (1, 3, 4) = (1, 3, 5)$ as all other $(a, b, c)$ yield $n$ larger than $1344$; turns out none of them work.

Let $a<b<c$ be three distinct natural numbers such that $a+ab+abc = 2022$ We need to minimize $LCM(a,b,c)$ which will be least when $b=ax$ and $c=axy$ $\implies a(1+x+xy) = 2022 = 2 \cdot 3 \cdot 337 \implies a = 1 , 2, 3, 6, 337$ $a = 337$ wont work because $x=2$ and $xy=3$ which is not possible $LCM(a,b,c) = L = axy \implies L = \frac{(2022-a)y}{1+y}$ where $1+y | 2022 -a$ Now $\frac{y}{1+y}$ is increasing as $y$ increases Now $(a,y) = (1, 42),(2,3) , (3,2), (6,2) \implies L = 1974, 1515, 1346, 1344$ Hence the least norwegian number is $1344$

How did you guys guess the answer?

The answer is $1344$, which works since $1344+ 672 + 7 = 2022$. To prove nothing smaller works, we use bounding. Let the number be $n$. If $n$ is not a part of the sum, the maximum sum possible is $\frac n2 + \frac n3 + \frac n4 = \frac{13}{12}n$, forcing $n \ge \frac{12}{13}(2022) > 1344$. Thus $n$ is part of the sum. If neither $\frac n2$ or $\frac n3$, the maximum possible sum is $n + \frac n4 + \frac n5 = \frac{29}{20}n$, forcing $n \ge \frac{20}{29}n > 1344$. Thus one of $\frac n3, \frac n2$ must be present in the sum. If $\frac n3$ is part of the sum but not $\frac n2$, we do casework on the third element. If it is $\frac n4$, we get $n + \frac n3 + \frac n4 = \frac{19}{12}n = 2022$, which does not have any integer solutions in $n$. If it is $\frac n5$, we get $n + \frac n3 + \frac n5 = \frac{23}{15}n = 2022$, which does not have any integer solutions in $n$. If it is $\frac n6$ or smaller, the maximum possible sum is $n + \frac n3 + \frac n6 = \frac 32 n$, forcing $n > 2022 (\frac 23) > 1344$. Thus there are no solutions when $\frac n2$ is not present. If both $n, \frac n2$ are present, we have $2022 - \frac 32 n \mid n$, so $4044 - 3n \mid 4044$, forcing $1348 - n \mid 1348$. Since $n$ is even, the only possible values of $1348 - n$ are $2,4, 2 \cdot 337$. The last one clearly fails due to size ($2n < 2022$), the first one would result in a larger value than $1344$, and the middle one gives $1344$. All options are exhausted, so we are done.

Infinite descent also works