NMTC Bhaskara ( junior level) 9th and 10th Aug 23 2025

Let the number be $$n$$

The number reduced by $$4$$ must be a common multiple of $$16, 24, 36$$, so it is a multiple of their LCM.

So, $$(n-4)=k\times L.C.M[16,24,36]$$ ,where $$k$$ is some natural number

or, $$n=k\times L.C.M[16,24,36]+4$$

Since $$16 = 2^4$$, $$24 = 2^3 \times 3$$ and $$36 = 2^2 \times 3^2$$, the LCM is $$2^4 \times 3^2 = 144$$. 

$$\therefore n=144k+4$$

Now, since $$n$$ is the greatest four-digit number that follows this pattern 

So, we should take $$k=69$$ as any other value greater than $$69$$ makes $$n$$ a five digit number

$$\therefore n=144*69+4=9940$$

Hence, Option B is correct

We are given that $$AP=8$$ units and $$PB=12$$ units

Now, in triangles $$POB$$ and $$DOC$$, we have:

$$\angle BOP=\angle COD$$ [vertically opposite angles]

$$\angle BPO=\angle CDO$$ [alternate interior angle between the parallel lines $$AB$$ and $$CD$$]

$$\angle PBO=\angle OCD$$ [alternate interior angle between the parallel lines $$AB$$ and $$CD$$]

So, triangles $$POB$$ and $$DOC$$ are similar triangles

$$\therefore \dfrac{BO}{OC}=\dfrac{PB}{CD}=\dfrac{3}{5}$$

Let us assume that $$BO$$ is $$3x$$ units long

Then, $$OC$$ is $$5x$$ units long

Now, the area of triangle $$ABC=0.5*20*10=100$$ square units

Also, the area of triangle $$BOP$$ : area of triangle $$ABC=0.5*PB*OB*Sin(\angle ABC):0.5*AB*BC*Sin(\angle ABC)$$

The area of triangle $$BOP$$ : area of triangle $$ABC=0.5*12*3x*Sin(\angle ABC):0.5*20*8x*Sin(\angle ABC)$$

The area of triangle $$BOP$$ : area of triangle $$ABC=9:40$$

We know that area of triangle $$ABC$$ is $$100$$ square units

So, area of triangle $$BOP: 100=9:40$$

Hence, the area of triangle $$BOP$$ is $$=\dfrac{45}{2}$$ square units

And the area of quadrilateral $$BPOC=$$area of triangle $$ABC-$$area of triangle $$BOP$$

The area of quadrilateral $$BPOC=100-\dfrac{45}{2}=\dfrac{155}{2}$$ square units

Hence, $$p=155$$ and $$q=2$$ 

This makes $$p+q=155+2=157$$

Hence, Option C is correct

$$\dfrac{\sqrt[7]{12+x}}{x} + \dfrac{\sqrt[7]{12+x}}{12} = \dfrac{64}{3}\left(\sqrt[7]{x}\right)$$

Taking $$\sqrt[7]{12+x}$$ as common we get

$$\sqrt[7]{12+x} \left( \dfrac{1}{x} + \dfrac{1}{12}\right) = \dfrac{64}{3}\left(\sqrt[7]{x}\right)$$

$$\sqrt[7]{12+x} \left( \dfrac{x+12}{12x}\right) = \dfrac{64}{3}\left(\sqrt[7]{x}\right)$$

Transposing $$12x$$ to the other side of the equation we get

$$\sqrt[7]{12+x} (12+x) = \dfrac{64}{3}\left(12x\sqrt[7]{x}\right)$$

Now, we can $$\sqrt[7]{12+x}$$ as $${(12+x)}^{\frac{1}{7}}$$. Using the exponent rule that $$x^a.x^b = x^{a+b}$$, we can write

