IOQM Previous Papers Sept 3 2023

The condition given is $$\log_{a}(b)+6\log_{b}(a)=5$$ with $$2\le a,b\lt 2023$$ and $$a,b\in\mathbb{N}$$.

Introduce the variable $$x=\log_{a}(b)$$. By the reciprocal property of logarithms, $$\log_{b}(a)=\frac{1}{x}$$ because $$\log_{b}(a)=\frac{1}{\log_{a}(b)}$$.

Substituting these into the equation gives
$$x+6\left(\frac{1}{x}\right)=5$$

Multiply by $$x$$ to clear the denominator:
$$x^{2}-5x+6=0$$

Factorisation gives
$$(x-2)(x-3)=0$$
so $$x=2$$ or $$x=3$$.

$$x=2\; \Rightarrow\;\log_{a}(b)=2\;\Rightarrow\;b=a^{2}$$
Constraints: $$2\le a\lt 2023$$ and $$b=a^{2}\lt 2023$$.
We need $$a^{2}\lt 2023$$, so $$a\lt\sqrt{2023}\approx 44.97$$.
Thus $$a$$ can take integer values $$2,3,\dots,44$$ — a total of $$44-2+1=43$$ choices, and each such $$a$$ gives exactly one corresponding $$b=a^{2}$$.

$$x=3\; \Rightarrow\;\log_{a}(b)=3\;\Rightarrow\;b=a^{3}$$
Constraints: $$b=a^{3}\lt 2023$$.
We need $$a^{3}\lt 2023$$, so $$a\lt\sqrt[3]{2023}\approx 12.63$$.
Thus $$a$$ can take integer values $$2,3,\dots,12$$ — a total of $$12-2+1=11$$ choices, each giving one $$b=a^{3}$$.

Adding both cases, the number of ordered pairs $$(a,b)$$ is $$43+11=54$$.

Hence the required number of elements in the set is 54.

The given relation is

$$x^{4}=(x-1)(y^{3}-23)-1 \qquad -(1)$$

Rewrite $$-(1)$$ so that the two factors involving $$x-1$$ become evident:

$$x^{4}+1=(x-1)(y^{3}-23)\qquad -(2)$$

To understand how $$x-1$$ can divide $$x^{4}+1$$, first calculate the greatest common divisor of the two numbers.

Since $$x\equiv 1 \pmod{x-1}$$, we have $$x^{4}\equiv 1 \pmod{x-1}$$, hence

$$x^{4}+1\equiv 2 \pmod{x-1}\,.$$

Therefore

$$\gcd(x-1,\;x^{4}+1)\in\{1,2\}\,.$$

Next perform polynomial division of $$x^{4}+1$$ by $$x-1$$:

$$x^{4}+1=(x-1)(x^{3}+x^{2}+x+1)+2 \qquad -(3)$$

Combine $$-(2)$$ and $$-(3)$$:

$$(x-1)(y^{3}-23)=(x-1)(x^{3}+x^{2}+x+1)+2$$

which simplifies to

$$(x-1)\bigl[y^{3}-23-(x^{3}+x^{2}+x+1)\bigr]=2 \qquad -(4)$$

Equation $$-(4)$$ shows that $$x-1$$ divides $$2$$, so

$$x-1\in\{1,2\}\quad\Longrightarrow\quad x\in\{2,3\}\,.$$

Substitute in $$-(1)$$:

$$2^{4}=(2-1)(y^{3}-23)-1$$ $$16 = 1\cdot(y^{3}-23)-1$$ $$y^{3}=40$$

Since $$40$$ is not a perfect cube, no integer $$y$$ exists in this case.

Substitute in $$-(1)$$:

$$3^{4}=(3-1)(y^{3}-23)-1$$ $$81 = 2(y^{3}-23)-1$$ $$2(y^{3}-23)=82$$ $$y^{3}=64$$ $$y=4$$

This is an admissible positive-integer solution.

The only solution in positive integers is $$(x,y)=(3,4)$$, giving

$$x+y=3+4=7\,.$$

Hence the maximum possible value of $$x+y$$ is 07.

For a regular polygon with $$k$$ sides $$\bigl(k \ge 3,\,k\in\mathbb{N}\bigr)$$ the measure of each interior angle is

$$\theta \;=\;180^\circ-\frac{360^\circ}{k}\quad -(1)$$

The question asks for all even positive integers $$n$$ that can equal $$\theta$$. Put $$\theta=n$$ in (1):

$$n \;=\;180-\frac{360}{k}\;\Longrightarrow\;\frac{360}{k}=180-n\quad -(2)$$

Let $$d=180-n$$. Then $$d=\dfrac{360}{k}$$, so $$d$$ must be a positive divisor of $$360$$. Also, because the polygon is convex, $$n\lt 180$$, hence $$d=180-n\gt 0$$.

From (2) we have $$k=\dfrac{360}{d}$$. The requirement $$k\ge 3$$ gives $$\dfrac{360}{d}\ge 3\;\Longrightarrow\;d\le 120\quad -(3)$$

Finally, the problem restricts $$n$$ to be even. Since $$180$$ is even, $$n=180-d$$ is even ⇔ $$d$$ is even. Therefore:

• $$d$$ is an even positive divisor of $$360$$.
• $$d\le 120$$ from (3).

List the divisors of $$360$$ (using $$360=2^{3}\!\cdot\!3^{2}\!\cdot\!5$$) and keep those that satisfy both conditions:

Even divisors of $$360$$ not exceeding $$120$$:
$$2,\,4,\,6,\,8,\,10,\,12,\,18,\,20,\,24,\,30,\,36,\,40,\,60,\,72,\,90,\,120$$

Total count = 16.

