Bound $n$, then check for remaining $n$.

lol, look at AMC 12B 2021 P25, it basically asks R(N)=R(N+1) for 2 digit numbers, did that problem a few months ago,

trivial, check n/2 + 1, n/2 + 2, n/2 + 3, ..., n - 1 do the same for n. odd with n-1/2

alexanderhamilton124 wrote: trivial, check n/2 + 1, n/2 + 2, n/2 + 3, ..., n - 1 Us bro.

Check remainders when $n$ is divided by $[n/2],[n/2]+1,.......n-1$ , prove their sum greater than $n$ for $n\geq 10$ Now case check to get $n=1,5$.

We see that $n=1$ works. Henceforth, suppose $n>1$. For even $n$, consider the positive integers $\frac{n}{2}+1, \frac{n}{2}+2, \frac{n}{2}+3$. The remainders when $n$ is divided by these integers are $\frac{n}{2}-1, \frac{n}{2}-2, \frac{n}{2}-3$, respectively. But then, for $n>10$, we have $R(n)\geq \frac{3n}{2}-6>n-1$, so $n\leq 10$ is forced. Checking the cases, we see that no such even $n$ exists. For odd $n$, consider the positive integers $\frac{n+1}{2}, \frac{n+1}{2}+1, \frac{n+1}{2}+2$. The remainders when $n$ is divided by these integers are $\frac{n-1}{2}, \frac{n-1}{2}-1, \frac{n-1}{2}-2$, respectively. But then, for $n>7$, we have $R(n)\geq \frac{3(n-1)}{2}-3>n-1$, so $n\leq 7$ is forced. Checking the cases, we see that only $n=5$ works. Therefore, the answer is $\boxed{n=1, 5}$ only.

From the OMC RMO livesolve. solved with rawlat_vanak, AdhityaMV, zvfrozel, and mathscrazy. Denote $k = \left\lfloor \frac{n}{2} \right\rfloor + 1$. We note that for all $k \leq x \leq n$ we have $n (\bmod x) = n-x$. As a result, we have \[R(n) = \sum \limits_{i = 1}^{n} {n (\bmod ~ i)} \geq \sum \limits_{i = k}^{n} (n-i) = \binom{\left \lceil \frac{n}{2} \right \rceil}{2}\]If $R(n) = n-1$, we therefore obtain $n \leq 9$. Checking all $n$ from $1$ through $9$, we obtain that the only two values that fit are $n \in \{1,5\}$. $\square$

Do we need to include the manual checking from 1 to 9 in the answer sheet?Otherwise how much marks would be deducted

In this question I found all value $R(n)$ from $n=1 to 20$ Then there was a special sequence formed in the second half of the sum and it went like $0+1+2+3+....$ So I wrote, "With observation in the sum of $R(n)$ in the second half of the sum is a sequence with sum $\frac{k(k+1)}{2}$. I proved that if n is even then the second half will be like $1+2+3+..........(\frac{n-2}{2})$ which gave me sum of this half as$( \frac{n-2}{2})(\frac{n}{2})$ Now, $R(n) >( \frac{n-2}{2})(\frac{n}{2})$ Got a quadratic solved it and got it as $n^2-2n+8<0$

Again I proved that for n is odd the second half is like $1,2,3...\frac{n-1}{2}$ Again got a quadractic $\frac{n^2}{7}-\frac{8n}{7}+1<0$ From here i got two values of n= ${{1,5}}$.

L_BarySaneTan37 wrote: Do we need to include the manual checking from 1 to 9 in the answer sheet?Otherwise how much marks would be deducted It depends on many factors. First is examiner. Second, If you somewhere mentioned, by observation we get this. SO it is always better to mention it. Well can't say anything here.

ez bounding problem got n >= 10 (not works) then n = 1 , 5

$OG!$ I will try to write as far as I remember the exact same quotation as I did in the contest We denote by $r_i(N)$, the remainder when $N$ is divided by $i$, so $R(n)= r_1(n)+r_2(n)+ \cdots r_n(n)$ Case1: $n>10$ Case1a: $n$ is even Set $n=2k$ observe that $r_{k+1}(2k) = k-1$ $r_{k+2}(2k) = k-2$ $\cdots$ $r_{2k-1}(2k) = 1$ So $2k-1=R(n)=R(2k)=r_1(2k)+r_2(2k)+ \cdots r_{2k}(2k) \geq r_{k+1}(2k)+ r_{k+2}(2k) + \cdots r_{2k-1}(2k) = k-1+k-2 \cdots 1 = \frac{k(k-1)}{2}$ as $R(n)=n-1$. So $2k-1 \geq \frac{k(k-1)}{2} \iff 0 \geq k^2-5k+2= k(k-5)+2$ But, since, $n>10 \implies 2k >10 \implies k>5 \implies 0 \geq k(k-5)+2>2 \implies 0>2.$ A contradiction! Hence, no solutions Case1b: $n$ is odd Set $n=2k+1$ observe that $r_{k+1}(2k+1) = k$ $r_{k+2}(2k+1) = k-1$ $\cdots$ $r_{2k}(2k+1) = 1$ So $2k=R(n)=R(2k+1)=r_1(2k+1)+r_2(2k+1)+ \cdots r_{2k+1}(2k+1) \geq r_{k+1}(2k+1)+ r_{k+2}(2k+1) + \cdots r_{2k}(2k+1)=k+k-1+k-2 \cdots 1 = \frac{k(k+1)}{2}$ as $R(n)=n-1$. So, $2k \geq \frac{k(k+1)}{2} \iff 0 \geq k^2-3k= k(k-3)$ But, since $n>10 \implies 2k+1 >10 \implies k>4.5 \implies k>4 \implies 0 \geq k(k-3)>0 \implies 0>0.$ A contradiction! Hence, no solutions Thus, $n \leq 10$ Now, on checking for $n=1,2,3 \cdots 10$ We easily get $n=1$ and $n=5$ as the only solutions.