IOQM Previous Papers Sept 7 2025

60% of $$x$$ is written as $$\frac{60}{100}\times x$$. We are told this equals $$40$$, so:

$$\frac{60}{100}\,x = 40$$

Solve for $$x$$ by multiplying both sides by $$\frac{100}{60}$$:

$$x = 40 \times \frac{100}{60} = \frac{4000}{60} = \frac{200}{3}$$

Now we need $$x\%$$ of $$60$$, that is $$\frac{x}{100}\times 60$$:

$$\frac{\frac{200}{3}}{100} \times 60 = \frac{200}{3} \times \frac{60}{100} = \frac{12000}{300} = 40$$

Hence, $$x\%$$ of $$60$$ is $$40$$.

Final Answer: 40

Multiples of $$3$$ that lie in $$[1,100]$$ are
$$3,6,9,\dots ,99$$.
The count of these numbers is given by $$\left\lfloor \frac{100}{3} \right\rfloor = 33$$.

A number divisible by both $$3$$ and $$2$$ must be divisible by their LCM, $$6$$. Multiples of $$6$$ in $$[1,100]$$ are
$$6,12,18,\dots ,96$$.
The count of these numbers is $$\left\lfloor \frac{100}{6} \right\rfloor = 16$$.

Numbers that are divisible by $$3$$ but not by $$2$$ are obtained by subtracting the above counts:
$$33 - 16 = 17$$.

Therefore, the required number of positive integers $$n \le 100$$ that satisfy the given condition equals $$17$$.

Answer: 17

Let the integer side lengths of the rectangle be $$l$$ and $$b$$, so that the area condition gives

$$l \times b = 20$$

Because $$l,b \in \mathbb{Z^+}$$, list all factor pairs of $$20$$:

$$(1,20),\;(2,10),\;(4,5)$$

For each pair, compute the perimeter $$P = 2(l+b)$$:

• $$l=1,\;b=20 \;\Rightarrow\; P = 2(1+20)=42$$
• $$l=2,\;b=10 \;\Rightarrow\; P = 2(2+10)=24$$
• $$l=4,\;b=5 \;\Rightarrow\; P = 2(4+5)=18$$

The smallest value among $$42, 24,$$ and $$18$$ is $$18$$.

Hence, the minimum possible perimeter is 18.

Let the equal sides be $$a,a$$ and the third side (base) be $$b$$, all in integers.

Perimeter condition: $$2a + b = 23$$ $$-(1)$$

Triangle inequality for an isosceles triangle:
$$a + a \gt b \;\;\Longrightarrow\;\; 2a \gt b$$ $$-(2)$$
The other two inequalities, $$a + b \gt a$$ and $$a + b \gt a$$, simply give $$b \gt 0$$, which is already implied by positive side lengths.

From $$-(1)$$, express $$b$$ in terms of $$a$$:
$$b = 23 - 2a$$ $$-(3)$$

Positive base: $$b \gt 0 \;\Longrightarrow\; 23 - 2a \gt 0 \;\Longrightarrow\; a \lt 11.5 \;\Longrightarrow\; a \le 11$$

Using $$-(2)$$ with $$-(3)$$:
$$2a \gt 23 - 2a$$
$$4a \gt 23$$
$$a \gt 5.75 \;\Longrightarrow\; a \ge 6$$

Therefore $$a$$ can take any integer value from $$6$$ to $$11$$ inclusive:
$$a = 6,\,7,\,8,\,9,\,10,\,11$$

Each admissible $$a$$ gives a unique integer $$b$$ via $$b = 23 - 2a$$, and all satisfy $$2a \gt b$$:
6 cases in total.

Hence, the number of isosceles integer-sided triangles with perimeter $$23$$ is 06.

Let the three-digit number be $$\overline{abc}$$, where $$a,b,c$$ are its decimal digits.

Digit conditions:
1. $$a \neq 0$$ because the number is three-digit, so $$a \in \{1,2,\dots ,9\}$$.
2. $$b,c \in \{0,1,\dots ,9\}$$ (any decimal digit).
3. Given condition: $$c = a + b$$.

Since $$c$$ itself must be a single digit, we need $$a + b \le 9$$. Thus the problem reduces to counting ordered pairs $$(a,b)$$ satisfying

$$a \in \{1,2,\dots ,9\}, \qquad b \in \{0,1,\dots ,9\}, \qquad a + b \le 9$$.

Fix a particular value of $$a$$ and determine how many $$b$$ are possible.

Continuing similarly, we get the following counts:

$$ \begin{aligned} a = 1 &\Rightarrow 9 \text{ choices for } b\\ a = 2 &\Rightarrow 8 \text{ choices for } b\\ a = 3 &\Rightarrow 7 \text{ choices for } b\\ &\ \vdots \\ a = 8 &\Rightarrow 2 \text{ choices for } b\\ a = 9 &\Rightarrow 1 \text{ choice for } b\\ \end{aligned} $$

The numbers of choices form an arithmetic progression $$9,8,7,\dots ,1$$.

Total number of valid pairs $$(a,b)$$ is therefore the sum of this progression:

$$ \text{Total} = 9 + 8 + 7 + \dots + 1 = \frac{9 \times 10}{2} = 45. $$

For every such pair $$(a,b)$$ the digit $$c$$ is fixed as $$c = a + b$$, so each pair gives exactly one valid three-digit number.

Hence the total count of three-digit numbers satisfying $$c = a + b$$ is $$\mathbf{45}$$.

Let the person’s age in the year $$2012$$ be $$n^2$$ years, where $$n$$ is a positive integer.

Thirteen years later, in $$2025$$, the age becomes $$n^2+13$$. We are told this is also a perfect square, say $$k^2$$. Hence

$$k^2 - n^2 = 13 \quad -(1)$$

Factorising the left side using the identity $$k^2-n^2=(k-n)(k+n)$$ gives

$$(k-n)(k+n)=13 \quad -(2)$$

Because $$13$$ is prime, the only pair of positive integers whose product is $$13$$ is $$1$$ and $$13$$. Thus

$$k-n = 1 \quad\text{and}\quad k+n = 13$$

Adding the two equations: $$2k = 14 \;\Longrightarrow\; k = 7$$.
Substituting back: $$n = k-1 = 6$$.

Therefore

Age in $$2012 = n^2 = 6^2 = 36 \text{ years}$$
Age in $$2025 = k^2 = 7^2 = 49 \text{ years}$$.

Let $$m$$ be the number of years after $$2025$$ when the age first becomes a perfect cube. We require

$$49 + m = p^3$$ for some integer $$p$$, with $$m \gt 0$$ and as small as possible.

The cubes just above $$49$$ are

$$4^3 = 64,\quad 5^3 = 125, \ldots$$

The first one that exceeds $$49$$ is $$64$$, and

$$64 - 49 = 15$$.

Thus the earliest perfect-cube age is $$64$$ years, reached $$m = 15$$ years after $$2025$$.

Smallest value of $$m = 15$$.

Answer: 15

Let the two real numbers be $$x$$ and $$y$$.

Their sum is given to be a positive integer $$n$$, so
$$x + y = n \qquad -(1)$$

The sum of their squares is
$$x^{2} + y^{2} = n + 1012 \qquad -(2)$$

For any two real numbers with a fixed sum, the expression $$x^{2}+y^{2}$$ attains its minimum value when the numbers are equal. Using the identity
$$x^{2}+y^{2} = (x+y)^{2} - 2xy,$$
and the AM-GM inequality $$xy \le \left(\frac{x+y}{2}\right)^{2},$$ we obtain

Minimum value of $$x^{2}+y^{2}$$ = $$\dfrac{(x+y)^{2}}{2} = \dfrac{n^{2}}{2}$$ when $$x = y = \dfrac{n}{2}$$.

