IOQM Previous Papers Sept 8 2024-25

The product $$1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$$ equals $$9! = 362{,}880$$.

To find the smallest positive integer that does not divide $$9!$$ we begin with natural numbers in increasing order.

Every integer from $$1$$ to $$9$$ itself appears as a factor in $$9!$$, so each of them divides $$9!$$.

Next is $$10$$. Since $$10 = 2 \times 5$$ and both $$2$$ and $$5$$ are among the factors of $$9!$$, $$10$$ also divides $$9!$$.

Consider $$11$$. The prime factor $$11$$ is absent from the prime-factorization of $$9!$$ (which contains only the primes $$2, 3, 5,$$ and $$7$$). Therefore $$11$$ does not divide $$9!$$.

Hence, the smallest positive integer that fails to divide the given product is $$11$$.

Answer: 11

A four-digit number formed from the digits $$1, 2, 3, 4$$ (each used exactly once) must satisfy two conditions:

1. All four positions (thousands, hundreds, tens, units) are filled with these digits without repetition.
2. The number is odd, so its units (last) digit must be an odd digit.

Step 1: Choose the units digit.
Among $$1, 2, 3, 4$$ the odd digits are $$1$$ and $$3$$. Hence there are $$2$$ possible choices for the units place.

Step 2: Arrange the remaining three digits.
After fixing the units digit, three distinct digits remain for the thousands, hundreds and tens places. The number of ways to arrange $$3$$ distinct objects is $$3! = 6$$.

Step 3: Total count.
Total four-digit odd numbers $$= (\text{choices for units digit}) \times (\text{arrangements of remaining digits})$$ $$= 2 \times 6 = 12$$.

Therefore, the required number of four-digit odd numbers is 12.

To extract the last two digits of any positive integer, reduce the number modulo $$100$$ because only the remainder upon division by $$100$$ affects the tens-and-units place.

Hence we need $$5^{2024} \bmod 100$$.

Start with the smallest powers of $$5$$ and inspect their residues modulo $$100$$:
$$5^1 = 5 \equiv 5 \pmod{100}$$
$$5^2 = 25 \equiv 25 \pmod{100}$$

Multiply once more to see whether the residue stabilises:
$$5^3 = 5^2 \cdot 5 = 25 \cdot 5 = 125 \equiv 25 \pmod{100}$$

The product $$25 \cdot 5$$ leaves the same remainder $$25$$ when divided by $$100$$ because $$125 - 25 = 100$$. Therefore each additional factor of $$5$$ keeps the residue at $$25$$:

For every $$n \ge 2$$,
$$5^{n} = 5^{n-1} \cdot 5 \equiv 25 \cdot 5 \equiv 25 \pmod{100}$$

Since $$2024 \gt 2$$, we are in this steady-state regime:
$$5^{2024} \equiv 25 \pmod{100}$$

Thus the last two digits of $$5^{2024}$$, written in the same order, are

25

Join the diagonal $$AC$$ so that the quadrilateral $$ABCD$$ is split into the two triangles $$\triangle ABC$$ and $$\triangle ACD$$.

Step 1: Work inside $$\triangle ACD$$
The two given angles at $$D$$ and $$C$$ are each $$70^\circ$$. Hence the third angle is

$$\angle CAD \;=\;180^\circ - (70^\circ+70^\circ)=40^\circ.$$

Step 2: Relate the angle at vertex $$C$$ of the quadrilateral
The interior angle of the quadrilateral at $$C$$, denoted $$\angle BCD$$, is the sum of the two angles that meet at $$C$$ along the diagonal:
$$\angle BCD=\angle BCA+\angle ACD = 10^\circ+70^\circ = 80^\circ.$$

Step 3: Use the angle-sum of the quadrilateral
For any quadrilateral,
$$\angle BAD+\angle ABC+\angle BCD+\angle CDA = 360^\circ.$$
Insert the known values: $$\angle BAD=110^\circ,\; \angle BCD=80^\circ,\; \angle CDA=\angle ADC=70^\circ.$$
$$110^\circ+\angle ABC+80^\circ+70^\circ = 360^\circ \\ \Longrightarrow \angle ABC = 360^\circ-260^\circ = 100^\circ.$$

Step 4: Angle-sum in $$\triangle ABC$$
Let $$\angle CAB = x$$. In $$\triangle ABC$$:
$$x + \angle ABC + \angle ACB = 180^\circ,$$
$$x + 100^\circ + 10^\circ = 180^\circ,$$
$$x = 180^\circ-110^\circ = 70^\circ.$$

Hence $$\angle CAB = 70^\circ.$

Final Answer: $$70^\circ$$

Because the exponents 20 and 24 are even, only the absolute values of the variables matter in the power sums.
Set $$|a|=x,\;|b|=y,\;|c|=z$$ with $$x,y,z\ge 0$$.

The two conditions become
$$x^{20}+y^{20}+z^{20}=1$$
$$x^{24}+y^{24}+z^{24}=1$$

Define $$f(t)=t^{24}-t^{20}=t^{20}(t^{4}-1).$$
Observe the sign of $$f(t)$$:

• If $$0\le t\le 1$$, then $$t^{4}-1\le 0$$, so $$f(t)\le 0$$.
• If $$t\gt 1$$, then $$t^{4}-1\gt 0$$, so $$f(t)\gt 0$$.

Sum the three expressions $$f(x),f(y),f(z)$$:

$$f(x)+f(y)+f(z)=\bigl(x^{24}+y^{24}+z^{24}\bigr)-\bigl(x^{20}+y^{20}+z^{20}\bigr)=1-1=0.$$

Suppose one of $$x,y,z$$ were greater than 1. Its twentieth power would already exceed 1, making $$x^{20}+y^{20}+z^{20}\gt 1,$$ contradicting the first condition. Hence every one of $$x,y,z$$ satisfies $$0\le t\le 1$$, so each $$f(t)\le 0$$.

The sum of three non-positive numbers is 0 only when each term is 0. Therefore
$$f(x)=f(y)=f(z)=0\;.$$

From $$f(t)=0$$ we get $$t^{20}(t^{4}-1)=0\implies t=0\ \text{or}\ t=1.$$
Consequently each of $$x,y,z$$ is either 0 or 1.