Each such $$d$$ gives a unique even interior angle $$n=180-d$$ in the open range $$(0,180)$$, so the set $$X$$ has exactly 16 elements.

Hence, the required number of even interior angles is 16.

The cube (standard dice) has $$24$$ possible rotational symmetries. Two dice are considered identical if one symmetry converts the first into the second.

Step 1 Fix an axis using the condition “1 opposite 2”
Because faces $$1$$ and $$2$$ must be on opposite sides, they determine one complete axis of the cube. Any dice can be rotated so that $$1$$ comes to the top and $$2$$ to the bottom. After this choice, only rotations about this vertical axis are still possible, and that residual symmetry is the cyclic group $$C_4$$ of order $$4$$ (turns by $$0^\circ,\,90^\circ,\,180^\circ,\,270^\circ$$).

Step 2 Arrange the numbers 3, 4, 5, 6 on the four side faces
Without considering further symmetry, there are $$4! = 24$$ ways to place $$3,4,5,6$$ around the belt. To eliminate the over-count coming from the surviving rotations of $$C_4$$ we apply Burnside’s lemma.

• Identity rotation fixes all $$24$$ placements.
• A $$90^\circ$$ or $$270^\circ$$ turn would force all four positions to be the same number, impossible because the numbers are distinct ⇒ $$0$$ fixed.
• A $$180^\circ$$ turn pairs front↔back and right↔left; it would require the two numbers in each pair to coincide, again impossible ⇒ $$0$$ fixed.

The number of distinct arrangements is therefore $$\frac{24 + 0 + 0 + 0}{4} = 6.$$

Step 3 Choose colours
Opposite faces must share the same colour, and there are two colours (red, yellow). With three opposite pairs $$(1,2),\;( \text{front,back}),\;( \text{right,left}),$$ each pair can independently be coloured red or yellow, giving $$2^3 = 8$$ possible colourings for every fixed numerical arrangement.

Step 4 Combine numbers and colours
Total distinct dice designs $$= 6 \text{ (numberings)} \times 8 \text{ (colourings)} = 48.$$

Answer: $$48$$ distinct dice can be designed.

A prime number has exactly two positive divisors, $$1$$ and itself. Hence the condition $$ab$$ is prime forces
either $$a=1,\;b=p$$ or $$b=1,\;a=p$$ where $$p$$ is some prime.

$$a=1,\;b=p$$ with $$p$$ prime.

The expression $$bc=p\,c$$ must be the product of exactly two primes. Because one prime factor is already $$p$$, $$c$$ itself must be a prime, say $$q$$. If $$q=p$$, then $$bc=p^2$$ and the square of $$p$$ divides $$abc$$, violating condition (c). Therefore $$q\neq p$$.

The size condition is $$abc=1\cdot p\cdot q=pq\le 30.$$

Count all ordered pairs of distinct primes $$(p,q)$$ with $$pq\le 30$$:

$$\begin{aligned} p=2:&\; q=3,5,7,11,13 &\Rightarrow& 5\\ p=3:&\; q=2,5,7 &\Rightarrow& 3\\ p=5:&\; q=2,3 &\Rightarrow& 2\\ p=7:&\; q=2,3 &\Rightarrow& 2\\ p=11:&\; q=2 &\Rightarrow& 1\\ p=13:&\; q=2 &\Rightarrow& 1 \end{aligned}$$

Total for Case 1: $$5+3+2+2+1+1=14$$ triples.

$$b=1,\;a=p$$ with $$p$$ prime.

Now $$bc=c$$ must be the product of exactly two primes. Write $$c=q\,r$$ where $$q,r$$ are primes.

Condition (c) forbids any square factor, so $$q\neq r$$. If $$p$$ coincided with $$q$$ or $$r$$, $$p^2$$ would divide $$abc=pqr$$, again breaking (c). Thus $$p,q,r$$ are three distinct primes.

The size limit gives $$abc=pqr\le 30.$$ The only product of three distinct primes not exceeding 30 is $$2\cdot3\cdot5=30.$$

For the set $$\{2,3,5\}$$ we can choose any of the three primes for $$a=p$$ (because $$b=1$$ is fixed); the remaining two form $$c$$. Hence Case 2 contributes $$3$$ triples.

Adding both cases: $$14+3=17.$$

Therefore, the required number of triples is 17.

Define the auxiliary quantity
$$\Delta_n \;=\;a_{n}^{\,2}\;-\;a_{n-1}\,a_{n+1},\qquad n\ge 1.$$

Step 1 : Establish a recurrence for $$\Delta_n$$
Using the given relation $$a_{n+2}=-4a_{n+1}-7a_{n}\,,$$ put $$n\to n-1$$ to get
$$a_{n+1} \;=\;-4a_{n}\;-\;7a_{\,n-1}$$ $$-(1)$$