Because $$x^{2}+y^{2}$$ in our problem equals $$n+1012,$$ it must be at least this minimum value:
$$n+1012 \;\ge\; \dfrac{n^{2}}{2} \qquad -(3)$$

Re-arrange inequality (3):
$$\dfrac{n^{2}}{2} - n - 1012 \le 0$$
$$n^{2} - 2n - 2024 \le 0 \qquad -(4)$$

Solve the quadratic equality $$n^{2}-2n-2024 = 0$$:
Discriminant $$\Delta = (-2)^{2} - 4(1)(-2024) = 4 + 8096 = 8100$$
$$\sqrt{\Delta} = 90$$

Roots are
$$n = \dfrac{2 \pm 90}{2} \;=\; 46 \text{ or } -44$$

Inequality (4) is satisfied for $$-44 \le n \le 46$$. Since $$n$$ is required to be a positive integer, the largest admissible value is $$n = 46$$.

Check attainability: take $$x = y = \dfrac{n}{2} = 23$$.
Then $$x+y = 46$$ and $$x^{2}+y^{2} = 23^{2}+23^{2} = 2 \times 529 = 1058 = 46 + 1012$$, satisfying both conditions.

Hence the maximum possible value of $$n$$ is 46.

Final Answer: 46

Let the four vertices be labelled in order as $$A,B,C,D$$ so that the sides are $$AB,\,BC,\,CD,\,DA$$ and the extra restriction is on the diagonal $$AC$$. A proper colouring demands that the two endpoints of every one of these five edges receive different colours.

Step 1 - Choose a colour for $$A$$. Any of the four colours (red, blue, green, yellow) may be used.
    Number of choices for $$A = 4$$.

Step 2 - Colour $$B$$. Edge $$AB$$ forces $$B$$ to be different from $$A$$, leaving three colours.
    Number of choices for $$B = 3$$.

Step 3 - Colour $$C$$. Vertex $$C$$ is adjacent to both $$B$$ (edge $$BC$$) and $$A$$ (diagonal $$AC$$), so its colour must differ from the colours on $$A$$ and $$B$$. The colours on $$A$$ and $$B$$ are already different, so exactly two colours remain available.
    Number of choices for $$C = 4-2 = 2$$.

Step 4 - Colour $$D$$. Vertex $$D$$ is adjacent to $$C$$ (edge $$CD$$) and to $$A$$ (edge $$DA$$). Hence $$D$$ must avoid the colours used on $$A$$ and $$C$$. Since those two colours are distinct, again two colours remain.
    Number of choices for $$D = 4-2 = 2$$.

Total colourings = $$4 \times 3 \times 2 \times 2 = 48$$.

Thus the number of ways to colour the quadrilateral so that every side and the diagonal $$AC$$ have differently coloured endpoints is 48.

The total surface area (TSA) of a closed right circular cylinder of radius $$r$$ and height $$h$$ is
$$\text{TSA}=2\pi r(r+h).$$

The volume is
$$\text{Volume}= \pi r^{2}h.$$

The condition in the problem states that these two are numerically equal:
$$2\pi r(r+h)=\pi r^{2}h.$$

Cancel $$\pi$$ from both sides:
$$2r(r+h)=r^{2}h \; \; -(1)$$

Divide by $$r\,(r\gt0)$$:
$$rh=2(r+h). \; \; -(2)$$

Rearrange $$-(2)$$ to isolate a factor of $$(r-2)$$:
$$rh-2h=2r \;\;\Longrightarrow\;\;h(r-2)=2r. \; \; -(3)$$

Let $$d=r-2$$. Because $$r$$ is a positive integer and $$r\ge3$$ (else $$h$$ would be non-positive), $$d$$ is a positive integer. Substitute $$r=d+2$$ into $$-(3)$$:
$$h\,d = 2(d+2)=2d+4.$$

Solve for $$h$$:
$$h = 2+\frac{4}{d}. \; \; -(4)$$

For $$h$$ to be an integer, $$\dfrac{4}{d}$$ must be an integer, so $$d$$ must be a positive divisor of $$4$$.
Possible $$d$$ values: $$1,2,4$$.

Compute the corresponding volumes $$V=\pi r^{2}h$$:

Case 1: $$V=\pi\,(3)^{2}(6)=54\pi$$
Case 2: $$V=\pi\,(4)^{2}(4)=64\pi$$
Case 3: $$V=\pi\,(6)^{2}(3)=108\pi$$

The smallest positive integer $$k$$ for which the volume equals $$k\pi$$ is therefore $$k=54$$.

Answer: 54

Let the two-digit number be $$\overline{ab}=10a+b=N\;(\;10\le N\le 99,\;a\neq 0\;).$$

The five-digit number is $$\overline{abcab}=10000a+1000b+100c+10a+b.$$

Simplify it in terms of $$N$$ and $$c$$.

$$\begin{aligned} \overline{abcab}&=10000a+1000b+100c+10a+b\\ &=100(100a+10b+c)+\bigl(10a+b\bigr)\\ &=100\bigl(10N+c\bigr)+N\\ &=1000N+100c+N\\ &=1001N+100c\;.\tag{1} \end{aligned}$$

Condition for divisibility by $$N$$:
From (1), $$\overline{abcab}=1001N+100c$$ is divisible by $$N$$ ⇔ $$N$$ divides $$100c$$ (because $$N$$ always divides $$1001N$$).

Thus we must have $$N\;|\;100c.\tag{2}$$

Prime factors available in $$100c$$:
$$100=2^{2}\,5^{2},\qquad c\in\{1,2,\ldots ,9\}\implies c$$ contributes only the primes $$2,3,5,7.$$
So $$N$$ can contain no prime other than $$2,3,5,7.$$ (We must also remember that the final number must not be divisible by $$13$$.)

Our goal is to maximise the digit-sum
$$S= a+b+c= \tfrac12\bigl(2a+2b\bigr)+c = 2(a+b)+c.\tag{3}$$

Check each value of $$c$$ (only two-digit divisors of $$100c$$ are listed).

Two-digit divisors of 900: $$10,12,15,18,20,25,30,36,45,50,60,75,90.$$ For each divisor compute $$S=2(a+b)+9.$$

The largest digit sum occurs for $$N=75\;(a=7,\;b=5):$$
$$S=2(7+5)+9=24+9=33.\tag{4}$$

Verify the number:
$$\overline{abcab}=75975.$$
Divisibility: $$75975\div 75=1013\;(\text{integer}),\qquad 75975\div 13=5844\text{ remainder }3\;(\text{not divisible}).$$

For $$c=8$$ the best divisor is $$64\;(a=6,b=4)\Rightarrow S=28\lt 33.$$br/> For $$c=7$$ the best divisor is $$35\;(a=3,b=5)\Rightarrow S=23\lt 33.$$br/> For $$c=6$$ the best divisor is $$75\;(a=7,b=5)\Rightarrow S=30\lt 33.$br/> For $$c=1,2,3,4,5$$ the maximum $$S$$ is even smaller.

If $$c=0$$, then (1) becomes $$$$\overline{ab0ab}$$=1001N.$$ Since $$1001=7$$\cdot$$ 11$$\cdot$$ 13,$$ every such number is automatically divisible by $$13,$$ so all $$c=0$$ cases are rejected.

