NMTC BHASKARA Junior Level 2026 Solved Question Paper

Since $$x, y, z$$ are distinct and each satisfies $$t^3 + t + 1 = 0$$, they are the three roots of that cubic, so $$x + y + z = 0$$, $$xy + yz + zx = 1$$ and $$xyz = -1$$. Writing $$p_k = x^k + y^k + z^k$$, Newton's identities give $$p_1 = 0$$, $$p_2 = -2$$, $$p_3 = -3$$ and the recurrence $$p_k = -p_{k-2} - p_{k-3}$$ for $$k \geq 4$$. This yields $$p_4 = 2$$, $$p_5 = 5$$, $$p_6 = 1$$, $$p_7 = -7$$, $$p_8 = -6$$ and finally $$p_9 = -p_7 - p_6 = 7 - 1 = 6$$.

For $$n = 1$$ the left side is $$\frac{8 - 2}{6 - 3} = 2$$, so $$n = 1$$ works. The equation is equivalent to $$8^n - 2^n = 2 \times 6^n - 2 \times 3^n$$, and for $$n = 2$$ the two sides are $$60$$ and $$54$$, with the left side growing faster afterwards because $$8^n$$ dominates. Hence the equality holds for exactly one natural number.

The cuts create cones similar to the whole cone with linear ratios $$\frac{1}{3}$$, $$\frac{2}{3}$$ and $$1$$, so the cumulative volumes from the vertex are in the ratio $$1^3 \colon 2^3 \colon 3^3 = 1 \colon 8 \colon 27$$. Subtracting successive cumulative volumes gives $$v_1 = 1$$, $$v_2 = 8 - 1 = 7$$ and $$v_3 = 27 - 8 = 19$$. Hence $$v_1 \colon v_2 \colon v_3 = 1 \colon 7 \colon 19$$.

Since $$OC = OD$$, triangle OCD is isosceles, so $$\angle OCD = \angle ODC = 32^\circ$$ and $$\angle COD = 116^\circ$$. The mid-point M of chord CD lies on the diameter AB, so AB is perpendicular to CD and C, D are symmetric about AB, making arc AC and arc AD each $$58^\circ$$. The inscribed angle $$\angle DBE = 35^\circ$$ gives arc DE $$= 70^\circ$$, so arc EB $$= 180^\circ - 58^\circ - 70^\circ = 52^\circ$$, i.e. $$\angle EOB = 52^\circ$$. As $$OE = OB$$, $$x = \frac{180 - 52}{2} = 64$$.

After 10 days the remaining food would have fed the original $$n$$ soldiers for 20 more days, which is $$20n$$ soldier-days. With $$n + 50$$ soldiers it lasts 16 days, so $$20n = 16(n + 50)$$. This gives $$4n = 800$$ and $$n = 200$$.

Take $$B = (0,0)$$, $$A = (0,5)$$, $$C = (12,0)$$, so $$AC = 13$$ and the line AC is $$5x + 12y - 60 = 0$$. The exterior squares put $$D = (-5,5)$$ and $$E = (12,-12)$$, so $$h_1 = \frac{|-25 + 60 - 60|}{13} = \frac{25}{13}$$ and $$h_2 = \frac{|60 - 144 - 60|}{13} = \frac{144}{13}$$. Their sum is $$\frac{169}{13} = 13$$, so $$\frac{h_1 + h_2}{AC} = \frac{13}{13} = 1$$.

The equation is $$2^{\frac{\sqrt{x}}{2}} - 2^{\frac{\sqrt{y}}{2}} = 48$$, and writing $$48 = 2^4(2^2 - 1)$$ shows the only way to express 48 as a difference of two powers of 2 is $$2^6 - 2^4$$. Hence $$\frac{\sqrt{x}}{2} = 6$$ and $$\frac{\sqrt{y}}{2} = 4$$, giving $$\sqrt{x} = 12$$ and $$\sqrt{y} = 8$$, so $$x = 144$$ and $$y = 64$$. Therefore $$x + y = 208$$.

Completing squares gives $$(x + 1)^2 + (y + 3)^2 = 0$$, and a sum of two real squares vanishes only when each is zero. Hence $$x = -1$$ and $$y = -3$$. Then $$\frac{x + y}{x - y} = \frac{-4}{2} = -2$$.

From $$2x - 1 = \sqrt{101}$$ we get $$4x^2 - 4x + 1 = 101$$, so $$x^2 = x + 25$$. Then $$x^3 = x \cdot x^2 = x^2 + 25x = 26x + 25$$, so $$x^3 - 26x - 23 = 25 - 23 = 2$$. Therefore the required value is $$2^5 = 32$$.