$${(12+x)}^{1+\frac{1}{7}} = (64 \times \dfrac{12}{3}) x^{1+\frac{1}{7}}$$

Simplifying the terms we get

$${(12+x)}^{\frac{8}{7}} =( 64 \times 4) x^{\frac{8}{7}}$$

$$64$$ can be expressed as $$2^6$$ and $$4 = 2^2$$. Thus, $$64 \times 4 = 2^8$$

$${(12+x)}^{\frac{8}{7}} = 2^8 x^{\frac{8}{7}}$$

Taking the eighth root we get

$${(12+x)}^{\frac{1}{7}} = 2 x^{\frac{1}{7}}$$

And raising the expression to the power of 7 we get

$$12+x = 128x \implies 127x = 12 \ \text{or}\ x = \dfrac{12}{127} $$

We can observe that $$12,127$$ have no common factors hence their $$gcd = 1$$

Thus $$ b-a = 127-12 = 115$$

We are given the expression: $$(52+6\sqrt{43})^{3/2} - (52-6\sqrt{43})^{3/2}$$

Now, the term $$52+6\sqrt{43}$$ can be rewritten as:

$$52+6\sqrt{43}=9+43+2*3*\sqrt{43}=(3+\sqrt{43})^2$$

Similarly, we can rewrite the term $$52-6\sqrt{43}$$ as:

$$52-6\sqrt{43}=9+43-2*3*\sqrt{43}=(3-\sqrt{43})^2$$ or $$(\sqrt{43}-3)^2$$

But whenever we have $$\sqrt{52-6\sqrt{43}}$$, the correct answer would be $$(\sqrt{43}-3)$$ as the square root is defined in mathematics to give only a positive answer due to functional constraints put on it.

Now, we had:  $$(52+6\sqrt{43})^{3/2} - (52-6\sqrt{43})^{3/2}$$

$$=(\sqrt{52+6\sqrt{43}})^3 - (\sqrt{52-6\sqrt{43}})^3$$

$$=(3+\sqrt{43})^3-(\sqrt{43}-3)^3$$

$$=(27+\sqrt{43}^3+27\sqrt{43}+9*43)-(-27+\sqrt{43}^3+27\sqrt{43}-9*43)$$

$$=2(27+9*43)$$

$$=828$$

Hence, Option D is correct.

TO BE FILLED - figure required. The positions of the points $$B, F, G$$ on the circle relative to the star-shaped chord pattern cannot be determined reliably from the extracted text alone, so the arcs needed for the chord-angle theorem cannot be assigned with confidence.

Using $$\frac{r}{R} = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ with $$B = 90^\circ$$ and $$\frac{r}{R} = \frac{2}{5}$$, solving for $$A$$ (with $$C = 90^\circ - A$$) gives a specific angle, and substituting back yields $$\cot^2\frac{A}{2} + \cot^2\frac{C}{2} = 13$$.

We are given the equations: $$|x-1|+|y-5|=1$$, $$y = 5+|x-1|$$

From the second equation, we get:

$$|x-1|=y-5$$

Substituting this into the first equation

$$|x-1|+|y-5|=1$$

$$y-5+|y-5|=1$$

Now, we will have to make cases

CASE 1: $$y\geq5$$

$$y-5+y-5=1$$

or, $$y=\dfrac{11}{2}$$

Then, $$|x-1|=y-5=\dfrac{1}{2}$$

Which gives $$x=\dfrac{3}{2}$$ or $$x=\dfrac{1}{2}$$

Hence, we get $$2$$ solutions from case 1 for $$(x,y)$$, which are $$\left(\dfrac{1}{2},\dfrac{11}{2}\right),\left(\dfrac{3}{2},\dfrac{11}{2}\right)$$

CASE 2: $$y<5$$

$$y-5-(y-5)=1$$

or, $$0=1$$

Which is never true

So, we cannot get any answer from here.

The common $$y = \beta = \frac{11}{2}$$. 

Taking $$\alpha = \frac{1}{2}$$ and $$\gamma = \frac{3}{2}$$ (or vice versa), we get:

$$\alpha+\beta+\gamma = 2 + \frac{11}{2} = \frac{15}{2}$$.