Hence the maximal attainable digit-sum is the value in (4), namely $$33.$$

Answer: 33

Let the quadrilateral be $$ABCD$$ in that order with the given consecutive sides

$$AB = a = 4\sqrt{3},\;\;BC = b = 9,\;\;CD = c = \sqrt{3}$$

and with the interior angles

$$\angle ABC = 30^{\circ},\;\;\angle BCD = 90^{\circ}.$$

We place the figure on the Cartesian plane by the following convenient choices.

Step 1: Assign vectors for the sides
Take vertex $$B$$ at the origin and draw $$\overrightarrow{BA}$$ along the positive $$x$$-axis.

$$\overrightarrow{BA}= (4\sqrt{3},\,0).$$

The side $$BC$$ makes an angle of $$30^{\circ}$$ with $$BA$$, so

$$\overrightarrow{BC}= \bigl(9\cos 30^{\circ},\,9\sin 30^{\circ}\bigr) =\left(\frac{9\sqrt3}{2},\,\frac{9}{2}\right).$$

Step 2: Draw $$CD$$ perpendicular to $$BC$$
Because $$\angle BCD = 90^{\circ},\;\overrightarrow{CD}$$ is perpendicular to $$\overrightarrow{BC}$$. The clockwise unit vector perpendicular to $$BC$$ is $$\bigl(\sin 30^{\circ},-\cos 30^{\circ}\bigr)=\left(\frac12,\,-\frac{\sqrt3}{2}\right).$$ Multiplying by $$|CD|=c=\sqrt3$$ gives

$$\overrightarrow{CD}= \left(\frac{\sqrt3}{2},\,-\frac{3}{2}\right).$$

(The anticlockwise direction would give a mirror image of the same convex quadrilateral; the acute angle between the diagonals is the same.)

Step 3: Express the two diagonals as vectors

Diagonal $$BD$$: $$\overrightarrow{BD}= \overrightarrow{BC}+\overrightarrow{CD} =\left(\frac{9\sqrt3}{2}+\frac{\sqrt3}{2},\,\frac92-\frac32\right) =\left(5\sqrt3,\,3\right).$$

Diagonal $$AC$$: $$\overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{BC} -\overrightarrow{BA} =\overrightarrow{BC}-\overrightarrow{BA} =\left(\frac{9\sqrt3}{2}-4\sqrt3,\,\frac{9}{2}-0\right) =\left(\frac{\sqrt3}{2},\,\frac{9}{2}\right).$$

Step 4: Use the dot-product to find the angle $$x$$ between the diagonals

Dot product $$\overrightarrow{BD}\,\cdot\,\overrightarrow{AC} = (5\sqrt3)\left(\frac{\sqrt3}{2}\right)+3\left(\frac{9}{2}\right) =\frac{15}{2}+\frac{27}{2}=21.$$

Magnitudes $$|\overrightarrow{BD}|=\sqrt{(5\sqrt3)^2+3^2} =\sqrt{75+9}=\sqrt{84}=2\sqrt{21},$$ $$|\overrightarrow{AC}|=\sqrt{\left(\frac{\sqrt3}{2}\right)^2+\left(\frac92\right)^2} =\sqrt{\frac34+\frac{81}{4}} =\sqrt{\frac{84}{4}}=\sqrt{21}.$$

Hence $$\cos x =\frac{\overrightarrow{BD}\cdot\overrightarrow{AC}} {|\overrightarrow{BD}|\;|\overrightarrow{AC}|} =\frac{21}{\bigl(2\sqrt{21}\bigr)\bigl(\sqrt{21}\bigr)} =\frac{21}{42}=\frac12.$$

Therefore $$x = \arccos\!\left(\frac12\right)=60^{\circ}.$$

The acute angle between the diagonals is $$\boxed{60}$$.

The functional equation is

$$f(mn+1)=f(m)\,f(n)-f(n)-m+2$$ for all integers $$m,n$$, with the initial value $$f(0)=1$$.

Step 1: Evaluate $$f(1)$$
Put $$m=0$$ in the equation:

$$f(0\cdot n+1)=f(0)\,f(n)-f(n)-0+2$$
$$\Longrightarrow\;f(1)=1\cdot f(n)-f(n)+2=2$$ (independent of $$n$$).

Step 2: Guess a simple candidate
The values $$f(0)=1$$ and $$f(1)=2$$ suggest the linear rule $$f(k)=k+1$$. Check it directly:

LHS: $$f(mn+1)=mn+1+1=mn+2$$
RHS: $$f(m)f(n)-f(n)-m+2=(m+1)(n+1)-(n+1)-m+2$$
$$=(mn+m+n+1)-n-1-m+2=mn+2$$

Both sides are equal, so $$f(k)=k+1$$ satisfies the functional equation.

Step 3: Show $$f(k)=k+1$$ for all non-negative integers
We prove by induction.

Base cases: $$f(0)=1=0+1$$ and $$f(1)=2=1+1$$.

Inductive step: Assume $$f(m)=m+1$$ for some $$m\ge 0$$. Take $$n=1$$ in the functional equation:

$$f(m\cdot1+1)=f(m)f(1)-f(1)-m+2$$
$$\Rightarrow f(m+1)=f(m)\cdot2-2-m+2=2f(m)-m$$

Substitute the induction hypothesis $$f(m)=m+1$$:

$$f(m+1)=2(m+1)-m=m+2=(m+1)+1$$

Thus the statement holds for $$m+1$$. By induction, $$f(k)=k+1$$ for every non-negative integer $$k$$.

Step 4: Sum of the first $$N$$ values
$$\sum_{k=1}^{N}f(k)=\sum_{k=1}^{N}(k+1)=\sum_{k=1}^{N}k+\sum_{k=1}^{N}1$$
$$=\frac{N(N+1)}{2}+N=\frac{N(N+3)}{2}$$

Step 5: Impose the condition
We need

$$\frac{N(N+3)}{2}\lt100$$ $$\Longrightarrow N(N+3)\lt200$$ $$\Longrightarrow N^{2}+3N-200\lt0$$

The positive root of $$N^{2}+3N-200=0$$ is

$$N=\frac{-3+\sqrt{9+800}}{2}=\frac{-3+\sqrt{809}}{2}\approx12.72$$

Therefore $$N$$ must be the greatest integer strictly less than $$12.72$$, i.e. $$N=12$$.

Hence the largest positive integer satisfying the given inequality is 12.

Think of the three coupon-pairs as three distinct “objects”:

$$P_1 = (1,2), \; P_2 = (3,4), \; P_3 = (5,6).$$

Each pair must be placed in one envelope and no two pairs may share an envelope, so we are choosing an injective function from $$\{P_1,P_2,P_3\}$$ to the six envelopes $$\{1,2,3,4,5,6\}$$.

First count the total number of injective assignments with no restriction. This is a permutation of 6 envelopes taken 3 at a time:

$$\text{Total} = {}^{6}P_{3} = 6 \times 5 \times 4 = 120.$$

Define events that violate the given condition “no coupon goes into the envelope bearing its own number”:

$$\begin{aligned} E_1 &: P_1 \text{ goes into envelope }1\text{ or }2,\\ E_2 &: P_2 \text{ goes into envelope }3\text{ or }4,\\ E_3 &: P_3 \text{ goes into envelope }5\text{ or }6. \end{aligned}$$

We require an arrangement in which none of $$E_1,E_2,E_3$$ occurs. Use the Principle of Inclusion-Exclusion (PIE).