Now enforce $$x^{20}+y^{20}+z^{20}=1.$$ Since $$1^{20}=1$$ and $$0^{20}=0,$$ exactly one of $$x,y,z$$ equals 1 and the other two equal 0.

This means exactly one of $$|a|,|b|,|c|$$ is 1 while the others are 0. Each non-zero variable can be $$+1$$ or $$-1$$. Counting possibilities:

• Choose which position (a, b, or c) carries the ±1: 3 ways.
• Choose the sign ( + or − ) for that position: 2 ways.

Total number of ordered triples: $$3\times 2 = 6.$$

Hence, the required number of triples is
06.

Let the two given vertices of the cube be $$P(1,2,0)$$ and $$Q(3,3,2)$$.
Their position-vector difference is

$$\overrightarrow{PQ}= \langle 3-1,\; 3-2,\; 2-0\rangle = \langle 2,1,2\rangle.$$

Hence the distance between the two vertices is

$$|\overrightarrow{PQ}| = \sqrt{2^{2}+1^{2}+2^{2}} = \sqrt{9}=3.$$

In a cube of edge length $$s$$, three distinct inter-vertex distances occur:

• edge length   $$=s$$
• face diagonal length   $$=s\sqrt{2}$$
• body diagonal length   $$=s\sqrt{3}$$

Thus the measured distance $$3$$ can match any one of the three possibilities:

For each case we can indeed construct a cube:

• For Case 1, take $$\overrightarrow{PQ}$$ as one edge; choose any vector of length 3 perpendicular to $$\overrightarrow{PQ}$$ as the second edge, and their cross product (scaled) as the third.
• For Case 2, write $$\overrightarrow{PQ}= \mathbf{u}+\mathbf{v}$$ with $$\mathbf{u}\perp\mathbf{v}$$ and $$|\mathbf{u}|=|\mathbf{v}|=s_2$$; these give the two edges of one face.
• For Case 3, simply regard $$\overrightarrow{PQ}$$ as the body diagonal.

Therefore all three edge lengths are possible, and the corresponding surface areas are

$$A_1 = 6s_1^{2}=6(3)^{2}=54,$$
$$A_2 = 6s_2^{2}=6\left(\dfrac{3}{\sqrt{2}}\right)^{2}=6\cdot\dfrac{9}{2}=27,$$
$$A_3 = 6s_3^{2}=6(\sqrt{3})^{2}=6\cdot3=18.$$

The sum of all possible surface areas is

$$A_1+A_2+A_3 = 54+27+18 = 99.$$

Hence the required sum is 99.

Let $$S(x)$$ denote the sum of the digits of an integer $$x$$.

Given: $$S(n)$$ is divisible by $$5$$ and $$S(n+1)$$ is also divisible by $$5$$.

If the last digit of $$n$$ is not $$9$$, then adding $$1$$ simply increases the digit-sum by $$1$$: $$S(n+1)=S(n)+1.$$ Two consecutive multiples of $$5$$ cannot differ by $$1$$, so the last digit must be $$9$$. Hence $$n$$ ends with at least one $$9$$.

Suppose $$n$$ ends with exactly $$k$$ consecutive $$9$$’s:
$$n=\dotsc\,d\,\underbrace{99\ldots9}_{k\text{ times}}$$ where $$d$$ (the digit just before the block of $$9$$’s) is $$0\le d\le 8$$.

When we add $$1$$:
• each of those $$k$$ digits $$9$$ becomes $$0$$ (loss of $$9k$$ in the digit-sum),
• the digit $$d$$ becomes $$d+1$$ (gain of $$1$$ in the digit-sum).
Thus

$$S(n+1)=S(n)+1-9k\tag{1}$$

Both $$S(n)$$ and $$S(n+1)$$ are multiples of $$5$$, so their difference must also be a multiple of $$5$$. Using (1):

$$1-9k \equiv 0 \pmod{5} \; \Longrightarrow\; 1-4k \equiv 0 \pmod{5} \; \Longrightarrow\; 4k \equiv 1 \pmod{5}.$$

The inverse of $$4$$ modulo $$5$$ is $$4$$ (since $$4\cdot4=16\equiv1$$), giving

$$k \equiv 4 \pmod{5}.$$

The smallest positive $$k$$ is therefore $$k=4$$. So the least possible $$n$$ must end with exactly four $$9$$’s:

$$n=\dotsc\,d\,9999.$$

Let the digit-sum of all more significant digits (if any) before $$d$$ be $$P$$. Then

$$S(n)=P+d+9\cdot4=P+d+36,$$ $$S(n+1)=P+(d+1).$$

Because $$36\equiv1\pmod5$$, the condition $$S(n)\equiv0\pmod5$$ becomes

$$P+d+1\equiv0\pmod5 \;\Longrightarrow\; P+d\equiv4\pmod5.\tag{2}$$

To obtain the smallest integer $$n$$ we keep the number of leading digits minimal. Thus take no extra digits before $$d$$, i.e. $$P=0$$. Equation (2) simplifies to $$d\equiv4\pmod5$$ with $$0\le d\le8$$. The smallest such $$d$$ is $$d=4$$.

Therefore

$$n=49999.$$

Verification: $$S(49999)=4+9+9+9+9=40$$ and $$S(50000)=5$$, both multiples of $$5$$. Hence $$49999$$ is indeed the least integer satisfying the given condition.

The first two digits of $$n$$ are $$49$$.

Answer: 49

All points lie on the $$4 \times 4$$ integer grid
$$X=\{(m,n)\mid 1\le m\le 4,\;1\le n\le 4\}$$
so there are $$16$$ points in total.

A knight step changes one co-ordinate by $$\pm 2$$ and the other by $$\pm 1$$.
For every point in the grid, count how many such steps keep the knight inside the grid. The number depends only on the “type’’ of square, i.e. on its distance from the edges.