Let the working rates of Ram, Ali and Peter be $$R,A$$ and $$P$$ respectively

$$R+A=\dfrac{1}{30}$$

$$A+P=\dfrac{1}{40}$$

$$R+P=\dfrac{1}{60}$$

Adding the three pairwise rates gives:

$$2(R+A+P) = \frac{1}{30}+\frac{1}{40}+\frac{1}{60} = \frac{3}{40}$$

So the combined rate is $$R+A+P = \frac{3}{80}$$ of the work per day. 

The time for all three working together to finish the whole house is $$=\frac{80}{3} = 26\frac{2}{3}$$ days.

We are given: $$x=\dfrac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$$

Rationalising the RHS

$$x=\dfrac{(\sqrt{a+3b}+\sqrt{a-3b})^2}{(\sqrt{a+3b}-\sqrt{a-3b})(\sqrt{a+3b}+\sqrt{a-3b})}$$

$$x=\dfrac{a+3b+a-3b+2\sqrt{a^2-9b^2}}{a+3b-a+3b}$$

$$x=\dfrac{a+\sqrt{a^2-9b^2}}{3b}$$ ....(1)

Now, squaring both sides

$$x^2=\dfrac{a^2+a^2-9b^2+2a\sqrt{a^2-9b^2}}{9b^2}$$

$$x^2=\dfrac{2a^2+2a\sqrt{a^2-9b^2}-9b^2}{9b^2}$$

$$x^2=\dfrac{2a^2+2a\sqrt{a^2-9b^2}}{9b^2}-1$$

$$x^2+1=\dfrac{2a^2+2a\sqrt{a^2-9b^2}}{9b^2}$$

$$3b(x^2+1)=\dfrac{2a^2+2a\sqrt{a^2-9b^2}}{3b}$$

$$\dfrac{3b(x^2+1)}{a}=\dfrac{2a+2\sqrt{a^2-9b^2}}{3b}$$

$$\dfrac{3b(x^2+1)}{a}=2\times\left(\dfrac{a+\sqrt{a^2-9b^2}}{3b}\right)$$

$$\dfrac{3b(x^2+1)}{a}=2\times x$$ [from 1]

$$\dfrac{3b(x^2+1)}{ax}=2$$

$$\dfrac{3bx^2+3b}{ax}=2$$

The inequality rearranges to $$|x-13| < \frac{9}{4} = 2.25$$ with $$x \neq 13$$. The integers satisfying $$10.75 < x < 15.25$$, excluding $$13$$ [as at $$x=13$$ the denominator of original inequation is undefined], are $$11, 12, 14, 15$$, giving $$4$$ integral solutions.

TO BE FILLED - figure required. The exact chord/tangent connections around points $$A, B, C, D, E$$ cannot be pinned down precisely enough from the extracted description to guarantee the correct angle-chase without the original diagram.

Let $$t = \sqrt{\log_2 x}$$

So $$\log_2 x = t^2$$ and $$\log_2 8x = 3+t^2$$.

Now, the equation becomes $$3t - (3+t^2) + 1 = 0$$

$$t^2-3t+2=0$$

$$t=1$$ or $$t=2$$

So $$x = 2$$ or $$x = 16$$. 

Thus $$\alpha = 2$$, $$\beta = 16$$, and $$\dfrac{\beta}{\alpha} = 8$$.

Let the number be $$n$$

If $$n$$ leaves a remainder of $$4$$ when divided by $$11$$, than we can write $$n$$ as:

$$n=11k+4$$ , where $$k$$ is some natural number

Now, squaring both sides

$$n^2=121k^2+16+88k$$

$$n^2=121k^2+88k+11+5$$

$$n^2=11(11k^2+8k+1)+5$$

So, we can see that now if $$n^2$$ is divided by $$11$$, the leftover $$5$$ becomes the new remainder.

For the first number, with tens digit $$t$$ and units digit $$2t$$, the number is $$12t$$ and the digit sum is $$3t$$, so $$36t^2 = 144$$ gives $$t=2$$ and the number is $$24$$. For the second, with units digit $$u$$ and tens digit $$2u$$, the number is $$21u$$ and the digit sum is $$3u$$, so $$63u^2=567$$ gives $$u=3$$ and the number is $$63$$. The sum of the two numbers is $$24+63 = 87$$.

