NMTC BHASKARA Junior Level 2019 Solved Question Paper PDF

Since $$ABCABC=1001\times ABC$$ and $$1001=13\times77$$, every number of this form is divisible by $$13$$. The digit $$A$$ can be chosen in $$4$$ ways from $$2,4,6,8$$. If $$C=0$$ there are $$8$$ choices for $$B$$, and if $$C$$ is nonzero there are $$3$$ choices for $$C$$ and $$8$$ choices for $$B$$. Thus the total is $$4\times8+4\times3\times8=128$$.

Let the vectors from the centroid to the vertices be $$\mathbf{a},\mathbf{b},\mathbf{c}$$, so $$\mathbf{a}+\mathbf{b}+\mathbf{c}=0$$. The median through $$B$$ is parallel to $$\mathbf{b}$$ and the median through $$C$$ is parallel to $$\mathbf{c}$$, so perpendicular medians give $$\mathbf{b}\cdot\mathbf{c}=0$$. Since $$\mathbf{a}=-(\mathbf{b}+\mathbf{c})$$, we get $$a^2=b^2+c^2+2\mathbf{b}\cdot\mathbf{c}$$ in centroid-vector lengths; converting these vector lengths to the corresponding side lengths gives $$b^2+c^2=5a^2$$.

Since $$AB=AD$$ and $$CB=CD$$, the quadrilateral is a kite with diagonal $$BD$$ as its symmetry axis. Also, $$AB+CD=AD+CB=23$$, which is exactly the condition for a convex quadrilateral to have an incircle. The angles at $$B$$ and $$D$$ are equal rather than supplementary, so the quadrilateral is not cyclic. Hence it has an in-circle but not a circum-circle.

If the integer side lengths are $$a,b,c$$, then $$6(a^2+b^2+c^2)=498$$, so $$a^2+b^2+c^2=83$$. The only positive integer solution, up to order, is $$1^2+1^2+9^2=83$$. Therefore the sum of the volumes is $$1^3+1^3+9^3=731$$.

If the diagonals are $$p$$ and $$q$$, then the half-diagonals and a side form a right triangle, so $$p^2+q^2=4\cdot5^2=100$$. Since one diagonal is at most $$6$$ and the other is at least $$6$$, the sum is maximized when the smaller diagonal is $$6$$. The other diagonal is then $$\sqrt{100-36}=8$$, giving a maximum sum of $$14$$.

The consecutive blocks whose sums are divisible by $$11$$ are $$4+8+10=22$$, $$8+10+16+21=55$$, $$8+10+16+21+25+30=110$$, and $$25+30=55$$. There are no other such consecutive blocks. Hence the number of blocks is $$4$$.

For each integer $$k$$, a subset either contains $$k$$ and contributes a factor $$1/k$$ or does not contain it and contributes a factor $$1$$. Thus the sum over all subsets, including the empty subset, is $$\prod_{k=1}^{17}\left(1+\frac{1}{k}\right)=2\cdot\frac{3}{2}\cdots\frac{18}{17}=18$$. Removing the empty subset contribution $$1$$ gives $$17$$.

Modulo $$x^2-1$$, we have $$x^2=1$$. Therefore every even power of $$x$$ is congruent to $$1$$, so the polynomial is congruent to $$1+1+1+1-6=-2$$. Hence the remainder is $$-2$$.

Every face of the rectangular box gives four acute triangles whose vertices are the four corners of that face. There are $$6$$ faces, giving $$24$$ such face-triangles, but each acute triangle is counted on three faces because its three edges lie on face diagonals. Thus the number of distinct acute triangles is $$24/3=8$$.

The subtraction gives a number consisting of $$50$$ leading digits equal to $$1$$ followed by $$50$$ digits forming the corresponding complement pattern. Direct subtraction gives the digit sum $$450$$. Therefore the correct option is $$450$$.

Cross multiplication gives $$49(m+n)=4(m^2+mn+n^2)$$. The positive integer solutions are $$\{m,n\}=\{6,10\}$$, which can be verified directly because $$49(16)=4(36+60+100)$$. Hence $$m+n=16$$.

Counting all connected triples in the two-block stamp arrangement gives $$28$$ triples contained in the four-cell blocks, $$12$$ additional triples involving the joining region, and $$2$$ further triples crossing the central connection. Hence the total number is $$28+12+2=42$$.

Let the winner receive $$x$$ votes. The other four candidates then receive $$x-9,x-13,x-18,x-25$$ votes. Thus $$5x-65=320$$, giving $$x=77$$. The lowest count is $$77-25=52$$.