Compute $$\Delta_{n+1}$$:
$$\Delta_{n+1}=a_{n+1}^{\,2}-a_{n}\,a_{n+2}.$$
Substitute $$a_{n+2}=-4a_{n+1}-7a_{n}$$:
$$\Delta_{n+1}=a_{n+1}^{\,2}-a_{n}\bigl(-4a_{n+1}-7a_{n}\bigr)$$ $$\qquad =a_{n+1}^{\,2}+4a_{n}a_{n+1}+7a_{n}^{\,2}$$ $$\qquad =a_{n+1}\bigl(a_{n+1}+4a_{n}+7a_{\,n-1}\bigr)\,+\,7\bigl(a_{n}^{\,2}-a_{\,n-1}a_{n+1}\bigr)$$ (using $$-(1)$$ for the term in parentheses). But $$a_{n+1}+4a_{n}+7a_{\,n-1}=0$$ by $$-(1)$$, so

$$\boxed{\;\Delta_{n+1}=7\,\Delta_n\;}\qquad\text{for all }n\ge 1.$$

Step 2 : Initial value of $$\Delta_n$$
First find $$a_2$$:
$$a_2=-4a_1-7a_0=-4(-4)-7(1)=16-7=9.$$
Hence
$$\Delta_1=a_{1}^{\,2}-a_{0}a_{2}=(-4)^{2}-1\cdot9=16-9=7.$$

Step 3 : Closed form for $$\Delta_n$$
From the recurrence $$\Delta_{n+1}=7\Delta_n$$ and the initial value $$\Delta_1=7$$, we get by induction
$$\Delta_n=7^{\,n}\qquad(n\ge 1).$$

Step 4 : Required expression
For $$n=50$$,
$$a_{50}^{\,2}-a_{49}a_{51}=\Delta_{50}=7^{50}.$$

Step 5 : Number of positive divisors
If $$N=p^{k}$$, the number of positive divisors of $$N$$ is $$k+1$$.
Here $$7^{50}$$ has prime factorisation with exponent $$50$$, so the divisor count is
$$50+1=51.$$

Answer : 51

Let the cubic be $$P(x)=x^{3}+ax^{2}+bx+c$$ with integers $$a,b,c$$ and odd $$c$$.

Define $$p_i=P(i)$$ for $$i=0,1,2,3$$.
Explicitly

$$\begin{aligned} p_0 &= c,\\ p_1 &= 1+a+b+c,\\ p_2 &= 8+4a+2b+c,\\ p_3 &= 27+9a+3b+c. \end{aligned}$$

The given condition is

$$p_{1}^{3}+p_{2}^{3}+p_{3}^{3}=3p_{1}p_{2}p_{3}\,.$$

Recall the factorisation

$$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\bigl(a^{2}+b^{2}+c^{2}-ab-bc-ca\bigr)\,.\;-(1)$$

Applying $$(1)$$ with $$a=p_1,\;b=p_2,\;c=p_3$$ gives

$$\bigl(p_1+p_2+p_3\bigr)\Bigl(p_1^{2}+p_2^{2}+p_3^{2}-p_1p_2-p_2p_3-p_3p_1\Bigr)=0.$$

Hence at least one of the following must hold:

Case 1: $$p_1+p_2+p_3=0$$

Sum the expressions for $$p_1,p_2,p_3$$:

$$p_1+p_2+p_3 = 36+14a+6b+3c.$$

Setting this to zero gives

$$14a+6b+3c=-36.\;-(2)$$

The left side of $$(2)$$ is the sum of two even numbers $$14a,6b$$ and an odd number $$3c$$ (because $$c$$ is odd). Therefore the left side is odd, whereas the right side $$-36$$ is even - a contradiction.
Thus Case 1 is impossible when $$c$$ is odd.

Case 2: $$p_1=p_2=p_3$$

Equating consecutive pairs:

$$\begin{aligned} p_1=p_2 &\Longrightarrow 1+a+b+c = 8+4a+2b+c \Longrightarrow b = -7-3a,\;-(3)\\[4pt] p_1=p_3 &\Longrightarrow 1+a+b+c = 27+9a+3b+c \Longrightarrow b = -13-4a.\;-(4) \end{aligned}$$

Solving $$(3)$$ and $$(4)$$ simultaneously:

$$-7-3a = -13-4a \;\;\Longrightarrow\;\; a=-6,$$

and then $$b = -7-3(-6)=11.$$

Therefore

$$a=-6,\qquad b=11,\qquad c \text{ any odd integer.}$$

Now evaluate the required expression

$$\begin{aligned} p_2 + 2p_1 - 3p_0 &= (8+4a+2b+c) + 2(1+a+b+c) - 3c \\[4pt] &= 10 + 6a + 4b \\[4pt] &= 10 + 6(-6) + 4(11) \\[4pt] &= 10 - 36 + 44 \\[4pt] &= 18. \end{aligned}$$

Notice that $$c$$ has cancelled out, so the value is the same for every odd $$c$$.

Hence the required value is 18.

Let the coordinates of the vertices be
$$A(x_A,\,y_A),\qquad B(0,0),\qquad C(x_C,\,y_C).$$

Step 1: Coordinates of the auxiliary points

1. $$BC:CD = 1:1$$ and $$D$$ lies beyond $$C$$ on $$BC$$.
 Vector form $$\vec{D} = \vec{C} + (\vec{C}-\vec{B}) = 2\vec{C}$$
 Hence $$D(2x_C,\,2y_C).$$

2. $$CA:AE = 1:2$$ with $$E$$ beyond $$A$$ on $$CA$$.
 Vector form $$\vec{E} = \vec{A} + 2(\vec{A}-\vec{C}) = 3\vec{A}-2\vec{C}$$
 Hence $$E(3x_A-2x_C,\,3y_A-2y_C).$$

3. $$AB:BF = 1:3$$ with $$F$$ beyond $$B$$ on $$AB$$.
 Vector form $$\vec{F} = \vec{B} + 3(\vec{B}-\vec{A}) = -3\vec{A}$$
 Hence $$F(-3x_A,\,-3y_A).$$