Letting $$\angle BAE = \angle CDE = \theta$$, the pentagon angle sum $$\theta+(\theta-4)+(\theta-6)+\theta+126=540$$ gives $$\theta=106^\circ$$, so $$\angle ABC=102^\circ$$ and $$\angle BCD=100^\circ$$. With $$P$$ on the extensions of $$AB$$ and $$DC$$, triangle $$PBC$$ gives $$\angle BPC = \angle ABC+\angle BCD-180^\circ=22^\circ$$; similarly triangle $$QED$$ gives $$\angle EQD = (180^\circ-\angle AED)+(180^\circ-\angle CDE)-180^\circ = 52^\circ$$. Halving both by the bisectors gives $$11^\circ$$ and $$26^\circ$$ at $$P$$ and $$Q$$ in triangle $$PRQ$$, so $$\angle PRQ = 180^\circ-11^\circ-26^\circ = 143^\circ$$.

Since $$(b+c)^2 = (b-c)^2+4bc = 64+4(-a^2-16) = -4a^2$$, and a square cannot be negative, we need $$a=0$$, forcing $$b+c=0$$ as well. Combined with $$b-c=8$$, this gives $$b=4, c=-4$$. Then $$a^{2025}+b^{2025}+c^{2025} = 0+4^{2025}+(-4)^{2025} = 0$$, since $$2025$$ is odd.

Substituting $$f(x)$$ into itself gives $$f(f(x)) = \frac{2025^2 x}{2026x+1}$$. Setting this equal to $$2025^2$$ gives $$x = 2026x+1$$, so $$-2025x = 1$$ and $$x = -\frac{1}{2025}$$.

Cubing both sides gives $$16-x^3 = (4-x)^3$$, and expanding causes the $$x^3$$ terms to cancel, leaving $$x^2-4x+4=0$$, i.e. $$(x-2)^2=0$$. By Vieta's formulas the sum of the (repeated) roots of this quadratic is $$4$$.

With the quadrant centred at $$A$$ of radius $$r_1$$ and the quadrant centred at $$D$$ of radius $$r_2$$ touching at $$B$$, requiring the midpoint $$F$$ of $$CE$$ to lie exactly on the second arc forces $$r_1 = r_2(\sqrt{2}-1)$$. Placing coordinates and computing the angle between $$EC$$ and $$ED$$ with this ratio gives $$\angle CED = 67.5^\circ$$.

For integer roots summing to $$6$$ with pairwise-product sum $$11$$, the only positive integer triple is $$1, 2, 3$$, since $$1+2+3=6$$ and $$1\cdot2+2\cdot3+3\cdot1=11$$. Comparing with Vieta's formulas, the product of roots is $$k-6$$, and $$1\times2\times3=6$$, so $$k=12$$.

The number $$100a+10b+5$$ is divisible by $$9$$ exactly when its digit sum $$a+b+5$$ is a multiple of $$9$$, i.e. $$a+b=4$$ or $$a+b=13$$ (with $$a\geq 1$$). For $$a+b=4$$ there are $$4$$ valid pairs ($$a=1,2,3,4$$), and for $$a+b=13$$ there are $$6$$ valid pairs ($$a=4$$ to $$9$$), giving $$4+6=10$$ numbers in total.

Since $$a^2 = 2025^3-2023^3 = (2024+1)^3-(2024-1)^3$$

$$a^2=2024^2+1+3*2024+3*(2024)^2-[2024^2-1+3*(2024)-3*(2024)^2]$$

$$a^2=6*(2024)^2+2$$

$$a^2-2=6*(2024)^2$$

$$\dfrac{a^2-2}{6}=2024^2$$

$$\sqrt{\dfrac{a^2-2}{6}}=2024$$

The $$14$$ students who solved fully and correctly represent $$100\%-12\%-32\%=56\%$$ of the class. So the total number of students is $$\frac{14}{0.56}=25$$.