Case 1: corner squares
Corners are $$(1,1),(1,4),(4,1),(4,4)$$. From a corner the legal moves are
$$\bigl(+2,+1\bigr)\quad\text{and}\quad\bigl(+1,+2\bigr)$$ (or their symmetric versions),
so each corner has $$2$$ moves.
Total contributions: $$4\times 2 = 8$$ ordered pairs.

Case 2: edge but not corner
Eight such squares exist (for example $$(1,2),(1,3),(2,1),(3,1)$$ and the symmetric ones). From any of these squares the knight can move in $$3$$ legal ways. Total contributions: $$8\times 3 = 24$$ ordered pairs.

Case 3: interior squares
The four central squares are $$(2,2),(2,3),(3,2),(3,3)$$. A knight located here has $$4$$ legal moves. Total contributions: $$4\times 4 = 16$$ ordered pairs.

Add all the contributions:
$$8+24+16 = 48$$

In this count every ordered pair $$\bigl((a,b),(c,d)\bigr)$$ that satisfies the knight-move condition appears exactly once, which is what the problem asks for. Hence,

Number of knight-move pairs $$= 48$$.

The given condition is
$$a^{2}+(p^{2}+9)b^{2}+9c^{2}-6ab-6pbc=0 \quad\quad -(1)$$
where $$a,\,b,\,c$$ are the three side-lengths of a triangle and $$p$$ is a positive integer.

Step 1 : Convert the left side of (1) into a sum of perfect squares

Group the terms involving $$a$$ and $$b$$ first:

$$a^{2}-6ab+9b^{2}=(a-3b)^{2}.$$

The remainder in (1) is $$p^{2}b^{2}+9c^{2}-6pbc$$ which factors similarly:

$$p^{2}b^{2}+9c^{2}-6pbc=(pb-3c)^{2}.$$

Hence (1) becomes

$$(a-3b)^{2}+(pb-3c)^{2}=0 \quad\quad -(2)$$

Step 2 : Deduce the equalities implied by (2)

Both squares in (2) are non-negative; their sum is zero only when each square is zero. Therefore

$$a-3b=0 \;\Rightarrow\; a=3b,$$

$$pb-3c=0 \;\Rightarrow\; c=\dfrac{p}{3}\,b.$$

Thus the three sides are proportional to

$$a:b:c = 3b:b:\dfrac{p}{3}b \; \Longrightarrow \; 3:1:\dfrac{p}{3}.$$

Step 3 : Impose the triangle inequalities

Let the three side lengths be $$3,\;1,\;\dfrac{p}{3}$$ (any common positive scale cancels).
For a triangle we need the sum of any two sides to exceed the third:

1. $$3+1 \gt \dfrac{p}{3} \;\Longrightarrow\; 4 \gt \dfrac{p}{3} \;\Longrightarrow\; p \lt 12.$$

2. $$1+\dfrac{p}{3} \gt 3 \;\Longrightarrow\; \dfrac{p}{3} \gt 2 \;\Longrightarrow\; p \gt 6.$$

3. $$3+\dfrac{p}{3} \gt 1$$ is automatically true for every positive $$p$$.

Combining (1) and (2):

$$6 \lt p \lt 12.$$

Step 4 : Count the admissible positive integers

The integers that satisfy $$6 \lt p \lt 12$$ are
$$p = 7,\,8,\,9,\,10,\,11.$$

There are exactly five such positive integral values of $$p$$.

Answer: 05

Let us rewrite the two given relations in a more symmetric way.

Put $$A = a+1,\; B = 2b+1,\; C = 3c+1$$.
Since $$a,b,c \gt 0$$, we have $$A,B,C \gt 1$$.

Then

$$a = A-1,\qquad 2b = B-1 \; \Longrightarrow\; b = \frac{B-1}{2},\qquad 3c = C-1 \; \Longrightarrow\; c = \frac{C-1}{3}.$$

Substituting in the two equations:

1. $$\frac{a}{2b+1}+\frac{2b}{3c+1}+\frac{3c}{a+1}=1$$ becomes
$$\frac{A-1}{B}+\frac{B-1}{C}+\frac{C-1}{A}=1.$$

2. $$\frac{1}{a+1}+\frac{1}{2b+1}+\frac{1}{3c+1}=2$$ becomes
$$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=2.$$

Introduce the reciprocals

$$x=\frac1A,\; y=\frac1B,\; z=\frac1C\qquad(0\lt x,y,z\lt 1).$$

With these, the second relation is simply

$$x+y+z=2 \; \; -(1).$$

The first relation changes to

$$\left(\frac1x-1\right)\!y+\left(\frac1y-1\right)\!z+\left(\frac1z-1\right)\!x =1,$$ which simplifies to $$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=3\; \; -(2).$$

Observe that for any positive numbers, the arithmetic-geometric mean inequality gives

$$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\;\ge\;3,$$ with equality only when $$\frac{y}{x}=\frac{z}{y}=\frac{x}{z}=1.$$

Because the left-hand side is exactly $$3$$ by (2), equality must occur, hence

$$\frac{y}{x}=1,\;\; \frac{z}{y}=1,\;\; \frac{x}{z}=1 \;\Longrightarrow\; x=y=z.$$

Using (1): $$3x=2\;\Longrightarrow\; x=y=z=\frac23.$$

This immediately gives

$$A=\frac1x=\frac32,\quad B=\frac1y=\frac32,\quad C=\frac1z=\frac32.$$

Recover $$a,b,c$$:

$$a=A-1=\frac32-1=\frac12,$$ $$b=\frac{B-1}{2}=\frac{\frac32-1}{2}=\frac14,$$ $$c=\frac{C-1}{3}=\frac{\frac32-1}{3}=\frac16.$$

Finally, compute the required sum:

$$\frac1a+\frac1b+\frac1c =\frac1{\frac12}+\frac1{\frac14}+\frac1{\frac16} =2+4+6=12.$$

Hence the value of $$\displaystyle\frac1a+\frac1b+\frac1c$$ is 12.