Example for $$E_1$$:
  • Choose an envelope for $$P_1$$ (two choices: 1 or 2).
  • Place $$P_2,P_3$$ in any two of the remaining five envelopes: $$^{5}P_{2}=5 \times 4=20$$ ways.
Thus $$|E_1| = 2 \times 20 = 40.$$ By symmetry $$|E_2| = 40,\; |E_3| = 40.$$

Sum over single events:

$$\sum |E_i| = 40+40+40 = 120.$$

Example for $$E_1\cap E_2$$:
  • Choose envelope for $$P_1$$: 2 ways (1 or 2).
  • Choose envelope for $$P_2$$: 2 ways (3 or 4). (These choices are in disjoint sets, so they never clash.)
  • Now $$P_3$$ can go to any of the remaining four envelopes: 4 ways.
Hence $$|E_1\cap E_2| = 2 \times 2 \times 4 = 16.$$ Similarly $$|E_1\cap E_3| = 16, \; |E_2\cap E_3| = 16.$

Sum over pairwise intersections:

$$$$\sum$$ |E_i$$\cap$$ E_j| = 16+16+16 = 48.$$

Each pair goes into one of its two “forbidden” envelopes, all sets being disjoint:

$$|E_1$$\cap$$ E_2$$\cap$$ E_3| = 2 $$\times$$ 2 $$\times$$ 2 = 8.$$

Apply PIE:

$$ $$\begin{aligned}$$ $$\text{Valid arrangements}$$ &= 120 \;-\; 120 \;+\; 48 \;-\; 8\\[4pt] &= 40. \end{aligned} $$

Therefore, the required number of ways is 40.

Let us compare the two quadratics through the difference
$$h(x)=f(x)-4\,g(x)$$

Because $$\frac{f(-2)}{g(-2)}=4$$ and $$\frac{f(3)}{g(3)}=4$$, we have
$$f(-2)=4g(-2) \quad\text{and}\quad f(3)=4g(3)$$

Hence $$h(-2)=0$$ and $$h(3)=0$$. The polynomial $$h(x)$$ is of degree at most $$2$$ and possesses the roots $$x=-2$$ and $$x=3$$, therefore

$$h(x)=k\,(x+2)(x-3)$$ for some constant $$k$$.

Express $$f(x)$$ in terms of $$g(x)$$:
$$f(x)=4\,g(x)+k\,(x+2)(x-3)$$

To determine $$k$$, use the data at $$x=7$$:
$$f(7)=4\,g(7)+k\,(7+2)(7-3)$$

Given $$f(7)=12$$ and $$g(7)=-6$$, substitute:
$$12 = 4(-6) + k\,(9)(4)$$
$$12 = -24 + 36k$$
$$36k = 36$$
$$k = 1$$

Now find $$f(5)$$ using $$g(5)=2$$:

$$f(5)=4\,g(5)+1\,(5+2)(5-3)$$
$$f(5)=4(2)+7\cdot2$$
$$f(5)=8+14$$
$$f(5)=22$$

Therefore, the required value is 22.

Let the adjacent sides of the parallelogram be represented by the vectors
$$\vec{u} \;=\; \overrightarrow{MT},\qquad \vec{v} \;=\; \overrightarrow{MI}.$$

Denote their lengths by
$$|\vec{u}| = MT = x,\qquad |\vec{v}| = MI = y.$$

The question gives the relation $$MI = \dfrac1{MT},$$ hence
$$y = \dfrac1x.$$

Let $$\theta$$ be the angle between the two adjacent sides (between $$\vec{u}$$ and $$\vec{v}$$). The area of a parallelogram is $$|\vec{u}\times\vec{v}| = xy\sin\theta$$, so

$$x\Bigl(\dfrac1x\Bigr)\sin\theta = \dfrac{40}{41}\;\;\Longrightarrow\;\;\sin\theta = \dfrac{40}{41}.$$

Because $$\sin\theta = \dfrac{40}{41},$$ the cosine may be either
$$\cos\theta = \pm\dfrac{9}{41}.$$

The diagonal whose length we need is $$\overrightarrow{MA} = \vec{u} + \vec{v},$$ so

$$d^2 = |\vec{u}+\vec{v}|^{\,2} = |\vec{u}|^{2} + |\vec{v}|^{2} + 2\,\vec{u}\!\cdot\!\vec{v} = x^{2} + \dfrac1{x^{2}} + 2\!\left(x\cdot\dfrac1x\right)\cos\theta = x^{2} + \dfrac1{x^{2}} + 2\cos\theta.$$

To find the least possible value of $$d^2$$ we must minimise both terms: 1. $$x^{2} + \dfrac1{x^{2}}$$, and 2. $$2\cos\theta$$ (choose the smaller of the two possible cosines).

$$d^{2}_{(1)} = x^{2} + \dfrac1{x^{2}} + \dfrac{18}{41}.$$ The minimum of $$x^{2} + \dfrac1{x^{2}}$$ occurs at $$x = 1$$ (by differentiating or AM ≥ GM), giving
$$d^{2}_{(1)\,\min} = 2 + \dfrac{18}{41} = \dfrac{100}{41}.$$

$$d^{2}_{(2)} = x^{2} + \dfrac1{x^{2}} - \dfrac{18}{41}.$$ Again the minimum of $$x^{2} + \dfrac1{x^{2}}$$ is attained at $$x = 1$$, so
$$d^{2}_{(2)\,\min} = 2 - \dfrac{18}{41} = \dfrac{64}{41}.$$

Since $$\dfrac{64}{41} \lt \dfrac{100}{41},$$ the least possible value of $$d^{2}$$ is

$$d^{2}_{\min} = \dfrac{64}{41}.$$

This fraction is already in its lowest terms, so $$a = 64,\; b = 41.$$ Therefore
$$|a - b| \;=\; |64 - 41| \;=\; 23.$$

Answer: 23

We have to arrange the multiset $$\{2,2,3,3,3,4,4,4,4\}$$ (two 2’s, three 3’s and four 4’s) in a row of nine positions so that at least one $$3$$ appears to the right of the right-most $$4$$.

Step 1 : Fix the positions of the two 2’s.
Choose any 2 of the 9 places for the identical 2’s.
Number of ways = $$\binom{9}{2}=36$$.

Step 2 : Work inside the remaining 7 positions.
After removing the two 2’s, seven places are left and must be filled with three 3’s and four 4’s.

The required condition “there is a 3 to the right of the right-most 4” means that, within these seven places, the last (right-most) of them must be a 3. Fix that last place as 3. The other six places have to accommodate the remaining
  • two 3’s
  • four 4’s

Number of distinct arrangements of those six symbols = $$\dfrac{6!}{2!\,4!}=15$$.

Step 3 : Combine the choices.
For each of the 36 ways to place the 2’s, there are 15 admissible ways to arrange the 3’s and 4’s.
Hence $$N = 36 \times 15 = 540$$.

Step 4 : Find the remainder mod 100.
$$540 \bmod 100 = 40$$.

Therefore the required remainder is 40.

Place the triangle on the coordinate plane so that $$B(0,0)$$, $$C(2,0)$$ and $$A(0,1)$$.
Then $$BC$$ is the $$x$$-axis, $$AB$$ is the $$y$$-axis and $$AC$$ has equation $$y = 1-\dfrac{x}{2}$$.

Points $$D$$ and $$E$$ lie on $$BC$$ with $$E$$ between $$C$$ and $$D$$, so their abscissae satisfy $$0 \lt d \lt e \lt 2$$.
Let $$D(d,0)$$ and $$E(e,0)$$.