The radii to Q and R are perpendicular to the two tangents, so the angle between the tangents and the central angle are supplementary, giving $$\angle QOR = 180^\circ - 72^\circ = 108^\circ$$, so arc QR is $$108^\circ$$. Q and R lie on the semicircle from A to B, so arc AQ plus arc RB $$= 180^\circ - 108^\circ = 72^\circ$$. The angle between the two intersecting chords equals half the sum of the intercepted arcs, so $$\angle RPB = \frac{72^\circ}{2} = 36^\circ$$.

The arithmetic mean 25 gives a sum of $$50$$ and the geometric mean 24 gives a product of $$24^2 = 576$$. A quadratic with these roots is $$x^2 - (\text{sum})x + (\text{product}) = 0$$. Hence the equation is $$x^2 - 50x + 576 = 0$$.

For $$x \neq 0$$ and $$x \neq 7$$ the left side simplifies as $$\frac{3x(x - 7)}{x(x - 7)} = 3$$, so the equation becomes $$3 = x - 4$$, giving $$x = 7$$. But $$x = 7$$ makes the original denominator zero, so it is not admissible. Hence there is no real solution.

The recorded total is $$6 \times 88 = 528$$, and correcting the entry gives $$528 - 86 + 68 = 510$$. The correct average is $$\frac{510}{6} = 85$$, so $$a = 8$$ and $$b = 5$$. Hence $$a^2 + b^2 = 64 + 25 = 89$$.

The only ways to write 25 as a sum of two squares are $$25 = 0 + 25 = 9 + 16$$. From $$0 + 25$$ we get the four points $$(\pm 5, 0)$$ and $$(0, \pm 5)$$, and from $$9 + 16$$ we get the eight points $$(\pm 3, \pm 4)$$ and $$(\pm 4, \pm 3)$$. Altogether there are 12 lattice points.

The recurrence is $$f(n + 1) = f(n) + \frac{1}{2}$$ with $$f(1) = \frac{1}{2}$$, so $$f(n) = \frac{n}{2}$$. Grouping the sum in pairs, each pair contributes $$f(2k) - f(2k - 1) = \frac{1}{2}$$. There are 50 such pairs, so the total is $$50 \times \frac{1}{2} = 25$$.

By the extended sine rule, $$\frac{AB}{\sin C} = 2R$$ where R is the circumradius. Here $$\frac{5}{\sin 30^\circ} = \frac{5}{\frac{1}{2}} = 10 = 2R$$. Hence $$R = 5$$ cm.

Put $$p = \sqrt{5}$$, $$q = \sqrt{12}$$, $$r = \sqrt{19}$$, so the numerator is $$pq + pr + qr + q^2 = (p + q)(q + r)$$ and the denominator is $$(p + q) + (q + r)$$. The quotient is therefore $$\frac{1}{\frac{1}{p + q} + \frac{1}{q + r}}$$, and rationalising gives $$\frac{1}{p+q} + \frac{1}{q+r} = \frac{2\sqrt{3} - \sqrt{5}}{7} + \frac{\sqrt{19} - 2\sqrt{3}}{7} = \frac{\sqrt{19} - \sqrt{5}}{7}$$. Hence the expression equals $$\frac{7}{\sqrt{19} - \sqrt{5}} = \frac{\sqrt{19} + \sqrt{5}}{2}$$, so $$a = 19$$, $$b = 5$$, $$c = 2$$ and $$a + b + 7c = 38$$.

Since $$x^2 - 5x + 6 = (x - 2)(x - 3)$$, clearing denominators gives $$3x - 25 = a(x - 2) + b(x - 3)$$. Putting $$x = 3$$ gives $$-16 = a$$, and putting $$x = 2$$ gives $$-19 = -b$$, so $$b = 19$$. Hence $$b - a = 19 + 16 = 35$$.

By Pythagoras, $$AC = \sqrt{18^2 + 24^2} = \sqrt{324 + 576} = 30$$ cm. The two known squares have sides $$\sqrt{225} = 15$$ cm and $$\sqrt{64} = 8$$ cm, and the three side lengths add up to AC, so the third square has side $$30 - 15 - 8 = 7$$ cm. Its area, the shaded region, is $$7^2 = 49$$ $$\text{cm}^2$$.

Write the digits as $$a, a + d, a + 2d, a + 3d$$ with $$a \geq 1$$ and every digit between 0 and 9. For $$d = 0, 1, 2$$ the leading digit can take 9, 6 and 3 values respectively, and $$d \geq 3$$ is impossible. For $$d = -1, -2, -3$$ there are 7, 4 and 1 choices, and $$d \leq -4$$ is impossible. The total is $$9 + 6 + 3 + 7 + 4 + 1 = 30$$.