The maximum score is $$15\cdot5+10\cdot6=135$$. A score of $$129$$ is obtained by leaving one six-mark question unanswered, $$128$$ by making one incorrect answer in the first negative-mark group, and $$127$$ by making one incorrect answer in the second negative-mark group. Thus $$126$$ is the only listed score that cannot be achieved.

The expression equals $$\left(1+A\right)\left(1+M\right)\left(1+T\right)\left(1+I\right)-1-\left(A+M+T+I\right)$$. Under the condition that the four positive integers sum to $$10$$, the maximum occurs at $$A=M=3$$ and $$T=I=2$$. The value is $$4\cdot4\cdot3\cdot3-1-10=133$$.

Write the number as $$100X+10Y+Z$$ and the quotient as $$10Z+X$$. The division condition gives $$100X+10Y+Z=8(10Z+X)+Y$$, hence $$92X+9Y=79Z$$. The only digit solution is $$X=4$$, $$Y=3$$, $$Z=5$$, so the number is $$435$$.

Let the digit sum of $$N$$ be $$s$$. Then $$N-s=s^2$$, so $$N=s^2+s$$. Checking possible digit sums up to $$27$$ gives $$s=12$$, which produces $$N=12^2+12=156$$. The digit sum of $$156$$ is indeed $$12$$, verifying the answer.

The known fourth column has sum $$30$$ and the known fourth row has sum $$31$$, so the ten line sums must be the consecutive values $$30$$ through $$39$$. Testing the remaining unused values in the diagonal and row constraints forces the entry at $$*$$ to be $$16$$. The completed construction then satisfies the required ten consecutive line sums.

Let the escalator move at $$r$$ steps per second and let the total number of steps be $$N$$. The two journeys give $$N=20(r+1)$$ and $$N=16(r+2)$$. Equating them gives $$20r+20=16r+32$$, so $$r=3$$ and therefore $$N=20(4)=80$$.

If the top row has $$m+1$$ coins and the bottom row has $$n$$ coins, the number of coins is $$\frac{n(n+1)-m(m+1)}{2}=2015$$. Hence $$\left(n-m\right)\left(n+m+1\right)=4030$$. The largest feasible value is $$n-m=62$$, obtained with $$n=63$$ and $$m=1$$, so the tallest tower has $$62$$ rows.

Let the radius of circles $$A$$ and $$B$$ be $$r$$, and let the relevant horizontal distance between the centres be $$x$$. From the tangency geometry, $$x^2=4(1-r)$$ and also $$x^2+2x=2r$$. Eliminating $$x$$ gives $$3x^2+4x-4=0$$, so the positive solution is $$x=\frac{2}{3}$$. Substitution gives $$r=\frac{8}{9}=0.8888888889$$.

For a power to equal $$1$$, either the exponent is zero, the base is one, or the base is negative one with an even exponent. The exponent zero gives $$x=5,6$$, the base one gives $$x=1,4$$, and the base negative one gives $$x=2,3$$ with an even exponent. Thus there are $$6$$ distinct integer solutions.

Because exactly seven moves are required, only the directed moves toward the target that can still reach $$B$$ in the remaining moves are relevant. Counting paths through the intermediate layers gives contributions $$1(1+3)+3(1+3+5)+5(3+5+4)+4(5+4)$$. This totals $$4+27+60+36=127$$ paths.

Let $$\angle BAC=x$$. Since $$AB=AC$$, the base angles give $$\angle1=\angle2=90^\circ-\frac{x}{2}$$. The equal-length conditions create the $$60^\circ$$ angles shown in the construction, giving $$\angle3=120^\circ-x$$ and $$\angle5=\angle4=30^\circ+\frac{x}{2}$$. Using $$\angle7+\angle5=\angle1$$ gives $$30^\circ+\frac{x}{2}+30^\circ+\frac{x}{2}=90^\circ-\frac{x}{2}$$, hence $$x=20^\circ$$.

Dividing the second equation by the first gives $$\frac{x^3+y^3}{xy(x+y)}=\frac{7}{6}$$. Since $$x^3+y^3=(x+y)(x^2-xy+y^2)$$, this gives $$6x^2+6y^2-13xy=0$$, or $$\left(3x-2y\right)\left(2x-3y\right)=0$$. Taking $$x=\frac{2y}{3}$$ in the first equation gives $$x^3=4$$, $$y^3=\frac{27}{2}$$ and $$x^3y^3=54$$, so the required value is $$2(4)+54+2\left(\frac{27}{2}\right)=89$$.