Step 2: Use the centroid of $$\triangle ABC$$

The centroid formula gives
$$\vec{G}=\frac{\vec{A}+\vec{B}+\vec{C}}{3} = \frac{\vec{A}+\vec{C}}{3}.$$
Given $$G(32,24),$$ so

$$\vec{A}+\vec{C}=3\vec{G}=(96,\,72). \quad -(1)$$

Step 3: Centroid of $$\triangle DEF$$

Add the position vectors of $$D,E,F$$:

$$\vec{D}+\vec{E}+\vec{F} = 2\vec{C} + (3\vec{A}-2\vec{C}) + (-3\vec{A}) = 0.$$

Therefore the centroid of $$\triangle DEF$$ is

$$\vec{K}= \frac{\vec{D}+\vec{E}+\vec{F}}{3} = \vec{0},$$
so $$K(0,0),$$ the origin.

Step 4: Distance between $$G(32,24)$$ and $$K(0,0)$$

$$GK = \sqrt{(32-0)^2 + (24-0)^2} = \sqrt{32^2 + 24^2} = \sqrt{1024 + 576} = \sqrt{1600} = 40.$$

Hence the required length is 40.

Final Answer: 40

All six sides of the hexagon are already red, so any triangle that uses at least one side of the hexagon automatically contains a red side.
Hence we only have to worry about triangles whose three sides are all diagonals.

In a regular hexagon the only triples of vertices that are pairwise non-adjacent (so that every side of the triangle joining them is a diagonal) are the alternate-vertex triples $$\{A_1,A_3,A_5\}$$ and $$\{A_2,A_4,A_6\}$$. Thus the only two “all-diagonal” triangles are $$\triangle A_1A_3A_5$$ and $$\triangle A_2A_4A_6$$.

Label the nine diagonals as follows:
Triangle $$A_1A_3A_5$$ (set $$X$$): $$A_1A_3,\;A_3A_5,\;A_5A_1$$  (3 diagonals)
Triangle $$A_2A_4A_6$$ (set $$Y$$): $$A_2A_4,\;A_4A_6,\;A_6A_2$$  (3 diagonals)
Remaining crossed diagonals (set $$Z$$): $$A_1A_4,\;A_2A_5,\;A_3A_6$$  (3 diagonals)

Total colourings of the nine diagonals (red/blue independently) $$=2^{9}=512.$$

The requirement “every triangle has at least one red side” is violated only if • all three diagonals of $$X$$ are blue, or • all three diagonals of $$Y$$ are blue (or both).

Let
$$A=\{\text{all edges of }X\text{ are blue}\},\qquad B=\{\text{all edges of }Y\text{ are blue}\}.$$

Count the forbidden colourings:

Case |A|: 3 edges fixed blue, the remaining 6 edges free $$\Longrightarrow |A|=2^{6}=64.$$

Case |B|: similarly $$|B|=64.$$

Case |A$$\cap$$ B|: both sets of 3+3=6 edges fixed blue, the 3 edges of $$Z$$ free $$\Longrightarrow |A\cap B|=2^{3}=8.$$

By the Principle of Inclusion-Exclusion, the number of forbidden colourings is $$|A|+|B|-|A\cap B|=64+64-8=120.$$

Therefore the required number of valid colourings is $$N=512-120=392.$$

The digits of $$N$$ are 3, 9, and 2. Sum of the squares of the digits: $$3^{2}+9^{2}+2^{2}=9+81+4=94.$$

Final answer: 94

Label the vertices of the convex $$50$$-gon $$\mathcal P$$, in counter-clockwise order, as $$1,2,\ldots ,50$$.

Pick the vertex $$1$$ and draw every diagonal emanating from it: $$\,(1,3),(1,4),\ldots ,(1,49)\,.$$ There are

$$50-3=47$$

such diagonals, and none of them intersect one another because they have the common endpoint $$1$$. They divide the polygon into

$$48$$

fan-shaped triangles $$T_1=[1,2,3],\;T_2=[1,3,4],\ldots ,T_{48}=[1,49,50].$$

Every diagonal that is not incident with the vertex $$1$$ must have its two endpoints on the boundary of two different fan triangles; call them the endpoint triangles of the diagonal. Such a diagonal necessarily crosses exactly one of the star diagonals, namely, the one that separates its two endpoint triangles.

No triangle can serve as an endpoint triangle for two different diagonals that are not incident with $$1$$. Indeed, if two diagonals shared the same endpoint triangle, then, inside that triangle, they would emanate from the same side and would have to cross each other before reaching their opposite endpoints, creating an additional interior intersection for at least one of them. That would violate the “at most one intersection per diagonal’’ condition.

Consequently, each diagonal that is not incident with $$1$$ occupies two exclusive endpoint triangles, and the $$48$$ triangles can supply at most

$$\left\lfloor\dfrac{48}{2}\right\rfloor=24$$

such diagonals. Call these the partner diagonals; each of them crosses exactly one star diagonal and is disjoint from every other diagonal in the set.