Every arrangement must be prime, so no digit may be even or equal to 5, leaving digits from $$\{1, 3, 7, 9\}$$. Testing the candidates between 150 and 500 in increasing order, numbers such as 173 and 179 fail because $$371 = 7 \times 53$$ and $$791 = 7 \times 113$$ are composite. The first number that survives is 199, since $$199$$, $$919$$ and $$991$$ are all prime.

Equal arc lengths give $$\frac{60}{360} \times 2\pi r_1 = \frac{45}{360} \times 2\pi r_2$$, so $$r_2 = \frac{4}{3} r_1$$. For orthogonal circles the tangents at a common point are perpendicular, so the radii there are perpendicular and $$d^2 = r_1^2 + r_2^2 = r_1^2 + \frac{16}{9}r_1^2 = \frac{25}{9}r_1^2$$. The required ratio of the two squares is $$\frac{d^2}{r_1^2} = \frac{25}{9}$$, that is $$25 \colon 9$$, which is about $$2.78$$.

The sum of the roots gives $$m + n = -2m$$, so $$n = -3m$$, and the product gives $$mn = 5n$$, so $$m = 5$$ because $$n \neq 0$$. Then $$n = -15$$ and the sum of the roots is $$m + n = 5 - 15 = -10$$, which indeed equals $$-2m$$.

By Viviani's theorem the three perpendicular distances add up to the altitude, so $$h = 6 + 8 + 10 = 24$$ cm. For an equilateral triangle of side $$a$$, $$h = \frac{\sqrt{3}}{2}a$$, so $$a = \frac{48}{\sqrt{3}} = 16\sqrt{3}$$ cm. The area is $$\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \times 768 = 192\sqrt{3}$$ $$\text{cm}^2$$, which is about $$332.55$$ $$\text{cm}^2$$.

In base $$b$$ the factors are $$b + 2$$, $$b + 5$$, $$b + 6$$ and the product is $$3b^3 + b^2 + 4b + 6$$, so $$b^3 + 13b^2 + 52b + 60 = 3b^3 + b^2 + 4b + 6$$. This reduces to $$b^3 - 6b^2 - 24b - 27 = 0$$, which is satisfied by $$b = 9$$. The sum is $$3b + 13 = 40$$ in decimal, and $$40 = 4 \times 9 + 4$$, so in base 9 the sum is written as 44.

The first equation gives $$2^{2x - x - y} = 2^3$$, so $$x - y = 3$$, and the second gives $$3^{2x + 2y - y} = 3^5$$, so $$2x + y = 5$$. Adding, $$3x = 8$$, so $$x = \frac{8}{3}$$ and $$y = -\frac{1}{3}$$. Then $$x^2 + xy + 7y^2 = \frac{64}{9} - \frac{8}{9} + \frac{7}{9} = \frac{63}{9} = 7$$.

To maximise a product with a fixed sum, the parts must be 2s and 3s, with as many 3s as possible, since $$3^2 > 2^3$$ for the same total 6 and a part of 4 or more can be split without loss. As $$1000 = 3 \times 332 + 4$$ and the leftover 4 is taken as $$2 + 2$$, the maximum product is $$3^{332} \times 2^2$$. This uses $$332 + 2 = 334$$ numbers.

Using $$(A + B)(C + D) + (A - B)(C - D) = 2(AC + BD)$$ with $$A = x^2$$, $$B = \frac{1}{y}$$, $$C = y^2$$, $$D = \frac{1}{x}$$, the expression equals $$2\left(x^2y^2 + \frac{1}{xy}\right)$$. Here $$xy = 2026 \times \frac{1}{1013} = 2$$, so $$x^2y^2 = 4$$ and $$\frac{1}{xy} = \frac{1}{2}$$. The value is $$2\left(4 + \frac{1}{2}\right) = 9$$.

The general term is $$\frac{1}{n^2 + n - 2} = \frac{1}{(n - 1)(n + 2)} = \frac{1}{3}\left(\frac{1}{n - 1} - \frac{1}{n + 2}\right)$$ for $$n = 3, 4, 5, \ldots$$. The series telescopes, leaving $$\frac{1}{3}\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = \frac{1}{3} \times \frac{13}{12} = \frac{13}{36}$$. Since 13 and 36 are coprime, $$a + b = 13 + 36 = 49$$.

Let $$n + 17 = p^2$$ and $$n - 40 = q^2$$, so $$p^2 - q^2 = 57$$ and $$(p - q)(p + q) = 57 = 3 \times 19$$. The factor pairs $$1$$ and $$57$$ give $$p = 29$$, $$q = 28$$, while $$3$$ and $$19$$ give $$p = 11$$, $$q = 8$$. The greatest value of $$n - 40$$ is $$28^2 = 784$$, whose square root is 28.