Since $$DEFG$$ is a square with $$DE$$ on $$BC$$, set the side length $$s = DE = e-d \;(\gt 0)$$.
Choose the square so that the remaining two vertices lie above $$BC$$ (inside the triangle):

$$F(e,s),\quad G(d,s).$$

1. Condition for $$F$$: it lies on $$AC$$.
 Using $$y = 1-\dfrac{x}{2}$$,

$$s = 1-\dfrac{e}{2}\; \Longrightarrow\; e = 2-2s.$$

2. From $$s=e-d$$ we get

$$d = e-s = (2-2s)-s = 2-3s.$$

3. Condition for $$G$$: it lies on the circle with centre $$A(0,1)$$ and radius $$AB=1$$, i.e.

$$(d)^2 + (s-1)^2 = 1.$$(1)

Substitute $$d = 2-3s$$ into (1):

$$(2-3s)^2 + (s-1)^2 = 1$$ $$\Longrightarrow 4-12s+9s^2 + s^2-2s+1 = 1$$ $$\Longrightarrow 10s^2 -14s +5 = 1$$ $$\Longrightarrow 10s^2 -14s +4 = 0$$ $$\Longrightarrow 5s^2 -7s +2 = 0.$$

4. Solve the quadratic:

$$s = \dfrac{7 \pm \sqrt{49-40}}{10} = \dfrac{7 \pm 3}{10} \; \Longrightarrow\; s = 1 \;\text{or}\; s = \dfrac{2}{5}.$$

Because $$0 \lt d = 2-3s$$ and $$d \lt e \lt 2$$, we must have $$s \lt \dfrac{2}{3}$$, so the only admissible value is

$$s = \dfrac{2}{5}.$$

5. Area of the square:

$$\text{Area} = s^2 = \left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}.$$

This is already in lowest terms, so $$m = 4,\; n = 25$$ and

$$m+n = 4+25 = 29.$$

Final answer: 29.

The mapping to be studied is
$$f(n)=n^{\,n}\pmod{7}, \qquad n\in\mathbb{N}.$$
We want the smallest positive integer $$T$$ such that

$$f(n+T)=f(n)\quad\text{for every positive integer }n.$$

The residue of $$n^{\,n}$$ (mod $$7$$) depends on

• the base $$n$$ modulo $$7$$, and
• the exponent $$n$$ modulo the Euler totient $$\phi(7)=6$$ (by Fermat’s Little Theorem, because any number that is not a multiple of $$7$$ satisfies $$a^{6}\equiv1\pmod 7$$).

Therefore the value of $$f(n)$$ is completely determined by the ordered pair

$$\bigl(n\bmod 6,\;n\bmod 7\bigr).$$

This pair repeats whenever $$n$$ is increased by a common multiple of $$6$$ and $$7$$, i.e. by $$\operatorname{lcm}(6,7)=42$$. Hence

$$f(n+42)=f(n)\quad\forall\,n\in\mathbb{N},$$

so $$T=42$$ is a period.

Next, we prove that no smaller $$T$$ can work.

Choose $$n=7k$$ for some $$k\ge1$$. Then $$n\equiv0\pmod7$$ and $$f(n)=0$$. Since $$7\nmid T$$, we have $$n+T\not\equiv0\pmod7$$, so the base of $$(n+T)^{\,n+T}$$ is not divisible by $$7$$; its remainder cannot be $$0$$. Thus $$f(n+T)\ne f(n)$$, contradicting the requirement. Hence $$7$$ must divide every admissible $$T$$.

Take $$n$$ that satisfies the simultaneous congruences

$$n\equiv0\pmod6,\qquad n\equiv3\pmod7.$$

(Such an $$n$$ exists by the Chinese Remainder Theorem.) Here $$n\equiv3\pmod7$$, so the base is $$3$$, which is coprime to $$7$$. Because $$n\equiv0\pmod6$$, we have
$$f(n)=3^{\,n}\equiv3^{\,0}\equiv1\pmod7.$$ Since $$6\nmid T$$, we get $$n+T\not\equiv0\pmod6$$, so $$(n+T)\bmod6\ne0$$ while $$n+T\equiv3\pmod7$$ (because $$7\mid T$$). Thus
$$f(n+T)=3^{\,n+T}\equiv3^{\,r}\pmod7,$$ where $$r=(n+T)\bmod6\ne0$$. The possible powers $$3^{\,r}\pmod7$$ for $$r=1,2,3,4,5$$ are $$3,2,6,4,5$$—none equals $$1$$. Hence $$f(n+T)\ne f(n)$$, contradicting the requirement. Therefore $$6$$ must also divide every admissible $$T$$.

Combining the two cases, any valid period $$T$$ must be a multiple of both $$6$$ and $$7$$, i.e. a multiple of $$42$$. The smallest such positive integer is $$42$$ itself.

Hence the least positive period is
$$\boxed{42}.$$

Call the game position $$(b,r)$$ when $$b$$ blue and $$r$$ red marbles remain, with $$r\ge 1$$ (the game ends as soon as $$r=0$$).

A position is winning if the player whose turn it is can force a win from there, otherwise it is losing. We build the table of positions by using the standard rule:

• A position is winning $$\Longleftrightarrow$$ it has at least one move to a losing position.
• A position is losing $$\Longleftrightarrow$$ every legal move goes to a winning position.

Step 1: Base positions
If $$r=1$$, the current player can simply pick that last red marble and wins immediately. Hence for every $$b\ge 0$$, the position $$(b,1)$$ is winning.

Step 2: Positions with $$r=2$$

• $$\,(0,2):$$ only move is $$\rightarrow(0,1)$$ (winning), so $$(0,2)$$ is losing.
• $$\,(1,2):$$ moves are $$\rightarrow(1,1)$$ (winning) and $$\rightarrow(0,2)$$ (losing).  There is a move to a losing position, so $$(1,2)$$ is winning.
• $$\,(2,2):$$ moves are to $$(2,1)$$ (winning) and $$(1,2)$$ (winning).  Both give the opponent a winning position, so $$(2,2)$$ is losing.
• Continuing likewise, one finds that for $$r=2$$ the position is losing when $$b$$ is even and winning when $$b$$ is odd.

Step 3: Positions with $$r=3$$
Work exactly as above beginning with $$(0,3)$$. The pattern reverses: $$(b,3)$$ is losing when $$b$$ is odd and winning when $$b$$ is even.

Step 4: General parity rule for $$r\ge 2$$
Induction on $$r$$ now shows the rule:

For every $$r\ge 2$$, the position $$(b,r)$$ is losing iff $$b$$ and $$r$$ have the same parity (either both even or both odd). Otherwise it is winning.

The induction step is simple: assume the rule for $$r-1$$. • If $$b$$ and $$r$$ differ in parity, the move “remove a red marble” goes to $$(b,r-1)$$ where parity matches, hence a losing position for the opponent, making $$(b,r)$$ winning.
• If $$b$$ and $$r$$ have the same parity, removing a blue marble keeps parity the same, and removing a red marble changes parity; both successor positions are winning for the opponent by the induction hypothesis, so $$(b,r)$$ is losing.

Step 5: Counting losing starting positions
Armaan starts from $$(m,n)$$ with $$1\le m,n\le 11$$.

• When $$n=1$$, the position is always winning (Step 1).
• For $$n\ge 2$$, the start is losing exactly when $$m$$ and $$n$$ have the same parity.