Place the square $$ABCD$$ on the coordinate plane with
$$A(0,0),\;B(16,0),\;C(16,16),\;D(0,16).$$

The side $$CD$$ is the segment from $$(16,16)$$ to $$(0,16).$$
Since $$CE=EF=FD=\dfrac{16}{3},$$ the trisection points are

$$E\left(16-\dfrac{16}{3},\,16\right)=\left(\dfrac{32}{3},\,16\right),\qquad F\left(16-\dfrac{32}{3},\,16\right)=\left(\dfrac{16}{3},\,16\right).$$

Equation of $$BF$$
Slope $$m_{BF}=\dfrac{16-0}{\tfrac{16}{3}-16} =\dfrac{16}{-\tfrac{32}{3}}=-\dfrac{3}{2}.$$ Hence $$BF: y=-\dfrac{3}{2}(x-16).$$

Equation of $$AE$$
Slope $$m_{AE}=\dfrac{16-0}{\tfrac{32}{3}-0} =\dfrac{16}{\tfrac{32}{3}}=\dfrac{3}{2}.$$ Thus $$AE: y=\dfrac{3}{2}x.$$

Intersection point $$M = BF \cap AE$$
Set the two expressions for $$y$$ equal: $$\dfrac{3}{2}x=-\dfrac{3}{2}x+24 \;\Longrightarrow\; 3x=24 \;\Longrightarrow\; x=8.$$ Then $$y=\dfrac{3}{2}\times8=12.$$ So $$M(8,12).$$

Area of $$\triangle MAB$$
Base $$AB$$ lies on the $$x$$-axis from $$x=0$$ to $$x=16,$$ so $$AB=16.$$ The perpendicular distance of $$M$$ from $$AB$$ is its $$y$$-coordinate $$12.$$ Therefore

$$ \text{Area}=\dfrac12\,(\text{base})\,(\text{height}) =\dfrac12\,(16)\,(12)=96. $$

Hence the required area is 96.

Each particle performs exactly $$80$$ moves. At every move it can go to
$$x+1,\;x,\;x-1$$ while $$y$$ definitely increases by $$1$$. Thus after $$80$$ moves every particle is somewhere on the line $$y=80$$ and its final $$x$$-coordinate is the algebraic sum of the $$80$$ chosen steps $$\left(+1,0,-1\right)$$.

The splitting rule ensures that every particle chooses one of the three options with equal proportion. Because the origin initially holds $$3^{80}$$ particles and the branching factor is $$3$$ at every step, the total number of distinct step-sequences of length $$80$$ is also $$3^{80}$$. Hence exactly one particle follows each possible sequence.
Consequently, the number of particles that finally reach any particular point $$(x,80)$$ equals the number of step-sequences that sum to that $$x$$.

Let
$$a=$$ number of $$+1$$ steps,  $$b=$$ number of $$0$$ steps,  $$c=$$ number of $$-1$$ steps.

We require two equations:
1. Total steps: $$a+b+c=80$$  $$-(1)$$
2. Final $$x$$-coordinate: $$a-c=79$$  $$-(2)$$

From $$-(2)$$, $$c=a-79\,. $$ Substitute this in $$-(1)$$:

$$a+b+(a-79)=80 \;\Longrightarrow\; b=159-2a.$$

All variables must be non-negative integers.

Because $$c=a-79\ge0$$, we need $$a\ge79$$. Trying $$a=79$$ gives $$c=0,\; b=159-2\!\times\!79=1,$$ which is admissible.
Trying $$a=80$$ gives $$c=1,\; b=159-160=-1,$$ which is impossible.

Therefore the only feasible distribution is
$$a=79,\; b=1,\; c=0.$$/p>

This means every valid sequence consists of exactly $$79$$ moves of $$+1$$ and exactly $$1$$ move of $$0$$, in some order. The position of the lone $$0$$ can be chosen in $$\binom{80}{1}=80$$ ways.

Hence, there are $$80$$ distinct step-sequences that end at $$(79,80)$$, and so the number of particles present there after $$80$$ steps is also $$80$$.

Final Answer: 80

Let the set be $$X=\{n,n+2,n+4,\dots ,n+38\}$$. These are 20 consecutive even (or odd) integers with common difference $$2$$.

Pick any three elements of $$X$$ and arrange them in non-decreasing order $$a\le b\le c$$. They must satisfy two conditions:

1. Triangle inequality $$a+b\gt c$$.
2. Acute-angled condition $$a^2+b^2\gt c^2$$.

The left-hand sides of both inequalities increase when we replace $$a$$ or $$b$$ by larger elements of $$X$$, while the right-hand side decreases if we replace $$c$$ by a smaller element. Therefore the “hardest” (most restrictive) triple is obtained by taking the two smallest and the largest numbers from the set:

$$a=b=n,\qquad c=n+38$$.

Checking the triangle inequality for this extreme triple:

$$n+n \gt n+38\; \Longrightarrow\; 2n\gt n+38\; \Longrightarrow\; n\gt 38.$$

Since we are ultimately going to need a much larger value of $$n$$ (see below), the triangle inequality will automatically be satisfied for all triples once the acute-angled condition is met.

Now impose the acute-angled condition on the same extreme triple:

$$n^2+n^2 \gt (n+38)^2$$ $$\Longrightarrow 2n^2 \gt n^2+76n+1444$$ $$\Longrightarrow n^2-76n-1444\gt0.$$

Solve the quadratic inequality $$n^2-76n-1444=0$$:

Discriminant $$\Delta=76^2+4\cdot1444=5776+5776=11552,$$ $$\sqrt{\Delta}\approx107.46.$$

Roots $$n=\frac{76\pm\sqrt{11552}}{2}\approx\frac{76\pm107.46}{2}.$$ The positive root is $$\approx91.73$$.

Because $$n$$ must be an integer, the smallest feasible value is

$$n=92.$$

Verification:

For $$n=92$$ the extreme triple is $$92,92,130$$.

Triangle inequality: $$92+92=184\gt130.$$ Acute condition: $$92^2+92^2=2\cdot8464=16928\gt130^2=16900.$$

If the extreme triple is acute, every other triple from $$X$$ has either larger $$a$$ or $$b$$, or a smaller $$c$$, causing $$a^2+b^2-c^2$$ to increase. Hence all triples are acute when $$n\ge92$$, and they fail to be acute when $$n\le91$$ (because the triple $$n,n,n+38$$ becomes right- or obtuse-angled).

