NMTC ramanujan ( inter level) 11th and 12th Aug 23 2025

The four numbers $$7-m, 7-n, 7-p, 7-q$$ are distinct integers whose product is 4, so they must be $$1, -1, 2, -2$$ in some order. This gives $$m, n, p, q$$ equal to $$6, 8, 5, 9$$ in some order, and all four are different positive integers as required. Hence $$m+n+p+q = 6+8+5+9 = 28$$.

Take each digit as the peak and multiply the number of smaller digits before it by the number of smaller digits after it. For the digits $$1, 3, 2, 5, 9, 7, 6, 8, 4$$ these products are $$0, 1 \times 1, 1 \times 0, 3 \times 1, 4 \times 4, 4 \times 2, 4 \times 1, 6 \times 1, 0$$. Adding them gives $$1+0+3+16+8+4+6 = 38$$.

A three digit number has digit product at most $$9^3 = 729$$, so $$N$$ has four digits. A leading digit 1 forces the product to be at most 729 again, so the leading digit is 2 and the remaining three digits must have product greater than 500, the smallest such ending being 789 with $$2 \times 7 \times 8 \times 9 = 1008$$. Hence $$N = 2789$$, so $$x = 27$$, $$y = 89$$ and $$x + y = 116$$.

For an odd number of consecutive odd integers the sum equals the count times the middle term, so the middle term is $$\frac{2025^{2025}}{2025} = 2025^{2024}$$. There are 1012 terms above the middle one and consecutive odd numbers differ by 2. Hence the largest term is $$2025^{2024} + 2 \times 1012 = 2025^{2024} + 2024$$.

The three corner triangles $$APR$$, $$BQP$$, $$CRQ$$ each have sides 1 and 5 enclosing an angle of $$60^\circ$$, so each has area $$\frac{1}{2} \times 1 \times 5 \times \sin 60^\circ = \frac{5\sqrt{3}}{4}$$. Since $$[ABC] = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3}$$, we get $$[PQR] = 9\sqrt{3} - 3 \times \frac{5\sqrt{3}}{4} = \frac{21\sqrt{3}}{4}$$. The required ratio is $$9\sqrt{3} \colon \frac{21\sqrt{3}}{4} = 36 \colon 21 = 12 \colon 7$$.

No digit can be 0, and at least two digits must be equal. Equilateral cases give 9 numbers. For a pair $$(x, x, y)$$ with $$y \neq x$$ the triangle inequality needs $$y < 2x$$, and counting valid $$y$$ for $$x = 1$$ to $$9$$ gives $$0+2+4+6+8+8+8+8+8 = 52$$ multisets, each arrangeable in 3 ways. The total is $$9 + 3 \times 52 = 165$$.

The numbers below 10000 with digit sum 7 correspond to solutions of $$d_1+d_2+d_3+d_4 = 7$$ with each digit at most 9, and there are $$\binom{10}{3} = 120$$ of them. So the list continues with the five digit numbers $$10006, 10015, 10024, 10033, 10042$$ as the 121st to 125th entries. Hence the 125th number is 10042.

Since $$\angle ABD = \angle ABC + \angle CBD = B + \frac{A}{2}$$, the chord $$AD = 2R \sin\left(B + \frac{A}{2}\right) = 2 \sin\left(B + \frac{A}{2}\right)$$ when $$R = 1$$. Then $$AD \cos \frac{A}{2} = 2 \sin\left(B+\frac{A}{2}\right)\cos\frac{A}{2} = \sin(A+B) + \sin B = \sin C + \sin B$$, and similarly for the other two terms. Adding gives $$2(\sin A + \sin B + \sin C)$$, so the required value is 2.

Put $$t = 10^{31}$$, so that $$\frac{10^{93}}{10^{31}+3} = \frac{t^3}{t+3} = t^2 - 3t + 9 - \frac{27}{t+3}$$. Since $$0 < \frac{27}{t+3} < 1$$, we get $$N = t^2 - 3t + 8$$. Both $$t^2$$ and $$3t$$ are multiples of 100, so $$N$$ leaves remainder 8 on division by 100.

Writing $$u = x - 199$$, the lattice points satisfy $$u^2 + y^2 = 199^2$$. Since 199 is a prime of the form $$4k+3$$, it is not a sum of two nonzero squares and neither is $$199^2$$, so the only solutions have $$u = 0$$ or $$y = 0$$. These give the four points $$(199, \pm 199)$$, $$(0, 0)$$ and $$(398, 0)$$, so the count is 4.