Hence the size $$|\mathcal F|$$ of any minimally friendly set satisfies

$$|\mathcal F| \le 47 \;+\;24 \;=\;71.$$

To see that $$71$$ can actually be attained, pair the triangles symmetrically:

$$T_1\!-\!T_{25},\;T_2\!-\!T_{26},\;\ldots ,\;T_{24}\!-\!T_{48}.$$

For each pair $$T_i,\;T_{i+24}$$ (indices mod $$48$$) draw the diagonal joining their outer vertices. Explicitly, for $$i=1,\dots ,24$$ join $$\bigl(i+1\bigr)\text{ with }\bigl(i+25\bigr).$$ These $$24$$ partner diagonals are mutually non-intersecting and every one of them crosses exactly the star diagonal $$(1,i+25).$$ Together with the original $$47$$ star diagonals we obtain

$$47+24=71$$

diagonals, and each diagonal intersects at most one other member of the family. Thus $$71$$ is both an upper bound and is achievable.

Therefore, the largest possible size of a minimally friendly set $$\mathcal F$$ in a convex $$50$$-gon is

71.

Let $$n$$ have $$d$$ digits. Write its decimal digits in some order as

$$\underbrace{1,1,\dots,1}_{k\text{ times}},\;a_1,a_2,\dots ,a_r\qquad(k\ge 0,\;r\ge 1).$$

The $$k$$ digits 1 only affect the digit-sum, whereas the $$r$$ digits $$a_1,\dots ,a_r\in\{2,3,4,5,6,7,8,9\}$$ determine the non-zero product.

Step 1: Digits permitted by “square free product”.
A square-free integer has no repeated prime factor. Factorising each possible digit gives

$$\begin{aligned} 2&=2, & 3&=3, & 4&=2^2, & 5&=5,\\ 6&=2\cdot3, & 7&=7, & 8&=2^3, & 9&=3^2. \end{aligned}$$

• 4, 8, 9 each contain a squared prime power ⇒ they are forbidden.
• 0 would make the product zero ⇒ forbidden.
Thus every allowed digit is chosen from $$\{1,2,3,5,6,7\}$$.

Step 2: Preventing repeated prime factors.
Write the overall product as $$P(n).$$ Each prime appearing in $$P(n)$$ must occur once only.

• If the digit 6 ( =2·3 ) is used, neither 2 nor 3 can appear elsewhere.
• Apart from that restriction we may choose at most one digit whose prime content is 2, at most one whose prime content is 3, and at most one 5 or 7.

Step 3: Choose the product as large as possible.
Adding many 1’s increases the digit-sum $$S(n)$$ by 1 each time but does not change $$P(n).$$ Because $$S(n)$$ must be a proper divisor of $$P(n),$$ we want $$P(n)$$ as large as possible so that many 1’s can be inserted without making the sum exceed the product.

The largest square-free product obtainable from single digits is obtained by taking the four mutually coprime digits 2, 3, 5, 7:

$$P(n)=2\cdot3\cdot5\cdot7=210.$$

Step 4: Express the digit-sum.
With these four digits fixed, let $$k$$ be the number of 1’s. Then

$$S(n)=2+3+5+7+k=17+k.$$

Step 5: Make $$S(n)$$ a proper divisor of 210.
The positive divisors of 210 are $$1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210.$$ The proper divisors that are >17 are

$$21,30,35,42,70,105.$$

Setting $$17+k$$ equal to each of these gives

$$\begin{aligned} 17+k=21 &\Rightarrow k=4,\\ 17+k=30 &\Rightarrow k=13,\\ 17+k=35 &\Rightarrow k=18,\\ 17+k=42 &\Rightarrow k=25,\\ 17+k=70 &\Rightarrow k=53,\\ 17+k=105&\Rightarrow k=88.\\ \end{aligned}$$

All these choices satisfy $$S(n)\lt P(n),$$ but the last one clearly gives the largest number of 1’s.

Step 6: Count the digits.
Taking $$k=88$$ gives $$S(n)=17+88=105,\qquad S(n)\,|\,P(n)\ (210/105=2).$$ Total number of digits

$$d=r+k=4+88=92.$$

Step 7: Why no longer number is possible.
Any other choice of digits yields a product that is either
• smaller than 210 (if one or more of 2, 3, 5, 7 is omitted), or
• repeats a prime factor (if the digit 6 is mixed with 2 or 3).
A smaller product places the upper bound $$S(n)\le P(n)-1$$ lower and therefore allows strictly fewer 1’s than the 88 obtained above.

Hence the maximum possible number of digits of a number in the set $$A$$ is

$$\boxed{92}.$$

Let $$|A|=m,\; \max(A)=p,\; |B|=n,\; \max(B)=q$$.
The two given relations translate to

$$p\,n = 12 \qquad -(1)$$ $$m\,q = 11 \qquad -(2)$$

Because all quantities are positive integers, we analyse the factorisations of the right-hand sides.

Since $$11$$ is prime, equation $$(2)$$ gives only two possibilities: Case 1: $$m=1,\; q=11$$   or   Case 2: $$m=11,\; q=1$$.

Case 1: $$m=1,\; q=11$$

Here $$A$$ has exactly one element, so $$p=\max(A)$$ equals that lone element. Equation $$(1)$$ becomes $$p\,n = 12$$. All factor pairs $$(p,n)$$ of $$12$$ are

$$(1,12),\; (2,6),\; (3,4),\; (4,3),\; (6,2),\; (12,1).$$

Because $$B$$ is a set (no repetition) whose largest element is $$11$$, its size cannot exceed $$11$$. Thus the pair $$(1,12)$$ is impossible. The remaining five pairs are valid.

For a fixed $$(p,n)$$, set $$A=\{p\}$$ is unique.
Set $$B$$ must contain $$11$$ and exactly $$n-1$$ other distinct numbers chosen from $$\{1,2,\dots,10\}$$. Therefore the number of such $$B$$ is $$\binom{10}{\,n-1\,}$$.