Therefore the smallest integer $$n$$ for which every choice of three elements from $$X$$ forms an acute-angled triangle is

92.

The given functional equation is

$$4f(3-x)+3f(x)=x^{2}\qquad\text{for all }x\in\mathbb{R}.\tag{-1}$$

Step 1: Write the same equation for $$3-x$$ instead of $$x$$.

Replace $$x$$ by $$3-x$$ in $$(\-1)$$:

$$4f\!\bigl(3-(3-x)\bigr)+3f(3-x)=(3-x)^{2}$$

Simplifying,

$$4f(x)+3f(3-x)=(3-x)^{2}.\tag{-2}$$

Step 2: Solve the linear system formed by $$(\-1)$$ and $$(\-2)$$.

From $$(\-1)$$: $$3f(x)+4f(3-x)=x^{2}.\tag{-1}$$

Multiply $$(\-1)$$ by $$4$$ and $$(\-2)$$ by $$3$$ to eliminate $$f(x)$$:

$$12f(x)+16f(3-x)=4x^{2}\tag{-3}$$

$$12f(x)+9f(3-x)=3(3-x)^{2}\tag{-4}$$

Subtract $$(\-4)$$ from $$(\-3)$$:

$$7f(3-x)=4x^{2}-3(3-x)^{2}.$$

Since $$(3-x)^{2}=x^{2}-6x+9$$,

$$7f(3-x)=4x^{2}-3(x^{2}-6x+9)=x^{2}+18x-27.$$

Therefore

$$f(3-x)=\frac{x^{2}+18x-27}{7}.\tag{-5}$$

Step 3: Obtain an explicit formula for $$f(x)$$.

Put $$(\-5)$$ back into $$(\-1)$$:

$$3f(x)=x^{2}-4f(3-x)=x^{2}-4\cdot\frac{x^{2}+18x-27}{7}.$$

Thus

$$3f(x)=\frac{7x^{2}-4x^{2}-72x+108}{7}=\frac{3x^{2}-72x+108}{7},$$

$$f(x)=\frac{3x^{2}-72x+108}{21}=\frac{x^{2}-24x+36}{7}.\tag{-6}$$

Step 4: Evaluate $$f(27)$$ and $$f(25)$$.

Using $$(\-6)$$:

$$f(27)=\frac{27^{2}-24\cdot27+36}{7}=\frac{729-648+36}{7}=\frac{117}{7}=16.7142857\ldots$$

$$f(25)=\frac{25^{2}-24\cdot25+36}{7}=\frac{625-600+36}{7}=\frac{61}{7}=8.7142857\ldots$$

Step 5: Compute the required difference.

$$f(27)-f(25)=\frac{117}{7}-\frac{61}{7}=\frac{56}{7}=8.$$

To the nearest integer, the value is $$\mathbf{08}$$.

Place $$\triangle ABC$$ on a Cartesian plane with its base $$BC$$ on the $$x$$-axis and the midpoint of $$BC$$ at the origin.

Thus
$$B(-15,0),\; C(15,0) \quad (\text{because } BC = 30).$$

Since $$AB = AC = 20$$, the height of the isosceles triangle is
$$h = \sqrt{20^{2} - 15^{2}} = \sqrt{400-225} = \sqrt{175}=5\sqrt7.$$
Hence
$$A(0,\,5\sqrt7).$$

The foot of the perpendicular from $$A$$ to $$BC$$ is $$D(0,0)$$ (altitude in an isosceles triangle meets the base at its midpoint). The midpoint of $$AD$$ is therefore
$$M\Bigl(0,\frac{5\sqrt7}{2}\Bigr).$$

Step 1: Circumcentre and circumradius of $$\triangle ABC$$
Because the triangle is isosceles, the circum-centre $$O$$ lies on the perpendicular bisector of $$BC$$, i.e. on the $$y$$-axis at $$(0,k).$$

Using the cosine rule in $$\triangle ABC$$:
$$BC^{2}=AB^{2}+AC^{2}-2(AB)(AC)\cos A$$
$$\Rightarrow 900 = 400+400-800\cos A \quad\Longrightarrow\quad \cos A=-\tfrac18,$$
so $$A$$ is obtuse. For an obtuse triangle the circum-centre lies outside the triangle, on the side of the obtuse angle, hence below the base. Therefore $$k$$ will be negative.

Circum-radius:
$$R=\frac{BC}{2\sin A}= \frac{30}{2\cdot\frac{3\sqrt7}{8}}=\frac{40}{\sqrt7}.$$

Coordinates of $$O$$ follow from $$OB = R$$:
$$(-15)^2 + (0-k)^2 = \Bigl(\tfrac{40}{\sqrt7}\Bigr)^2$$
$$225 + k^{2} = \frac{1600}{7}\quad\Longrightarrow\quad k^{2}=\frac{25}{7}.$$ Taking the negative root (centre below the base),
$$O\Bigl(0,\;-\frac{5}{\sqrt7}\Bigr).$$

Step 2: Equation of the circum-circle
$$x^{2} +\Bigl(y+\frac{5}{\sqrt7}\Bigr)^{2}= \Bigl(\frac{40}{\sqrt7}\Bigr)^{2}= \frac{1600}{7}.$$

Step 3: Horizontal chord through $$M$$
Chord $$PQ$$ is parallel to $$BC$$, so it is horizontal. Its equation is the horizontal line through $$M$$:
$$y = \frac{5\sqrt7}{2}.$$

Substitute this value of $$y$$ in the circle equation to find the $$x$$-coordinates of $$P$$ and $$Q$$: $$x^{2} + \Bigl(\frac{5\sqrt7}{2} +\frac{5}{\sqrt7}\Bigr)^{2}= \frac{1600}{7}.$$

Simplify the bracket:
$$\frac{5\sqrt7}{2} + \frac{5}{\sqrt7}= \frac{5\sqrt7}{2} + \frac{5\sqrt7}{7}=5\sqrt7\Bigl(\frac12+\frac17\Bigr)=5\sqrt7\cdot\frac{9}{14}= \frac{45\sqrt7}{14}.$$

