NMTC ramanujan ( inter level) 11th and 12th Aug 31 2024

Using denominator $$60$$, the sum is $$\frac{60+30+20+15+12+10}{60}=\frac{147}{60}=\frac{49}{20}$$. Thus $$m=49=7^2$$. Therefore, the highest power of $$7$$ dividing $$m$$ is $$2$$.

Each ball has radius $$5$$, while its centre can be at most $$8-5=3$$ units from the cylinder axis. Two consecutive centres can therefore be separated horizontally by at most $$6$$, so their vertical separation is at least $$\sqrt{10^2-6^2}=8$$. Five centres require four such gaps, and adding the top and bottom radii gives $$5+4\mathbin{\times}8+5=42$$.

If the expression equals an integer $$k$$, then $$2k=7+\frac{15}{2n+3}$$. Hence $$2n+3$$ must be a signed divisor of $$15$$, namely $$\pm1,\pm3,\pm5,\pm15$$. The corresponding values of $$k$$ are $$11,6,5,4,-4,1,2,3$$, whose sum is $$28$$.

Split the quadrilateral into triangles $$PAB,PBC,PCD$$ and $$PDA$$. Their total area is at most $$\frac{1}{2}(2\mathbin{\times}4+4\mathbin{\times}5+5\mathbin{\times}6+6\mathbin{\times}2)=35$$ because each included sine is at most $$1$$. This value is attained when the four angles at $$P$$ are right angles, so the maximum area is $$35$$.

The general term is $$(3^{(j+3)/3}-3^{j/3})^3=(2\mathbin{\times}3^{j/3})^3=8\mathbin{\times}3^j$$ for $$j=1,2,\ldots,7$$. Therefore, the sum is $$8(3+3^2+\cdots+3^7)$$. Using the geometric-series formula gives $$8\mathbin{\times}\frac{3(3^7-1)}{2}=12(3^7-1)$$.

In each pass, remove every fifth person from the current list and retain the others in their order. Near the end, the successive survivor lists become $$\{1,2,3,4,52,64,79,98\}$$, then $$\{1,2,3,4,64,79,98\}$$, then $$\{1,2,3,4,79,98\}$$ and then $$\{1,2,3,4,98\}$$. The next person removed is therefore the person originally in position $$98$$.

For $$n=1$$ and $$n=3$$, the sums are $$1$$ and $$9$$, which are perfect squares, while the sums for $$n=2$$ and $$n=4$$ are $$3$$ and $$33$$. For every $$n\geq5$$, all later factorials are divisible by $$120$$, so the sum is congruent to $$33\pmod{120}$$. Since $$33$$ is not a quadratic residue modulo $$120$$, no further values work, giving exactly $$2$$ values.

Put $$a=2025$$. The divisor is $$a^2+a+1$$, and $$(a-1)(a^2+a+1)=a^3-1$$, so $$a^3\equiv1$$ modulo the divisor. Since $$2026\equiv1\pmod3$$, we get $$a^{2026}\equiv a$$, and hence the required remainder is $$0$$.

Let $$u=\sqrt{27}+\sqrt{23}$$ and $$v=\sqrt{27}-\sqrt{23}$$, so $$0<v<1$$. The conjugate sum $$u^6+v^6$$ is an integer, and using the recurrence with $$u^2+v^2=100$$ and $$u^2v^2=16$$ gives $$u^6+v^6=995200$$. Therefore, $$N=995200-1=995199$$, whose remainder modulo $$1000$$ is $$199$$.

Let $$\angle BAD=\angle ADB=x$$ because $$AB=BD$$, and let $$\angle DBC=\angle BCD=y$$ because $$BD=CD$$. The straight line at $$D$$ gives $$x=2y$$. Hence $$\angle ABC=(180^\circ-2x)+y=180^\circ-3y$$, which is largest for the smallest positive integer $$y=1$$, giving $$177^\circ$$.

Write the sides as $$2,2r,2r^2$$ with $$r>1$$. The Pythagorean theorem gives $$4+4r^2=4r^4$$, so $$r^4-r^2-1=0$$ and $$r^2=\frac{1+\sqrt5}{2}$$. The hypotenuse is therefore $$2r^2=1+\sqrt5$$.

An angle formed by two intersecting diagonals of a regular $$n$$-gon is an integer multiple of $$\frac{180^\circ}{n}$$. Thus $$50^\circ=k\frac{180^\circ}{n}$$ for some positive integer $$k$$, so $$5n=18k$$. The least possible $$n$$ is therefore $$18$$, and suitable diagonals can be chosen to produce the required angle.

Let $$PQ=x$$ and $$\angle P=\theta$$. The sine rule gives $$\frac{x}{\sin2\theta}=\frac{4}{\sin\theta}$$, so $$x=8\cos\theta$$. Applying the cosine rule at $$P$$ gives $$16=25+x^2-10x\cos\theta=25-\frac{x^2}{4}$$, and hence $$x=6$$.

Let the chosen numbers be $$x_1,x_2,\ldots,x_{10}$$ cyclically. The announcements give $$x_{i-1}+x_{i+1}=2i$$ for every $$i$$. Solving these cyclic equations gives $$(x_1,\ldots,x_{10})=(6,-3,-2,9,10,1,2,13,14,5)$$, so the person announcing $$6$$ chose $$1$$.