If the roots are positive integers $$r, s, t$$ then $$r+s+t = p+1$$, $$rs+st+tr = \frac{71p-1}{5}$$ and $$rst = \frac{66p-1}{5}$$. Eliminating $$p$$ and searching the resulting integer conditions gives the single possibility $$r, s, t = 1, 17, 59$$ with $$p = 76$$, since $$1+17+59 = 77 = p+1$$, $$1079 = \frac{71 \times 76 - 1}{5}$$ and $$1003 = \frac{66 \times 76 - 1}{5}$$. As this is the only such value, the required sum is 76.

Substituting $$x - 4 = 1$$, that is $$x = 5$$, makes the right side equal to $$a_0 + a_1 + \cdots + a_{20}$$. The left side is a geometric series equal to $$\frac{1+x^{21}}{1+x}$$. At $$x = 5$$ this gives $$\frac{1+5^{21}}{6}$$.

Fix $$a$$ and count the pairs $$b < c$$ with $$b + c = 15 - a$$ and $$b > a$$. For $$a = 1$$ the pairs are $$(2,12), (3,11), (4,10), (5,9), (6,8)$$, for $$a = 2$$ there are 4, for $$a = 3$$ there are 2 and for $$a = 4$$ there is 1, while $$a \ge 5$$ gives none. The total is $$5+4+2+1 = 12$$.

Adding 1 to both sides factorises the left side as $$(6m-1)(5n-1) = 2020$$. Since $$6m-1$$ must leave remainder 5 on division by 6, the divisors of $$2020 = 2^2 \times 5 \times 101$$ allow only $$6m-1 = 5$$ with $$5n-1 = 404$$, giving $$m = 1$$ and $$n = 81$$. Then $$30mn - 5m - 6n = 2430 - 5 - 486 = 1939$$.

The maximum score is $$1+2+3+4+5+6 = 21$$, and a student who misses a set of questions with total value $$k$$ scores $$21 - 2k$$. Every value of $$k$$ from 0 to 21 is attainable as a subset sum of $$\{1,2,3,4,5,6\}$$, so there are 22 possible scores. By the pigeonhole principle $$\left\lceil \frac{100}{22} \right\rceil = 5$$ students must share a score.

The $$n$$th term has numerator $$\frac{n(n+1)(2n+1)}{6}$$ and denominator $$n(n+1)$$, so it simplifies to $$\frac{2n+1}{6}$$. Summing for $$n = 1$$ to 60 gives $$\frac{1}{6}\left(2 \times \frac{60 \times 61}{2} + 60\right) = \frac{3660+60}{6}$$. This equals $$\frac{3720}{6} = 620$$.

Using the sum of cubes factorisation, $$3^{21}+1 = (3^7+1)(3^{14} - 3^7 + 1) = 2188 \times 4780783$$. Here $$2188 = 2^2 \times 547$$ and $$4780783 = 7^2 \times 43 \times 2269$$. The largest prime factor is therefore 2269.

Colouring the sectors of a circle so that neighbours differ is the same as properly colouring a cycle with 10 vertices. The number of ways with $$k$$ colours is $$(k-1)^n + (-1)^n (k-1)$$. With $$n = 10$$ and $$k = 3$$ this is $$2^{10} + 2 = 1026$$.

There are $$9 + 9^2 + 9^3 = 819$$ special numbers with at most three digits, so the required number is the $$2025 - 819 = 1206$$th four digit special number. Writing $$1206 - 1 = 1205$$ in base 9 gives the digits $$1, 5, 7, 8$$, and adding 1 to each digit converts it to the special number. This gives 2689.

With $$e_1 = a+b+c = 0$$, Newton's identities give $$p_2 = -2e_2$$, $$p_3 = 3e_3$$ and $$p_5 = -5e_2e_3$$. The condition $$p_3 = p_5$$ with $$e_3 = abc \neq 0$$ forces $$3 = -5e_2$$, so $$e_2 = -\frac{3}{5}$$ and $$a^2+b^2+c^2 = -2e_2 = \frac{6}{5}$$. Hence $$\frac{5}{a^2+b^2+c^2} = \frac{25}{6}$$, which is approximately 4.1667.

Multiplying by $$x$$ gives the quartic $$x^4 - 4x - 1 = 0$$, which factorises as $$\left(x^2 + \sqrt{2}x + 1 + \sqrt{2}\right)\left(x^2 - \sqrt{2}x + 1 - \sqrt{2}\right) = 0$$. The first factor has negative discriminant while the second has discriminant $$4\sqrt{2} - 2 > 0$$, so the two real roots come from $$x^2 - \sqrt{2}x + 1 - \sqrt{2} = 0$$. Thus $$\alpha + \beta = \sqrt{2}$$ and $$(\alpha+\beta)^2 = 2$$.