Hence
$$x^{2} + \Bigl(\frac{45\sqrt7}{14}\Bigr)^{2}= \frac{1600}{7}$$
$$\Rightarrow x^{2}= \frac{1600}{7}-\frac{(45)^{2}\cdot7}{14^{2}} =\frac{1600}{7}-\frac{2025\cdot7}{196} =\frac{44800-14175}{196} =\frac{30625}{196}.$$

Therefore $$x = \pm\sqrt{\frac{30625}{196}} =\pm\frac{175}{14} =\pm\frac{25}{2}.$$

Step 4: Length of the chord $$PQ$$
The endpoints are $$P\bigl(-\tfrac{25}{2},\tfrac{5\sqrt7}{2}\bigr)$$ and $$Q\bigl(\tfrac{25}{2},\tfrac{5\sqrt7}{2}\bigr).$$ Since the chord is horizontal, its length is simply twice the absolute $$x$$-coordinate: $$PQ = 2\left(\frac{25}{2}\right)=25.$$

Hence the required length of the chord is 25.

Let the two-digit numbers be written in decimal form as
$$p = 10a + b,\qquad q = 10c + d$$
with digits $$a,b,c,d \in \{0,1,\dots ,9\}$$, where $$a,c \neq 0$$ because $$p,q$$ have two digits and $$b,d \neq 0$$ because they are not divisible by $$10$$.

When the computer writes the four-digit number $$r$$ by appending the digits of $$p$$ followed by those of $$q$$, we obtain
$$r = 100p + q\;.\quad -(1)$$

The conditions given are

1. $$\gcd(p,q)=1$$,

2. $$p+q$$ divides $$r$$.

Denote $$s = p+q$$. Using $$(1)$$, the divisibility requirement is

$$s \mid (100p + q).$$

Because $$q = s-p$$, substitute to get

$$100p + q \;=\; 100p + (s-p) \;=\; 99p + s.$$ Thus
$$s \mid (99p + s)\;\Longrightarrow\; s \mid 99p.$$

Now, $$\gcd(p,s)=\gcd\bigl(p,p+q\bigr)=\gcd(p,q)=1$$ by condition 1. Since $$p$$ is coprime to $$s$$ and $$s\mid 99p$$, we must have

$$s \mid 99.\quad -(2)$$

The positive divisors of $$99$$ are $$1,3,9,11,33,99$$. But $$s=p+q$$ is the sum of two two-digit numbers, so $$20 \le s \le 198$$. Hence, from (2), the only possibilities are

$$s = 33 \quad \text{or} \quad s = 99.$$

Here $$q = 99 - p$$. To keep $$q$$ two-digit we need $$10 \le q \le 89\;\Longrightarrow\; 10 \le p \le 89$$.

The four-digit number equals
$$r = 100p + q = 100p + (99-p) = 99p + 99.$$
Because the coefficient of $$p$$ is positive, $$r$$ increases with $$p$$, so we search downward from the largest admissible $$p=89$$ until all conditions are met.

The first admissible value is $$p=86,\;q=13$$, giving
$$r = 100\,(86)+13 = 8613.$$ Any smaller $$p$$ yields a smaller $$r$$, so this is the largest $$r$$ in Case 1.

Now $$q = 33 - p$$ with $$10 \le p \le 23$$ to keep both two-digit. Here
$$r = 100p + q = 99p + 33,$$ again increasing with $$p$$. Testing downwards:

This gives $$r = 100\,(19)+14 = 1914,$$ which is far smaller than $$8613$$ from Case 1.

Therefore the largest four-digit number printed by the computer is $$N = 8613.$$ The last two digits of $$N$$ are $$\boxed{13}$$.

Label the five points as $$\{1,2,3,4,5\}$$. Join every pair of points; the resulting graph is the complete graph $$K_5$$ with $$10$$ edges. We must colour every edge red or blue so that no three vertices form a triangle whose three sides have the same colour.

Step 1: Degree restriction at each vertex
Pick any vertex, say $$1$$. Let $$d_R(1)$$ and $$d_B(1)$$ be the number of red and blue edges incident at $$1$$; clearly $$d_R(1)+d_B(1)=4$$.

Assume $$d_R(1)\ge 3$$. Take three red neighbours of $$1$$, say $$2,3,4$$. Among the three edges $$\{2,3\},\{3,4\},\{4,2\}$$ at least one is red or blue:
- If any of them is red, say $$\{2,3\}$$, then $$1,2,3$$ create a red triangle - forbidden.
- Otherwise all three are blue, giving blue triangle $$2,3,4$$ - also forbidden.
Thus $$d_R(1)\not\ge 3\Rightarrow d_R(1)\le 2$$. The same argument with colours swapped gives $$d_B(1)\le 2$$.

Because $$d_R(1)+d_B(1)=4$$ and both quantities are $$\le 2$$, we must have $$d_R(1)=d_B(1)=2$$. Since the choice of vertex was arbitrary, $$d_R(v)=d_B(v)=2 \quad\text{for every vertex }v.$$

Step 2: Structure of the red (and blue) subgraphs
The red edges form a $$2$$-regular graph on five vertices. A $$2$$-regular graph is a disjoint union of cycles. Because triangles are forbidden, no cycle may have length $$3$$. Hence the only possibility is a single cycle of length $$5$$, i.e. the red subgraph is the $$5$$-cycle $$C_5$$.

The blue edges are the complement of the red edges inside $$K_5$$, so they also form the complement of a $$5$$-cycle. But the complement of $$C_5$$ on five vertices is again $$C_5$$. Thus the blue subgraph is another $$5$$-cycle, automatically free of triangles as required.

Step 3: Counting admissible colourings
Every admissible colouring is obtained by choosing which $$5$$-cycle will be coloured red; the remaining five edges automatically become blue.
Conversely, any red $$5$$-cycle gives a colouring with no monochromatic triangle, as neither colour contains a triangle.