Count these losing pairs:

Even $$n$$ in the range 2-11: $$2,4,6,8,10$$ (5 values). For each, even $$m$$ can be $$2,4,6,8,10$$ (5 choices). Total = $$5\times5 = 25$$.

Odd $$n$$ in the range 3-11: $$3,5,7,9,11$$ (5 values). For each, odd $$m$$ can be $$1,3,5,7,9,11$$ (6 choices). Total = $$5\times6 = 30$$.

Total losing starts = $$25+30 = 55$$.

Step 6: Winning starts for Armaan
Total possible pairs = $$11\times11 = 121$$.
Winning positions = $$121-55 = 66$$.

Hence Armaan can force a win for exactly 66 ordered pairs $$(m,n)$$ with $$1\le m,n\le 11$$.

Final Answer: 66

Let the rectangle be $$ABCD$$ with $$A(0,0),\;B(b,0),\;C(b,c),\;D(0,c)$$.
Thus $$AB=b$$ and $$BC=c$$.

Take $$M(x,0)$$ on $$AB$$ and $$N(b,y)$$ on $$BC$$ with $$0\lt x\lt b,\;0\lt y\lt c$$.

Step 1: Using $$MC = CD$$
$$MC^{2} = (b-x)^{2}+c^{2},\quad CD^{2}=b^{2}$$
$$(b-x)^{2}+c^{2}=b^{2}\;\Rightarrow\;c^{2}=2bx-x^{2}\;-(1)$$

Step 2: Using $$MD = MN$$
$$MD^{2}=x^{2}+c^{2},\quad MN^{2}=(b-x)^{2}+y^{2}$$
$$x^{2}+c^{2}=(b-x)^{2}+y^{2}\;\Rightarrow\;y^{2}=4bx-x^{2}-b^{2}\;-(2)$$

Step 3: Circle through $$C,D,M$$
Because $$C(b,c)$$ and $$D(0,c)$$ have the same $$y$$-coordinate, the perpendicular bisector of $$CD$$ is the vertical line $$x=\dfrac{b}{2}$$.
Hence the centre of the circle is $$\left(\dfrac{b}{2},\,k\right)$$ for some $$k$$.

Equal radii to $$D$$ and $$M$$ give
$$\left(\dfrac{b}{2}\right)^{2}+(k-c)^{2}=\left(\dfrac{b}{2}-x\right)^{2}+k^{2}$$ $$\Rightarrow\;2ck-c^{2}=bx-x^{2}\;-(3)$$

Step 4: Point $$N$$ is concyclic
$$N(b,y)$$ lies on the same circle, so $$\left(\dfrac{b}{2}\right)^{2}+(y-k)^{2}=\left(\dfrac{b}{2}\right)^{2}+(k-c)^{2}$$ $$\Rightarrow\;(y-k)^{2}=(k-c)^{2}$$

Rejecting $$y=c$$ (which would force $$N=C$$ and violate $$MC=CD$$), take $$y=2k-c\;-(4)$$.

Step 5: Eliminate $$k,y$$ and solve

From (3): $$k=\dfrac{3bx-2x^{2}}{2c}$$.
Substitute this in (4): $$y=\dfrac{bx-x^{2}}{c}$$.
Insert $$y$$ in (2): $$\left(\dfrac{bx-x^{2}}{c}\right)^{2}=4bx-x^{2}-b^{2}$$.
Use (1) to replace $$c^{2}$$ and let $$t=\dfrac{x}{b}\;(0\lt t\lt 1)$$.

With $$c^{2}=b^{2}(2t-t^{2})$$, the equality becomes $$\dfrac{t^{2}(1-t)^{2}}{2t-t^{2}}=4t-t^{2}-1$$ $$\Longrightarrow\;-2t\left(2t^{2}-4t+1\right)=0$$ $$\Rightarrow\;2t^{2}-4t+1=0$$ $$\Rightarrow\;t=1-\dfrac{\sqrt{2}}{2}\;\;(\text{the root }0\lt t\lt1).$$

Step 6: Compute $$(AB/BC)^{2}$$
$$\left(\dfrac{AB}{BC}\right)^{2}=\left(\dfrac{b}{c}\right)^{2}=\dfrac{1}{2t-t^{2}}.$$ For the obtained $$t$$, $$2t-t^{2}=0.5=\dfrac12$$ $$\therefore\;\left(\dfrac{AB}{BC}\right)^{2}=2=\dfrac{2}{1}.$$

Thus $$m=2,\;n=1,\;m+n=3$$.

Answer: 03.

We have to find the remainder $$R(x)$$ when $$P(x)=x^{2025}$$ is divided by $$Q(x)=x^{4}+x^{3}+2x^{2}+x+1$$ and then evaluate $$R(3)$$.

1. Factorise the divisor
Notice that
$$(x^{2}+1)(x^{2}+x+1)=x^{4}+x^{3}+2x^{2}+x+1=Q(x).$$
Hence $$Q(x)$$ splits into two coprime quadratic factors $$A(x)=x^{2}+1,\qquad B(x)=x^{2}+x+1.$$ Because they are coprime, the Chinese Remainder Theorem (CRT) lets us determine the unique remainder of degree at most 3 from its residues modulo $$A(x)$$ and $$B(x).$$

2. Remainder of $$P(x)$$ modulo $$A(x)=x^{2}+1$$
Inside the quotient ring $$\mathbb{R}[x]/(x^{2}+1)$$ we have $$x^{2}\equiv-1,$$ so the powers of $$x$$ repeat every 4:

$$x^{0}\equiv1,\;x^{1}\equiv x,\;x^{2}\equiv-1,\;x^{3}\equiv -x,\;x^{4}\equiv1,\ldots$$

Since $$2025=4\cdot506+1,$$
$$x^{2025}\equiv x^{1}\equiv x\pmod{x^{2}+1}.$$ Therefore $$r_{1}(x)=x.$$

3. Remainder of $$P(x)$$ modulo $$B(x)=x^{2}+x+1$$
Inside $$\mathbb{R}[x]/(x^{2}+x+1)$$ we have $$x^{2}\equiv -x-1,$$ which gives $$x^{3}\equiv1.$$ Hence the powers of $$x$$ repeat every 3:

$$x^{0}\equiv1,\;x^{1}\equiv x,\;x^{2}\equiv -x-1,\;x^{3}\equiv1,\ldots$$

Because $$2025=3\cdot675,$$
$$x^{2025}\equiv x^{0}\equiv1\pmod{x^{2}+x+1}.$$ Thus $$r_{2}(x)=1.$$

4. Construct the remainder $$R(x)$$ of degree $$\le3$$
Let $$R(x)=ax^{3}+bx^{2}+cx+d.$$ It must satisfy simultaneously $$R(x)\equiv x\pmod{A(x)},\qquad R(x)\equiv1\pmod{B(x)}.$$ Case 1: Modulo $$A(x)=x^{2}+1$$
Use $$x^{2}\equiv-1,\;x^{3}\equiv -x:$$ $$R(x)\equiv a(-x)+b(-1)+cx+d=(-a+c)x+(-b+d).$$ To match $$x,$$ we need $$-a+c=1,\qquad -b+d=0\;\Rightarrow\;d=b.$$ Case 2: Modulo $$B(x)=x^{2}+x+1$$
Use $$x^{2}\equiv -x-1,\;x^{3}\equiv1:$$ $$R(x)\equiv a + b(-x-1)+cx+d=(c-b)x+(a-b+d).$$ To match the constant $$1,$$ we need $$c-b=0\;\Rightarrow\;c=b,\qquad a-b+d=1.$$

From $$c=b$$ and $$d=b$$, the last equation becomes $$a=1.$$ Plugging $$a=1$$ into $$-a+c=1$$ yields $$c=2,$$ hence $$b=c=2$$ and $$d=2.$$

Therefore $$R(x)=x^{3}+2x^{2}+2x+2.$$

5. Evaluate $$R(3)$$
$$R(3)=3^{3}+2\cdot3^{2}+2\cdot3+2=27+18+6+2=53.$$

Hence the required value is 53.