Counting for each valid $$n$$:

$$n=6:\; \binom{10}{5}=252$$
$$n=4:\; \binom{10}{3}=120$$
$$n=3:\; \binom{10}{2}=45$$
$$n=2:\; \binom{10}{1}=10$$
$$n=1:\; \binom{10}{0}=1$$

Total pairs in Case 1: $$252+120+45+10+1 = 428$$.

Case 2: $$m=11,\; q=1$$

Since $$\max(B)=1$$, the only possible set is $$B=\{1\}$$, giving $$n=1$$.
Equation $$(1)$$ now yields $$p\,\cdot 1 = 12 \implies p=12$$.

Set $$A$$ must have size $$11$$ with maximum $$12$$, so it must be $$\{12\}$$ together with any $$10$$ distinct numbers from $$\{1,2,\dots,11\}$$. Number of choices for $$A$$: $$\binom{11}{10}=11$$. With $$B$$ fixed, pairs in Case 2 = $$11$$.

Total ordered pairs: $$N = 428 + 11 = 439$$.

Express $$N$$ as $$100a + b$$ with $$a,b \lt 100$$: $$439 = 100 \times 4 + 39 \implies a = 4,\; b = 39$$.

Hence $$a + b = 4 + 39 = 43$$.

Final answer: 43

Let $$n\in\mathbb{N}$$ and let the required function be $$f:\{1,2,\dots ,n\}\rightarrow\mathbb{Z}_{\ge 0}$$ with $$f(1)\le f(2)\le\cdots\le f(n)$$.

The condition on the sum gives
$$\sum_{i=1}^{n}\bigl(i+f(i)\bigr)=\frac{n(n+1)}{2}+\sum_{i=1}^{n}f(i)=2023.$$ Hence
$$\sum_{i=1}^{n}f(i)=2023-\frac{n(n+1)}{2}\;.\quad -(1)$$

Because every $$f(i)$$ is non-negative, the right-hand side of $$(1)$$ must be $$\ge 0$$, so $$\dfrac{n(n+1)}{2}\le 2023$$. To minimise $$\sum f(i)$$ we must maximise $$\dfrac{n(n+1)}{2}$$ subject to this inequality.

Compute the largest such $$n$$:
$$n(n+1)\le 4046.$$ Checking successive integers, $$63\cdot 64=4032\le 4046,\qquad 64\cdot 65=4160\gt 4046.$$ Therefore $$n_{\text{max}}=63.$

With $$n=63$$, equation $$(1)$$ gives
$$$$\sum_{i=1}^{63}f(i$$)=2023-$$\frac{63\cdot 64}{2}$$=2023-2016=7.$$ Thus the least possible value of $$$$\sum$$ f(i)$$ is $$7$$ and it is achieved when $$n=63$$.

We now have to count the non-decreasing sequences $$0\le f(1)\le f(2)\le$$\cdot$$s\le f(63),\qquad$$\sum_{i=1}^{63}f(i$$)=7.$$

Such a sequence is obtained by writing the integer $$7$$ as an unordered sum of non-negative integers (a partition) and then appending enough zeros to reach length $$63$$. Hence the number of admissible sequences equals the number of partitions of $$7$$ into any number of parts (at most $$63$$, but $$63\gt 7$$ so this restriction is irrelevant).

The partitions of $$7$$ are:
$$7;\;6+1;\;5+2;\;5+1+1;\;4+3;\;4+2+1;\;4+1+1+1;\;3+3+1;\;3+2+2;\; 3+2+1+1;\;3+1+1+1+1;\;2+2+2+1;\;2+2+1+1+1;\;2+1+1+1+1+1;\; 1+1+1+1+1+1+1.$$ There are $$15$$ of them.

Therefore exactly $$15$$ functions satisfy all the conditions.

Answer: $$\boxed{15}$$

Let the six given integral lengths be arranged in increasing order as
$$5,\;6,\;7,\;8,\;9,\;10.$$

We have to pick four distinct numbers and arrange them as sides of a non-degenerate trapezium (one pair of opposite sides parallel) in such a way that no two admissible trapezia are congruent.

Step-1 : Combinatorics of choosing sides
Choosing any four different numbers out of the six can be done in $$\binom{6}{4}=15$$ ways.
Fix such a 4-tuple and denote its members (in increasing order) by $$a\lt b\lt c\lt d.$$

Step-2 : Which two of the four will be the parallel bases?
Out of the four numbers, any unordered pair can be declared to be the bases; the remaining two then become the legs.
Number of unordered pairs of a 4-set = $$\binom{4}{2}=6.$$
Hence each 4-tuple offers 6 possible $$(\text{bases},\text{legs})$$ assignments.

Step-3 : Existence condition for a trapezium
Let $$B_1\;(\gt B_2)$$ be the two bases and $$L_1,\,L_2$$ the two legs. Draw $$B_1B_2$$ perpendiculars to obtain two triangles glued along a segment of length $$B_1-B_2$$. Non-degeneracy is possible ⇔ that segment can really be spanned by $$L_1,L_2$$;
in other words

$$\lvert L_1-L_2\rvert \; \lt\; B_1-B_2\; \lt\; L_1+L_2\qquad -(1)$$

When either inequality becomes equality the height of the trapezium is zero, i.e. it degenerates into a straight line. Therefore strict inequalities are needed.