The number of distinct Hamiltonian (length-5) cycles in $$K_5$$ is well known:
For $$K_n$$ it is $$\dfrac{(n-1)!}{2}$$ (the factor $$2$$ divides out the two directions of traversal). Hence for $$n=5$$ we get $$\frac{(5-1)!}{2}=\frac{4!}{2}=12.$$

Step 4: Conclusion
There are exactly $$12$$ ways to colour the $$10$$ edges of $$K_5$$ red or blue so that no monochromatic triangle is formed.

Answer: 12

Let us denote the two admissible moves on a number $$n$$ as
(1) $$n \rightarrow 2n$$   (multiply by $$2$$)
(2) $$n \rightarrow n-3$$   (subtract $$3$$).

The target $$121$$ is odd, therefore it cannot be obtained from the previous number by doubling (operation 1). Hence the last move must be operation 2, coming from $$124$$:

$$124 \;-\;3\;=\;121$$

To minimise the total number of steps we now work backwards. For any integer $$x$$ the possible predecessors are

• $$x+3$$   (because subtracting $$3$$ would take that predecessor to $$x$$),
• $$\dfrac{x}{2}$$ if $$x$$ is even (because doubling that predecessor would give $$x$$).

Starting from $$121$$ we repeatedly generate all predecessors that have not appeared before. Doing this level by level guarantees the first time we meet $$11$$ corresponds to the shortest route.

Backward search

$$\begin{array}{c|l} \text{Depth} & \text{Numbers obtained at this depth} \\\hline 0 & 121 \\ 1 & 124 \\ 2 & 62,\;127 \\ 3 & 31,\;65,\;130 \\ 4 & 34,\;68,\;133 \\ 5 & 17,\;37,\;71,\;136 \\ 6 & 20,\;40,\;74,\;139 \\ 7 & 10,\;23,\;43,\;77,\;142 \\ 8 & 5,\;13,\;26,\;46,\;80,\;145 \\ 9 & 8,\;16,\;29,\;49,\;83,\;148 \\ 10 & \boxed{11},\,4,\,\ldots \end{array}$$

The first appearance of $$11$$ is at depth $$10$$, so a minimum of 10 moves is necessary.

Constructing the forward path
Tracing predecessors from $$11$$ back to $$121$$ and reversing the order gives an explicit optimal sequence:

$$11 \xrightarrow{-3} 8 \xrightarrow{-3} 5 \xrightarrow{\times2} 10 \xrightarrow{\times2} 20 \xrightarrow{-3} 17 \xrightarrow{\times2} 34 \xrightarrow{-3} 31 \xrightarrow{\times2} 62 \xrightarrow{\times2} 124 \xrightarrow{-3} 121$$

The list of operations is
$$-3,\; -3,\; \times2,\; \times2,\; -3,\; \times2,\; -3,\; \times2,\; \times2,\; -3$$ - exactly $$10$$ steps.

Therefore, starting from $$11$$ the least number of steps required to reach $$121$$ is 10.

Let the perpendicular sides of the right-angled triangle be $$AB = x$$ and $$AC = y$$. Then $$BC = \sqrt{x^{2}+y^{2}}$$ by the Pythagorean theorem.

The point $$D$$ divides $$BC$$ internally in the ratio $$BD : DC = 2 : 1$$. Write $$BD = 2k$$ and $$DC = k$$, so $$BC = BD + DC = 3k \;\Rightarrow\; k = \dfrac{\sqrt{x^{2}+y^{2}}}{3}$$.

The given condition is $$AB + BD = AC + CD.$$ Substituting the lengths,

$$x + 2k = y + k \;\Longrightarrow\; x - y + k = 0 \;\Longrightarrow\; y - x = k.$$

Replace $$k$$ with its value:

$$y - x = \dfrac{\sqrt{x^{2}+y^{2}}}{3} \quad -(1)$$

Introduce the ratio $$r = \dfrac{y}{x}$$. Then $$y = rx$$ and equation $$-(1)$$ becomes

$$rx - x = \dfrac{\sqrt{x^{2}+r^{2}x^{2}}}{3}.$$

Simplify each side:

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

Because $$x \gt 0$$, divide both sides by $$x$$:

$$3(r-1) = \sqrt{1+r^{2}}.$$

Square both sides to remove the square root:

$$9(r-1)^{2} = 1 + r^{2}.$$

Expand and collect like terms:

$$9(r^{2} - 2r + 1) = 1 + r^{2}$$ $$9r^{2} - 18r + 9 = 1 + r^{2}$$ $$8r^{2} - 18r + 8 = 0.$$

Divide by 2 for simpler coefficients:

$$4r^{2} - 9r + 4 = 0.$$

Solve the quadratic using the discriminant method:

$$r = \dfrac{9 \pm \sqrt{81 - 64}}{8} = \dfrac{9 \pm \sqrt{17}}{8}.$$

We need $$y \gt x$$ (because $$y - x = k \gt 0$$), so $$r \gt 1$$. Choose the positive root:

$$\dfrac{AC}{AB} = r = \dfrac{9 + \sqrt{17}}{8}.$$

Thus $$m = 9$$, $$n = 8$$ and $$p = 17$$ (where $$p$$ is prime). Therefore, $$m + n + p = 9 + 8 + 17 = 34$$.

Final answer: 34

We need the least natural number $$n$$ for which the fourteen fourth-powers
$$1^4,\,2^4,\,3^4,\,\dots ,\,14^4$$
give fourteen different remainders on division by $$n$$. In other words, for every pair $$i\neq j$$ in $$\{1,2,\dots ,14\}$$ we must have $$n \nmid i^4-j^4$$.

Step 1 : A trivial lower bound
There are fourteen numbers. If all remainders are to be different, we obviously need $$n\ge 14$$ (otherwise the pigeon-hole principle forces a repetition).

Step 2 : The range $$15\le n\le 28$$ always fails
For any such $$n$$ let $$a=n-14$$. Because $$15\le n\le 28$$ we get $$1\le a\le 14$$ and $$a\neq 14$$.
Now observe the identity$$(n-a)^4\equiv a^4\pmod n.\tag{-1}$$Choosing $$a=n-14$$ gives the pair $$a$$ and $$14$$ inside the set $$\{1,\dots ,14\}$$ with the same remainder by (-1).
Hence no $$n$$ in this interval can work.