Subtracting the first equation from the second gives $$xyz - (xy+yz+zx) + (x+y+z) - 1 = 0$$, which is exactly $$(x-1)(y-1)(z-1) = 0$$, so one variable equals 1. Taking $$x = 1$$ reduces the first equation to $$y + z + yz = 2024$$, that is $$(1+y)(1+z) = 2025$$. The factorisation $$2025 = 3 \times 675$$ gives $$y = 2$$ and $$z = 674$$, the largest value the maximum can take.

If all three primes were odd then $$pqr - (pq+qr+rp)$$ would be even, but 2025 is odd, so one prime is 2. Putting $$p = 2$$ gives $$qr - 2q - 2r = 2025$$, that is $$(q-2)(r-2) = 2029$$. Since 2029 is prime the only possibility is $$q = 3$$ and $$r = 2031$$, so $$p+q+r = 2 + 3 + 2031 = 2036$$.

For a cyclic quadrilateral with sides $$a, b, c, d$$ and area $$K$$, the circumradius satisfies $$R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$$. Here $$s = \frac{21}{2}$$ and $$K = \sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{\frac{9075}{16}}$$, while $$(ab+cd)(ac+bd)(ad+bc) = 55 \times 55 \times 49 = 148225$$. Hence $$R^2 = \frac{148225}{9075} = \frac{49}{3}$$ and $$3R^2 = 49$$.

The term with $$k$$ blocks of 53 is the digit string 2 followed by $$k$$ copies of 53 followed by 4, and it has $$2k+2$$ digits. Splitting the even length string into two digit groups gives $$25, 35, 35, \ldots, 35, 34$$, and a number is divisible by 99 exactly when these groups add to a multiple of 99, that is $$59 + 35(k-1) \equiv 0 \pmod{99}$$. Since $$35 \times 17 \equiv 1 \pmod{99}$$, this gives $$k - 1 \equiv 17 \times 40 \equiv 86$$, so $$k = 87$$ and the number of digits is $$2 \times 87 + 2 = 176$$.

With equal sides $$a$$ and base $$b = 16 - 2a$$, the triangle inequality needs $$b < 2a$$, leaving $$(5,5,6)$$, $$(6,6,4)$$ and $$(7,7,2)$$. Heron's formula with $$s = 8$$ gives areas $$\sqrt{8 \times 3 \times 3 \times 2} = 12$$, $$\sqrt{8 \times 2 \times 2 \times 4} = 8\sqrt{2}$$ and $$\sqrt{8 \times 1 \times 1 \times 6} = 4\sqrt{3}$$. The largest of these is 12.

Pairing the factors gives $$(a+b+c)(a+b-c) = (a+b)^2 - c^2 = 2ab$$ and $$(a-b+c)(a-b-c) = (a-b)^2 - c^2 = -2ab$$, using $$a^2+b^2 = c^2$$. Their product is $$-4a^2b^2$$, and since $$ab = c$$ this equals $$-4c^2$$. Dividing by $$c^2$$ gives $$-4$$.

For a right triangle with legs $$a, b$$ and hypotenuse $$c$$ the inradius is $$r = \frac{a+b-c}{2}$$, so $$a+b-c = 24$$. Substituting $$c = a+b-24$$ into $$a^2+b^2 = c^2$$ gives $$(a-24)(b-24) = 288$$, and then $$c = 24 + d + \frac{288}{d}$$ where $$d = a - 24$$. This is largest when $$d = 1$$, giving $$a = 25$$, $$b = 312$$ and $$c = 313$$.

Since $$M = \frac{A+B+C}{3}$$ and $$N = \frac{A+B+D}{3}$$, we get $$MN = \frac{CD}{3}$$. Both triangles are right angled, at $$C$$ and at $$D$$, so with $$A = (0,0)$$ and $$B = (25,0)$$ the feet give $$C = (1.96, 6.72)$$ and $$D = (16, -12)$$. Then $$CD = \sqrt{14.04^2 + 18.72^2} = 23.4$$ and $$MN = \frac{23.4}{3} = 7.8$$.

Adding $$\frac{1}{2}$$ to both sides shows $$a_n + \frac{1}{2} = 3\left(a_{n-1} + \frac{1}{2}\right)$$, so $$a_n = \frac{3^{n+1}-1}{2}$$. Hence $$a_{11} = \frac{3^{12}-1}{2} = \frac{531440}{2} = 265720$$. Since $$265720 = 97 \times 2739 + 37$$, the remainder is 37.