Let $$n$$ range over $$\{1,2,3,\dots ,37\}$$. For each admissible $$n$$ we must divide the integers $$1,2,\dots ,2n$$ into $$n$$ unordered pairs $$\{a_i,b_i\}$$ in such a way that the product $$\displaystyle\prod_{i=1}^{n}(a_i+b_i)$$ is a perfect square.

We treat three separate cases.

The only pairing is $$\{1,2\}$$ whose sum is $$3$$, so the product is $$3$$, which is not a square. Hence $$n=1$$ does not work.

Pair each integer with its “mirror image’’ about $$2n+1$$: $$\{1,2n\},\ \{2,2n-1\},\dots ,\{n,\,n+1\}.$$ Every pair has the same sum $$1+2n=2n+1,\;2+2n-1=2n+1,\;\dots ,n+(n+1)=2n+1.$$ Thus

$$\prod_{i=1}^{n}(a_i+b_i)=(2n+1)^n.$$

Because the exponent $$n=2m$$ is even, each prime factor of $$2n+1$$ occurs an even number of times, so the product is a perfect square. Therefore every even $$n$$ (that is, $$n=2,4,6,\dots ,36$$) satisfies the requirement.

Step I - fix the first six numbers.
Pair the numbers $$1$$ to $$6$$ as $$\{1,5\},\quad \{2,4\},\quad \{3,6\}.$$ The corresponding sums are $$6,\;6,\;9,$$ and their product is $$6\cdot6\cdot9=36\cdot9=18^2,$$ already a perfect square.

Step II - handle the remaining numbers.
After removing $$1$$ through $$6$$, the numbers left are $$7,8,9,\dots ,2n.$$ Their count is $$2n-6=2(2k+1)-6=4(k-1),$$ which is a multiple of $$4$$ because $$k-1$$ is an integer. Split them into consecutive blocks of four: $$\bigl(7,8,9,10\bigr),\ \bigl(11,12,13,14\bigr),\ \dots$$ In each block $$\bigl(x,x+1,x+2,x+3\bigr)$$ form the two pairs

$$\{x,x+3\},\qquad \{x+1,x+2\}.$$

Both pairs have the same sum $$x+(x+3)=x+1+(x+2)=2x+3.$$ Hence the contribution of this block to the overall product is $$(2x+3)\times(2x+3)=(2x+3)^2,$$ a perfect square.

Step III - combine everything.
The product from Step I is a square, and every four-number block from Step II contributes another square. The product of several perfect squares is again a perfect square, so the required condition is fulfilled for every odd $$n\ge 3$$ (that is, $$n=3,5,7,\dots ,37$$).

Combining all three cases:

• $$n=1$$ fails. • All other $$n\in\{1,2,\dots ,37\}$$ succeed.

Therefore the number of values of $$n$$ that work is $$37-1 = 36.$$

Final Answer: 36

Let the finite sequence be $$a_1,a_2,\dots ,a_n$$.

Denote the sliding four-term averages by $$S_k=\dfrac{a_k+a_{k+1}+a_{k+2}+a_{k+3}}{4},\;1\le k\le n-3$$.
Because these averages are strictly increasing,

$$S_{k+1}\gt S_k \;\Longrightarrow\; a_{k+1}+a_{k+2}+a_{k+3}+a_{k+4}\gt a_k+a_{k+1}+a_{k+2}+a_{k+3} \;\Longrightarrow\; a_{k+4}\gt a_k \quad -(1)$$

Similarly, let the seven-term averages be $$T_k=\dfrac{a_k+a_{k+1}+\dots +a_{k+6}}{7},\;1\le k\le n-6.$$ Because these are strictly decreasing,

$$T_{k+1}\lt T_k \;\Longrightarrow\; a_{k+1}+a_{k+2}+\dots +a_{k+7}\lt a_k+a_{k+1}+\dots +a_{k+6} \;\Longrightarrow\; a_{k+7}\lt a_k \quad -(2)$$

Thus, for every admissible index $$k$$ we have the pair of inequalities

$$a_{k+4}\gt a_k,\qquad a_{k+7}\lt a_k.$$

Case 1: $$n\ge 11$$ (we shall derive a contradiction).

Consider the eleven indices in the order
$$1,\;8,\;4,\;11,\;7,\;3,\;10,\;6,\;2,\;9,\;5.$$

Using (1) and (2) successively we obtain the strict chain

$$\begin{aligned} a_1 &\gt a_8 &&\bigl(\text{from }(2)\text{ with }k=1\bigr)\\ a_8 &\gt a_4 &&\bigl((1)\text{ with }k=4\bigr)\\ a_4 &\gt a_{11} &&\bigl((2)\text{ with }k=4\bigr)\\ a_{11}&\gt a_7 &&\bigl((1)\text{ with }k=7\bigr)\\ a_7 &\gt a_3 &&\bigl((1)\text{ with }k=3\bigr)\\ a_3 &\gt a_{10} &&\bigl((2)\text{ with }k=3\bigr)\\ a_{10}&\gt a_6 &&\bigl((1)\text{ with }k=6\bigr)\\ a_6 &\gt a_2 &&\bigl((1)\text{ with }k=2\bigr)\\ a_2 &\gt a_9 &&\bigl((2)\text{ with }k=2\bigr)\\ a_9 &\gt a_5 &&\bigl((1)\text{ with }k=5\bigr) \end{aligned}$$

Finally, applying (1) with $$k=1$$ gives $$a_5\gt a_1$$, so altogether we have

$$a_1\gt a_8\gt a_4\gt a_{11}\gt a_7\gt a_3\gt a_{10}\gt a_6\gt a_2\gt a_9\gt a_5\gt a_1,$$

a strict cycle of inequalities that is impossible for real numbers. Hence no sequence can satisfy the given conditions when $$n\ge 11$$, and therefore

$$n\le 10.$$

Case 2: Construction of a sequence of length $$10$$ that fulfils the requirements.

Take

$$a_1=10,\; a_2=20,\; a_3=30,\; a_4=0,\; a_5=12,\; a_6=22,\; a_7=35,\; a_8=8,\; a_9=15,\; a_{10}=25.$$

Check the nine inequalities that must hold:

Four-term type (1):
$$a_5\gt a_1\;(12\gt10),\; a_6\gt a_2\;(22\gt20),\; a_7\gt a_3\;(35\gt30),\; a_8\gt a_4\;(8\gt0),$$ $$a_9\gt a_5\;(15\gt12),\; a_{10}\gt a_6\;(25\gt22).$$

Seven-term type (2):
$$a_8\lt a_1\;(8\lt10),\; a_9\lt a_2\;(15\lt20),\; a_{10}\lt a_3\;(25\lt30).$$

All conditions are satisfied, so a length-10 sequence exists.

Combining Case 1 and Case 2, the maximal possible length of the sequence is

10.