Step 3 : The value $$n=14$$ also fails
A short table suffices:

$$\begin{array}{c|cccccccccccccc} k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\\hline k^4\bmod 14 & 1 & 2 & 11 & 4 & 9 & 8 & 7 & 8 & 9 & 4 & 11 & 2 & 1 & 0 \end{array}$$
The remainder $$8$$ appears for both $$k=6$$ and $$k=8$$, so $$n=14$$ is ruled out.

Step 4 : Eliminate $$n=29$$ and $$n=30$$

• $$n=29$$ (prime)
Compute a few fourth powers modulo 29: $$1^4\equiv1,\;2^4\equiv16,\;5^4\equiv16\pmod{29}.$$ The pair $$2,5$$ already coincid​e, so 29 is impossible.

• $$n=30$$
Because $$30=2\cdot3\cdot5,$$ equal remainders mod 30 appear as soon as they coincide modulo any prime factor. Indeed, $$1^4\equiv1,\;7^4=2401\equiv1\pmod{30}$$, giving a repetition. Hence 30 is impossible.

Step 5 : Show that $$n=31$$ works

31 is prime and $$31\equiv3\pmod4$$. In such a prime field, the equation$$x^4\equiv y^4\pmod{31}$$implies $$x\equiv\pm y\pmod{31}$$ for all non-zero residues, because the remaining factor $$x^2+y^2$$ cannot vanish (−1 is a quadratic non-residue when the prime is 3 mod 4).

Among the integers $$1,2,\dots ,14$$ no two are negatives of each other modulo 31 (their negatives lie in $$17,\dots ,30$$). Therefore their fourth powers are all distinct modulo 31.

Thus 31 is admissible, and every smaller $$n\ge14$$ has already been excluded.

Final answer: the smallest such natural number is
$$\boxed{31}$$.

Write $$n$$ in its prime-factor form:
$$n = 2^{19}\,3^{12}$$.

1. Total divisors of $$n^{2}$$

The exponent of 2 in $$n^{2}$$ is $$2\times19 = 38$$ and that of 3 is $$2\times12 = 24$$. Using the divisor-count formula $$\prod (e_i+1)$$,

$$\tau(n^{2}) = (38+1)(24+1)=39\times25 = 975.$$

2. Divisors of $$n^{2}$$ that are < $$n$$

For every divisor $$d$$ of $$n^{2}$$ the number $$\dfrac{n^{2}}{d}$$ is also a divisor. Exactly one of the pair $$\{d,\dfrac{n^{2}}{d}\}$$ is smaller than $$n$$ (except when $$d=n$$). Hence the divisors of $$n^{2}$$ fall into $$\dfrac{975-1}{2}$$ such pairs plus the single divisor $$d=n$$ itself.

Number of divisors of $$n^{2}$$ that are strictly less than $$n$$:

$$\frac{975-1}{2}= \frac{974}{2}=487.$$

3. Among these, how many already divide $$n$$?

Divisor count of $$n$$ itself:

$$\tau(n) = (19+1)(12+1)=20\times13 = 260.$$

Exactly one of those 260 divisors equals $$n$$; the remaining 259 are less than $$n$$. Therefore 259 of the 487 numbers obtained in step 2 are common to both $$n^{2}$$ and $$n$$.

4. Required count $$M$$

$$M = 487-259 = 228.$$

5. Last two digits of $$M$$

The number formed by the last two digits of 228 is 28.

Answer: 28

Let $$AB=x$$ and $$BC=y$$ be the two perpendicular sides and let the hypotenuse be $$AC=c$$.

1. Right-triangle relation: $$x^{2}+y^{2}=c^{2}$$ $$-(1)$$

2. Altitude formula (altitude from the right angle to the hypotenuse): $$BD=\dfrac{xy}{c}$$. Given $$BD=12$$,

$$\dfrac{xy}{c}=12 \;\Longrightarrow\; xy=12c$$ $$-(2)$$

3. Let the perimeter be $$P=x+y+c$$. Both $$P$$ and $$c$$ are required to be integers, so we must have

$$x+y=S\in\mathbb{Z}$$ $$-(3)$$

4. Relate $$S$$, $$c$$ using (1) and (2):

$$S^{2}=(x+y)^{2}=x^{2}+y^{2}+2xy=c^{2}+24c$$ $$\Longrightarrow\; S^{2}=c(c+24)$$ $$-(4)$$

Because $$S$$ is an integer, the right-hand side must be a perfect square. Hence we need an integer $$c$$ such that $$c(c+24)$$ is a perfect square.

5. Feasibility condition for real, positive $$x,\,y$$: Treat $$x,\,y$$ as the roots of $$t^{2}-St+12c=0$$. Its discriminant must be non-negative:

$$\Delta=S^{2}-48c=(c^{2}+24c)-48c=c^{2}-24c=c(c-24)\ge 0$$ Thus $$c\ge 24$$ $$-(5)$$

6. Search for the smallest integer $$c\ge 24$$ making $$c(c+24)$$ a square.

Try $$c=24$$: $$24\cdot48=1152\neq \text{perfect square}$$

Try $$c=25$$: $$25\cdot49=1225=35^{2}$$ ― perfect square found.

Therefore the minimum possible hypotenuse is $$c=25$$.

7. Verification: For $$c=25$$, equation (4) gives $$S=35$$. With $$S=35$$ and $$xy=12c=300$$, the quadratic $$t^{2}-35t+300=0$$ yields $$t=\dfrac{35\pm\sqrt{35^{2}-4\cdot300}}{2}=\dfrac{35\pm5}{2}$$, giving $$x=20$$ and $$y=15$$.

This is the familiar $$15\text{-}20\text{-}25$$ right triangle, and indeed $$BD=\dfrac{15\cdot20}{25}=12$$ as required, while the perimeter $$15+20+25=60$$ is an integer.

Hence the minimum possible length of $$AC$$ is 25.