$$|2a^3-3a^2-2a+1| - |2a^3-3a^2-3a-2025|$$

or, $$|2a^3-3a^2-2a+1| - |2a^3-3a^2-3a-a|$$

or, $$|2a^3-3a^2-2a+1| - |2a^3-3a^2-4a|$$

Let us assume that $$t=2a^3-3a^2-2a$$

or, $$t=a(2a^2-3a-2)$$

or, $$t=a(2a+1)(a-2)$$

or, $$t=(2a+1)(a)(a-2)$$

Since, $$a=2025$$, we will get $$t>2a$$

Now, we have: $$|2a^3-3a^2-2a+1| - |2a^3-3a^2-4a|$$

Substituting value of $$t$$, we get:

$$|t+1|-|t-2a|$$

Since both of the quantities inside the modulus are positive

So, we get:

$$t+1-t+2a$$

$$=2a+1=4051$$

TO BE FILLED - figure required. The exact tangency relationship between the rectangular frame and the circular hoop (which side of the frame touches the hoop, and where) cannot be determined with confidence from the extracted figure, and different reasonable assumptions give very different, non-matching radii.

We are given the expression: $$\frac{16(n^2-n-1)^2}{2n-1}$$

Now, let's try to rewrite the numerator as:

$$16(n^2-n-1)^2=(2n-1)Q(n)+r$$ where $$Q(n)$$ is the quotient and $$r$$ is the remainder

So, now to find the remainder, lets put $$n=\dfrac{1}{2}$$ on both sides

$$16(\dfrac{1}{4}-\dfrac{1}{2}-1)^2=(0)Q(\dfrac{1}{2})+r$$

$$25=r$$

So, $$16(n^2-n-1)^2=(2n-1)Q(n)+25$$

Now, $$\dfrac{16(n^2-n-1)^2}{2N-1}=Q(x)+\dfrac{25}{2n-1}$$

Now, $$Q(n)$$ is going to be a cubic function, which will give integral results

Only, term $$\dfrac{25}{2n-1}$$ will be troublesome

So $$2n-1$$ must divide $$25$$. The positive divisors of $$25$$ are $$1, 5, 25$$, giving $$n=1, 3, 13$$, so there are exactly $$3$$ such values of $$n$$

But we still need to verify these values so that we get natural numbers as a solution

Putting $$n=1$$ in $$\frac{16(n^2-n-1)^2}{2n-1}$$, we get :$$16$$

Putting $$n=3$$ in $$\frac{16(n^2-n-1)^2}{2n-1}$$, we get :$$80$$

Putting $$n=13$$ in $$\frac{16(n^2-n-1)^2}{2n-1}$$, we get :$$15,376$$

All of these are natural numbers, hence there are $$3$$ possible values of $$n$$

The first equation gives $$\log_2 x = 2\log_2 y$$, i.e. $$x = y^2$$ 

Now, both $$x,y>0$$ as they are the arguments of the log function in the given equations

Substituting into the second equation gives $$y^4-2y^2-8=0$$,

Which can be factored as:

$$(y^2-4)(y^2+2)=0$$

So $$y^2=4$$, 

$$y=2$$ (taking the positive root) 

$$x=4$$. 

This gives exactly $$1$$ valid solution pair which is $$(4,2)$$

TO BE FILLED - figure required. It is not possible to determine reliably from the extracted figure which points are collinear with $$Q$$ (chord $$SR$$ versus line $$BR$$ through the tangent $$PA$$), so the similar-triangle relation coming from $$AR \parallel PB$$ cannot be set up with confidence.

The non-water (solid) content is $$2\%$$ of $$20$$ kg, i.e. $$0.4$$ kg, and this stays constant. After evaporation the solid content is $$5\%$$ of the new weight, so the new weight is $$\frac{0.4}{0.05}=8$$ kg. The reduction in weight is $$20-8=12$$ kg.

With first term $$a$$ and common ratio $$r$$, the conditions give $$ar^2(r-1)=24$$ and $$ar(1+r)=6$$, which combine to $$r^2-5r-4=0$$, so $$r=\frac{5+\sqrt{41}}{2}$$ for the increasing case. Solving for $$a$$ and computing $$ar+ar^2+ar^3$$ with this ratio gives a sum of approximately $$35.1$$.