Let each removed triangle have equal legs $$x$$. For the remaining octagon to be regular, its horizontal side and sloping side must be equal, so $$1-2x=x\sqrt2$$. Thus $$x=\frac{1}{2+\sqrt2}$$, and the required area is $$1-4\left(\frac{x^2}{2}\right)=1-2x^2=2(\sqrt2-1)$$.

From the net, the opposite face pairs are $$a$$ with $$2026$$, $$b$$ with $$c$$, and $$d$$ with $$1$$. Thus the averaging conditions give $$4a=b+c+d+1$$, $$4b=a+d+2027$$, $$4c=a+d+2027$$ and $$4d=a+b+c+2026$$. Solving this system gives $$a=811$$, so the requested value is $$811$$.

Write the selected elements as $$x_1<x_2<x_3<x_4$$ with $$x_{i+1}-x_i\geq2$$. Define $$y_i=x_i-(i-1)$$, which gives $$1\leq y_1<y_2<y_3<y_4\leq12$$. Therefore, the number of choices is $$\binom{12}{4}=495$$.

The final selected card is equally likely to be any one of the original $$52$$ cards because the discard and selection process is symmetric. There are $$4$$ multiples of $$13$$ among these cards. Hence the probability is $$\frac{4}{52}=\frac{1}{13}$$, so $$m+n=14$$.

Fix one chosen object. Of the remaining positions, its two neighbours and its opposite position are forbidden, and counting the allowable pairs among the other six positions gives $$9$$ choices. Rotating the fixed choice gives $$10\mathbin{\times}9$$ counts, but every valid triple is counted once for each of its three objects. Thus the number of triples is $$\frac{10\mathbin{\times}9}{3}=30$$.

From the second congruence, $$a\equiv3+b+7c\pmod{13}$$. Substitution into the other two gives $$9b+8c\equiv2\pmod{13}$$ and $$b\equiv0\pmod{13}$$. Hence $$c\equiv10$$ and $$a\equiv8$$, so $$a+b+c\equiv18\equiv5\pmod{13}$$.

Since $$N=13s$$, where $$s$$ is the digit sum of $$N$$, the bound on a three-digit digit sum gives $$1\leq s\leq27$$. Checking these possible values in the condition that the digit sum of $$13s$$ equals $$s$$ gives $$s=9,12,15$$. The corresponding numbers are $$117,156,195$$, whose sum is $$468$$.

If a smaller hexagon has side length $$s$$, the side length of the outer regular hexagon is $$3s$$. Areas of similar figures are proportional to the squares of their side lengths, so one small hexagon has $$\frac{1}{9}$$ of the outer area. Therefore, the six smaller hexagons have area $$6\mathbin{\times}\frac{900}{9}=600$$.

For every integer below $$7$$, the sum of five remainders is at most $$30$$. For $$7$$, the possible largest patterns are $$0,7,7,7,7$$ or $$7,7,7,7,7$$, whose sums are $$28$$ and $$35$$, not $$32$$. For $$8$$, division by $$8,9,10,11,12$$ gives remainders $$0,8,8,8,8$$, whose sum is $$32$$.

Let the number be $$100h+10t+u$$. The condition becomes $$99h+10t=t^2+u^3$$ for digits $$h,t,u$$. Checking the digit possibilities gives the solutions $$135,175,518,598$$, of which the largest is $$598$$.

Factorising or multiplying shows $$8\mathbin{\times}9\mathbin{\times}10\mathbin{\times}11\mathbin{\times}12\mathbin{\times}13\mathbin{\times}14=17297280$$. Also, $$63\mathbin{\times}64\mathbin{\times}65\mathbin{\times}66=17297280$$. Thus the alternate consecutive product begins with $$63$$.

The gap $$\frac{1}{i}$$ occurs in the distance between $$P_r$$ and $$P_s$$ exactly when $$r\leq i<s$$. There are $$i(100-i)$$ such pairs, so this gap contributes $$\frac{1}{i}\mathbin{\times}i(100-i)=100-i$$. Summing over $$i=1$$ to $$99$$ gives $$99+98+\cdots+1=4950$$.

Checking the positive integers below $$38$$ gives no total of $$21$$ for the three divisor counts. At $$n=38$$, the numbers are $$36,38,40$$. Their divisor counts are $$d(36)=9$$, $$d(38)=4$$ and $$d(40)=8$$, whose sum is $$21$$.

Let the slower bug travel a fraction $$t$$ of $$HF$$, so before the faster bug reaches $$C$$ it travels a fraction $$2t$$ of $$AC$$, where $$0\leq t\leq\frac12$$. With suitable cube coordinates, the squared separation divided by the edge length squared is $$(3t-1)^2+t^2+1=10t^2-6t+2$$. This is minimised at $$t=\frac{3}{10}$$ with value $$\frac{11}{10}$$, so the minimum distance is $$4\sqrt{110}\sqrt{\frac{11}{10}}=44$$.

Suppose $$k$$ angles are $$163^\circ$$ and the other $$n-k$$ angles are $$171^\circ$$. Equating their sum to $$(n-2)180^\circ$$ gives $$9n+8k=360$$. This forces $$n$$ to be a multiple of $$8$$, and the first feasible choice is $$n=24$$ with $$k=18$$.

The root-sum condition gives $$-\frac{b}{a}=40$$, so $$b=-40a$$. Solving this relation together with the three given function values yields $$P(x)=\frac{1}{10}x^3-4x^2+\frac{241}{10}x-26$$. Substitution of $$x=35$$ gives $$P(35)=205$$.