Step-4 : Exhaustive check for every 4-tuple
Because the numbers are only from 5 to 10, the verification can be done case-wise very quickly. The table below lists, for every 4-tuple, those unordered pairs that satisfy (1). (The larger of the two numbers of a pair is written first; this does not double count anything.)

5 6 7 8 : (8,5)   (1)
5 6 7 9 : (9,7), (9,6), (9,5)   (3)
5 6 7 10: (10,7), (10,6), (10,5)  (3)
5 6 8 9 : (9,5)   (1)
5 6 8 10: (10,8), (10,6), (10,5)  (3)
5 6 9 10: (10,5)   (1)
5 7 8 9 : (9,5), (8,5), (7,5)  (3)
5 7 8 10: (10,5)   (1)
5 7 9 10: (10,5), (9,5), (7,5)  (3)
5 8 9 10: (10,5), (9,5), (8,5)  (3)
6 7 8 9 : (9,6)   (1)
6 7 8 10: (10,8), (10,7), (10,6)  (3)
6 7 9 10: (10,6)   (1)
6 8 9 10: (10,6), (9,6), (8,6)  (3)
7 8 9 10: (10,7)   (1)

The right-hand margin of each line shows how many of the 6 unordered pairs fulfil condition (1). Adding them up

$$1+3+3+1+3+1+3+1+3+3+1+3+1+3+1 = 31.$$

Step-5 : Congruence check
All the counted trapezia have four mutually different side lengths. Interchanging the two legs merely produces a mirror image, which is congruent. Any two trapezia appearing in different rows or with different pairs of parallel sides have different ordered 4-tuples of side lengths and therefore are not congruent. Thus every entry above corresponds to exactly one mutually non-congruent non-degenerate trapezium.

Hence the required number of mutually non-congruent non-degenerate trapeziums is 31.

Let the number of notes of each denomination be

$$\begin{aligned} a &\;:\;64 \text{-Ben notes}\\[-2pt] b &\;:\;32 \text{-Ben notes}\\[-2pt] c &\;:\;16 \text{-Ben notes}\\[-2pt] d &\;:\; 8 \text{-Ben notes}\\[-2pt] e &\;:\; 4 \text{-Ben notes}\\[-2pt] f &\;:\; 2 \text{-Ben notes}\\[-2pt] g &\;:\; 1 \text{-Ben notes} \end{aligned}$$

Government rule: each of $$a,b,c,d,e,f,g$$ can take the values $$0,1,2$$, except $$a$$ which can be only $$0$$ or $$1$$ (because $$2\times64=128\gt100$$).

All possible combinations satisfy the Diophantine equation

$$64a + 32b + 16c + 8d + 4e + 2f + g = 100 \qquad -(1)$$

We count the integral solutions of (1) under the above limits by systematic case-work.

Equation (1) becomes

$$32b + 16c + 8d + 4e + 2f + g = 36 \qquad -(2)$$

Here $$b\in\{0,1\}$$ (two 32-Ben notes would overshoot the remaining 36).

Then (2) reduces to

$$16c + 8d + 4e + 2f + g = 4 \qquad -(3)$$

The left side now involves only denominations $$\le 4$$, whose total value is at most $$2\times4 + 2\times2 + 2\times1 = 14$$, so every variable can still be $$0,1,2$$.

Because $$4$$ itself is a denomination, list the possibilities for (3):

• one 4-Ben note : $$(c,d,e,f,g)=(0,0,1,0,0)$$
• two 2-Ben notes : $$(0,0,0,2,0)$$
• one 2-Ben and two 1-Ben notes : $$(0,0,0,1,2)$$

Hence, Case 1.1 contributes $$3$$ valid combinations.

Equation (2) now reads

$$16c + 8d + 4e + 2f + g = 36 \qquad -(4)$$

Choose $$c \in \{0,1,2\}$$.

Sub-case 1.2.1: $$c = 2$$  ⇒ $$8d + 4e + 2f + g = 4$$ Exactly the same equation as (3), therefore contributes $$3$$ more solutions.

Sub-case 1.2.2: $$c = 1$$  ⇒ $$8d + 4e + 2f + g = 20$$ Take $$d \in \{0,1,2\}$$.

• $$d = 2$$ ⇒ $$4e + 2f + g = 4$$ ⇒ again yields the same 3 solutions.
• $$d = 1$$ ⇒ $$4e + 2f + g = 12$$.
  Pick $$e = 2$$ ⇒ $$2f + g = 4$$ gives     - $$(f,g)=(2,0)$$, $$(1,2)$$  ⇒ 2 solutions.
• $$d = 0$$ is impossible because the maximum from lower denominations is 14.

So Sub-case 1.2.2 adds $$3 + 2 = 5$$ solutions.

Sub-case 1.2.3: $$c = 0$$ The maximum attainable value with the remaining denominations is $$14 \lt 36$$, so no solution.

Adding the contributions of Case 1:

$$3\;(\text{Case 1.1}) + 3 + 5\;(\text{Case 1.2}) = 11$$ solutions.

Equation (1) becomes

$$32b + 16c + 8d + 4e + 2f + g = 100 \qquad -(5)$$

Now $$b \in \{0,1,2\}$$.

Case 2.1: $$b = 2$$  ⇒ remaining amount $$= 36$$.

Exactly the same equation (4) analysed above, and we saw it has $$8$$ solutions (3 when $$c=2$$ and 5 when $$c=1$$). Hence Case 2.1 contributes $$8$$ solutions.

Case 2.2: $$b = 1$$  ⇒ remaining amount $$= 68$$. The maximum possible with $$16,8,4,2,1$$ is $$62 \lt 68$$, so no solution.

