POMO Problems

Solution

We note that $\lfloor\sqrt{i}\rfloor = k$ exactly when $k^2 \le i \le (k+1)^2 - 1$, which is a range of $2k+1$ consecutive integers. We compute the contribution of each $k$ to the sum (noting $\lfloor\sqrt{100}\rfloor = 10$, so $i=100$ contributes its own term):

Therefore $x = 3+10+21+36+55+78+105+136+171+10 = 625$. Since $625 = 25^2$, we get $\lfloor\sqrt{625}\rfloor = \boxed{25}$.

Solution

Let $v_2(m)$ denote the 2-adic valuation of $m$ (the largest power of 2 dividing $m$). Since $n = 1024 = 2^{10}$, we apply Kummer's theorem: the number of carries when adding $k$ and $2^{10}-k$ in base 2 equals $v_2\!\left(\dbinom{2^{10}}{k}\right)$.

Since $2^{10}-1$ is all 1s in binary, adding any $k-1$ to $2^{10}-k$ in base 2 produces no carries, so $v_2\!\left(\dbinom{2^{10}-1}{k-1}\right) = 0$. Using the identity $\dbinom{2^{10}}{k} = \dfrac{2^{10}}{k}\dbinom{2^{10}-1}{k-1}$, we obtain: \[v_2\!\left(\binom{2^{10}}{k}\right) = 10 - v_2(k), \quad 1 \le k \le 1023.\]

It follows that $f(j)$ counts the number of $k \in \{1,\dots,1023\}$ satisfying $v_2(k) = 10-j$. For $1 \le j \le 10$ this gives $f(j) = 2^{j-1}$, and $f(j) = 0$ for all other $j$.

Since $f(j) = 0$ outside $\{1,\dots,10\}$, choosing indices beyond this range contributes nothing to the sum. So it suffices to consider nonempty subsets of $\{1,\dots,10\}$. The attainable sums are exactly the nonempty subset sums of \[\{f(1), f(2), \dots, f(10)\} = \{1, 2, 4, \dots, 512\}.\] Since these are the first 10 powers of 2, every integer from $1$ to $1+2+\cdots+512 = 1023$ is achieved by a unique subset. The score is $\boxed{1023}$.

Solution

We use the identity $OH^2 = 9R^2 - (a^2+b^2+c^2)$, where $a,b,c$ are the side lengths and $R$ is the circumradius. We have $a^2+b^2+c^2 = 25+49+9 = 83$.

Finding $R$. By the Law of Cosines applied to $\angle ABC$: \[AC^2 = AB^2 + BC^2 - 2\cdot AB\cdot BC\cdot\cos(\angle ABC)\] \[49 = 9 + 25 - 30\cos(\angle ABC) \implies \cos(\angle ABC) = -\tfrac{1}{2} \implies \angle ABC = 120°.\] By the Extended Law of Sines: \[2R = \frac{AC}{\sin(\angle ABC)} = \frac{7}{\sin 120°} = \frac{7}{\frac{\sqrt{3}}{2}} = \frac{14}{\sqrt{3}} \implies R^2 = \frac{49}{3},\] so $9R^2 = 147$.

Therefore: \[OH^2 = 147 - 83 = 64 \implies OH = 8 = \frac{8}{1},\] giving $a+b = 8+1 = \boxed{9}$.