The relation to be satisfied is

$$\frac{\operatorname{lcm}(a,c)+\operatorname{lcm}(b,c)}{a+b}=\frac{26c}{27}\qquad (1)$$

Write $$\operatorname{lcm}(u,v)=\dfrac{uv}{\gcd(u,v)}.$$
Let $$d=\gcd(a,c),\;e=\gcd(b,c).$$
Then

$$\operatorname{lcm}(a,c)=\dfrac{ac}{d},\qquad \operatorname{lcm}(b,c)=\dfrac{bc}{e}.$$

Substituting in (1) and clearing the denominator $$a+b$$ gives

$$27\!\left(\dfrac{ac}{d}+\dfrac{bc}{e}\right)=26c(a+b).$$

Because $$c\gt0$$, divide by $$c$$ to obtain

$$27\!\left(\frac{a}{d}+\frac{b}{e}\right)=26(a+b). \qquad (2)$$

Write $$a=d\,x,\;b=e\,y\; (x,y\in\mathbb{Z}^{+}).$$ Substituting in (2)

$$27(x+y)=26\bigl(dx+ey\bigr).\qquad (3)$$

Rearrange (3) to

$$x\,(27-26d)+y\,(27-26e)=0.\qquad (4)$$

With $$x,y\gt0$$, the two coefficients in (4) must have opposite signs. Hence exactly one of $$d,e$$ equals $$1$$ and the other is at least $$2$$.

Equation (4) becomes $$x+(\,27-26e\,)y=0\Rightarrow x=(26e-27)\,y.$$ Take the smallest positive solution $$x_0=26e-27,\;y_0=1$$ and write the general solution $$x=k(26e-27),\;y=k\;(k\in\mathbb{Z}^{+}).$$ The corresponding $$a,b$$ are

$$a=x=k(26e-27),\qquad b=e\,y=ek.$$

Require $$a,b\le50$$.

Since $$26e-27\ge25$$ for $$e\ge2$$, the inequality $$a\le50$$ forces $$e=2$$ and $$k=1,2$$. The choice $$k=2$$ gives $$a=50$$, $$b=4$$, but then $$\gcd(50,c)=1$$ (needed because $$d=1$$) is impossible for any even $$c$$. Hence only $$k=1$$ remains, giving

$$a=25,\;b=2,\;e=2,\;d=1.$$

Now $$d=1\Rightarrow\gcd(a,c)=1,$$ and $$e=2\Rightarrow 2\mid c,\;\gcd\!\bigl(b,c\bigr)=2$$. Thus $$c$$ must be even ($$2\mid c$$) and not divisible by $$5$$ (so that $$\gcd(25,c)=1$$). Among $$1\le c\le50$$ there are $$25$$ even numbers; five of them (10,20,30,40,50) are divisible by 5. Therefore there are $$25-5=20$$ admissible values of $$c$$.

Hence Case 1 contributes $$20$$ ordered triples $$\bigl(a,b,c\bigr)=(25,2,c).$$

By symmetry of (4) we get $$y=x(26d-27).$$ Taking $$x=k,\;y=k(26d-27)$$ produces

$$a=d\,x=dk,\qquad b=y=k(26d-27).$$

Again require $$a,b\le50$$.

With $$d\ge2$$ we must have $$d=2$$ (otherwise $$b\gt50$$). For $$d=2$$ we get $$a=2k,\;b=25k$$. The condition $$b\le50$$ forces $$k=1,2$$, but $$k=2$$ would give $$b=50$$ and $$\gcd(50,c)=1$$ (needed because $$e=1$$) impossible when $$a$$ is even. Thus only $$k=1$$ is feasible, giving

$$a=2,\;b=25,\;d=2,\;e=1.$$

Now $$d=2\Rightarrow\gcd(a,c)=2$$, so $$c$$ must be even; $$e=1\Rightarrow\gcd(b,c)=1,$$ so $$c$$ must be coprime to $$25$$. Exactly the same description as in Case 1 (even & not divisible by 5) applies, giving again $$20$$ admissible values of $$c$$.

So Case 2 also contributes $$20$$ ordered triples $$\bigl(a,b,c\bigr)=(2,25,c).$$

No other choices of $$(d,e)$$ can satisfy (4), hence no further solutions exist.

Total number of ordered triples $$=20\;(\text{from }25,2,c)+20\;(\text{from }2,25,c)=40.$$

Therefore the required number of ordered triples is 40.

The given relation is
$$\sqrt{x-\sqrt{x+a}}=\sqrt{a}-y,$$
where $$a$$ is a positive integer that is not a perfect square and $$x,y$$ are non-negative integers.

Both square-roots must be real, so
$$x-\sqrt{x+a}\ge 0 \quad\text{and}\quad \sqrt{a}-y\ge 0.$$
Hence $$y\le\sqrt{a}.$$

Square the equation once:

$$x-\sqrt{x+a}=a-2y\sqrt{a}+y^{2}. \; -(1)$$

Isolate the remaining root and square again. From (1)

$$\sqrt{x+a}=x-a-y^{2}+2y\sqrt{a}.$$

Square a second time:

$$x+a=(x-a-y^{2}+2y\sqrt{a})^{2}.$$

Write the right side as “rational part + irrational part”:
Let $$s=\sqrt{a}$$ (irrational because $$a$$ is not a perfect square).
Put $$A=x-a-y^{2}\,(\text{rational}), \; B=2ys\,(\text{contains }s).$$
Then

$$x+a=A^{2}+4y^{2}a+4Ay\,s.$$

The left side $$x+a$$ is purely rational, so the coefficient of the irrational term $$s$$ must vanish:

$$4Ay=0.$$

Then $$A=0\;\Rightarrow\;x=a+y^{2}.$$
Put this in $$x+a=4y^{2}a$$ obtained above to give

$$2a+y^{2}=4ay^{2}\;\Rightarrow\;y^{2}=\dfrac{2a}{4a-1}.$$

Because $$4a-1>2a$$ for every positive $$a$$, the right side is <1, contradicting $$y\gt0$$ (an integer).
Hence no solution exists when $$y\gt0$$.

Equation (1) becomes $$x-\sqrt{x+a}=a.$$ Rearrange and square only once (no irrational term now):

$$x-\sqrt{x+a}=a\;\Longrightarrow\;(x-a)^{2}=x+a.$$

Expand and collect terms in $$x$$:

$$x^{2}-(2a+1)x+a(a-1)=0.$$

The discriminant must be a perfect square for $$x$$ to be integral:

$$\Delta=(2a+1)^{2}-4a(a-1)=8a+1.$$

Therefore $$8a+1=m^{2}$$ for some odd integer $$m$$, and

$$a=\dfrac{m^{2}-1}{8}.$$

List all odd $$m$$ that keep $$a\lt 100$$ and give non-square $$a$$:

$$\begin{array}{c|c} m & a=\dfrac{m^{2}-1}{8}\\\hline 3 & 1\;(\text{square, reject})\\ 5 & 3\\ 7 & 6\\ 9 & 10\\ 11& 15\\ 13& 21\\ 15& 28\\ 17& 36\;(\text{square, reject})\\ 19& 45\\ 21& 55\\ 23& 66\\ 25& 78\\ 27& 91\\ 29&105\;(\gt100,\text{ stop}) \end{array}$$

The admissible values of $$a$$ below 100 are
$$3,\,6,\,10,\,15,\,21,\,28,\,45,\,55,\,66,\,78,\,91.$$

The largest among them is $$a=91$$.

Hence the greatest possible value of $$a$$ satisfying the given condition is
91.