Case 2.3: $$b = 0$$  ⇒ need full 100 with denominations $$\le16$$. Again impossible because their maximum total is $$62$$.

Therefore Case 2 contributes only $$8$$ solutions.

Adding both main cases:

$$\boxed{11 + 8 = 19}$$

Hence, the change for 100 Bens can be given in exactly 19 different ways following the Binary Government’s rules.

Let the four integers be written in increasing order as $$x_1\lt x_2\lt x_3\lt x_4$$.
With this notation we have $$a=x_1,\; b=x_2,\; d=x_3,\; c=x_4$$ (because the required order is $$a\lt b\lt d\lt c$$), and every choice of four distinct integers from $$1\!-\!20$$ gives exactly one admissible quadruple.

Total number of admissible quadruples
    $$= \binom{20}{4}=4845$$.

A quadruple is balanced when $$a+c=b+d$$, i.e. when $$x_1+x_4=x_2+x_3$$.
Re-writing the equality gives $$x_4-x_3 = x_2-x_1$$.
Thus the outer two gaps must be equal. Set that common gap to $$d\,(d\ge 1)$$:

Case d fixed
Choose $$x_1$$ and $$x_3$$ such that

$$\begin{aligned} x_2 &= x_1+d, & x_4 &= x_3+d,\\ 1 &\le x_1, & x_4 &\le 20,\\ x_2&=x_1+d &\lt &x_3,\\ x_3&\lt x_4=x_3+d. \end{aligned}$$

The inequalities reduce to

$$\begin{cases} 1\le x_1\le 19-2d,\\[4pt] x_1+d+1\le x_3\le 20-d. \end{cases}$$

For a given $$d$$ the number of possible $$x_3$$ for each $$x_1$$ is
$$\bigl(20-d\bigr)-\bigl(x_1+d+1\bigr)+1 = 20-2d-x_1.$$

Hence the number of balanced quadruples for that $$d$$ is

$$ N_d=\sum_{x_1=1}^{19-2d}\bigl(20-2d-x_1\bigr) =(19-2d)\bigl(20-2d\bigr)-\frac{(19-2d)(20-2d)}{2} =\frac{(19-2d)(18-2d)}{2}. $$

Valid $$d$$ values satisfy $$19-2d\ge 1\;\Longrightarrow\;d\le 9$$. Computing $$N_d$$:

$$\begin{array}{c|ccccccccc} d & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline N_d & 153 & 120 & 91 & 66 & 45 & 28 & 15 & 6 & 1 \end{array}$$

Total balanced quadruples
$$N=\sum_{d=1}^{9} N_d = 153+120+91+66+45+28+15+6+1 = 525.$$

Therefore

$$\text{unbalanced quadruples}=4845-525=4320.$$

We must choose $$4411$$ quadruples for the set $$S$$. Even if we take all unbalanced ones (4320 of them), we still need
$$4411-4320=91$$ more, and every extra choice will necessarily be balanced.

Hence the least possible number of balanced quadruples in any such set $$S$$ is $$91$$.

Answer: 91

Write the running divisor-sum in parity (odd/even) form:

For any integer $$m$$, the divisor count $$d(m)$$ is odd ⇔ $$m$$ is a perfect square.
Reason: divisors come in pairs $$a,\;m/a$$; a divisor is unpaired only when $$a=m/a$$, i.e. when $$a^2=m$$.

Hence for every $$i$$,

$$d(i)\equiv\begin{cases}1 &\text{(mod 2) if } i \text{ is a square}\\ 0 &\text{(mod 2) otherwise}\end{cases}$$

Consider the partial sum

$$S(n)=\sum_{i=1}^{n}d(i)\pmod{2}.$$

Its parity equals the number of perfect squares ≤ $$n$$, counted modulo 2. The count of squares ≤ $$n$$ is $$\lfloor\sqrt{n}\rfloor$$, so

$$S(n)\text{ is odd}\;\Longleftrightarrow\;\lfloor\sqrt{n}\rfloor\text{ is odd}.$$(1)

Let $$k=\lfloor\sqrt{n}\rfloor$$. For each odd $$k$$, condition (1) holds for every $$n$$ in the block

$$k^2\le n\le (k+1)^2-1.$$

The length of this block is

$$(k+1)^2-k^2=(2k+1).$$(2)

All $$n$$ under consideration satisfy $$n\le2023$$. Since $$44^2=1936\lt2023$$ and $$45^2=2025\gt2023$$, we have $$\lfloor\sqrt{n}\rfloor\le44.$$ Thus $$k$$ runs over the odd integers

$$1,\,3,\,5,\dots,\,43.$$

There are $$\dfrac{43+1}{2}=22$$ such values. Using (2), the required count is

$$r=\sum_{\substack{k=1\\k\text{ odd}}}^{43}(2k+1).$$(3)

Write $$k=2j-1\;(j=1\text{ to }22)$$. Then

$$2k+1=2(2j-1)+1=4j-1,$$

and (3) becomes

$$r=\sum_{j=1}^{22}(4j-1)=4\sum_{j=1}^{22}j-\sum_{j=1}^{22}1.$$

Compute the two sums:

$$\sum_{j=1}^{22}j=\frac{22\cdot23}{2}=253,$$

$$\sum_{j=1}^{22}1=22.$$

Therefore

$$r=4(253)-22=1012-22=990.$$

The problem asks for the sum of the digits of $$r$$:

$$9+9+0=18.$$

Answer: 18