IOQM Previous Papers Sept 28 2025

Coordinates of the given vertices are
$$A(0,0),\; B(10,0),\; C(8,5),\; D(a,b).$$

1. Condition $$AB \parallel CD$$:
  • The slope of $$AB$$ is $$0$$ (horizontal line).
  • Therefore the slope of $$CD$$ must also be $$0$$, which gives
$$b-5 = 0 \;\Longrightarrow\; b = 5.$$

2. Condition $$AD = BC$$:
  • Compute $$BC$$:
$$BC = \sqrt{(8-10)^{2} + (5-0)^{2}} = \sqrt{(-2)^{2}+25} = \sqrt{29}.$$
  • Write $$AD$$ in terms of $$a$$ and $$b$$:
$$AD = \sqrt{(a-0)^{2} + (b-0)^{2}} = \sqrt{a^{2}+b^{2}}.$$
  • Equate the two lengths:
$$\sqrt{a^{2}+b^{2}} = \sqrt{29}\;\Longrightarrow\; a^{2}+b^{2}=29.$$

3. Substitute $$b = 5$$ into the length equation:
$$a^{2}+25 = 29 \;\Longrightarrow\; a^{2}=4 \;\Longrightarrow\; a = \pm 2.$$

4. Required value $$a^{2}b$$:
$$a^{2}b = 4 \times 5 = 20.$$
(The result is the same for $$a=2$$ or $$a=-2$$ because $$a^{2}$$ is positive.)

Hence, the value of $$a^{2}b$$ is 20.

First write the definition of the function.

If $$n$$ is odd, $$f(n)=n-1$$ (which is even).
If $$n$$ is even, $$f(n)=n^{2}-1$$ (which is odd).

We want all positive integers $$n$$ for which $$f(f(n))=99$$.

$$n$$ is odd.
Then $$f(n)=n-1$$ (even).
Apply $$f$$ again: $$f(f(n))=(n-1)^{2}-1$$ because the input is now even.
Set equal to $$99$$:

$$(n-1)^{2}-1 = 99 \;\;\Longrightarrow\;\; (n-1)^{2}=100 \;\;\Longrightarrow\;\; n-1=\pm10.$$

Since $$n$$ must be positive, $$n-1=10$$ gives $$n=11$$, while $$n-1=-10$$ gives the negative integer $$n=-9$$, which is rejected. Hence the only solution in this case is $$n=11$$.

$$n$$ is even.
Then $$f(n)=n^{2}-1$$ (odd).
Apply $$f$$ again: $$f(f(n))=(n^{2}-1)-1 = n^{2}-2.$$ Set equal to $$99$$:

$$n^{2}-2 = 99 \;\;\Longrightarrow\;\; n^{2}=101.$$

But $$101$$ is not a perfect square, so there is no even integer $$n$$ satisfying the equation.

Combining both cases, the only valid positive integer is $$n=11$$.

Required sum of all possible values of $$n$$ = $$11$$.

Answer: 11

A positive integer contains no prime factors other than $$2$$ or $$3$$ precisely when it can be written in the form $$n = 2^{a}\,3^{b}$$ with non-negative integers $$a,b$$.

We must list all such numbers not exceeding $$100$$.

Upper limits for the exponents:
Because $$2^{6}=64 \lt 100 \lt 2^{7}$$, we need only $$0 \le a \le 6$$.
Because $$3^{4}=81 \lt 100 \lt 3^{5}$$, we need only $$0 \le b \le 4$$.

$$n = 2^{a}$$ gives $$1,\,2,\,4,\,8,\,16,\,32,\,64$$ → 7 numbers.

$$n = 3\cdot2^{a} \le 100$$ Allowed $$a=0\text{ to }5$$ (because $$3\cdot2^{5}=96 \le 100$$). Numbers: $$3,\,6,\,12,\,24,\,48,\,96$$ → 6 numbers.

$$n = 9\cdot2^{a} \le 100$$ Allowed $$a=0\text{ to }3$$ (since $$9\cdot2^{4}=144 \gt 100$$). Numbers: $$9,\,18,\,36,\,72$$ → 4 numbers.

$$n = 27\cdot2^{a} \le 100$$ Allowed $$a=0,1$$. Numbers: $$27,\,54$$ → 2 numbers.

$$n = 81\cdot2^{a} \le 100$$ Only $$a=0$$ works. Number: $$81$$ → 1 number.

Adding all cases: $$7 + 6 + 4 + 2 + 1 = 20$$.

Therefore, the count of positive integers $$n \le 100$$ with no prime divisors other than $$2$$ or $$3$$ is $$\mathbf{20}$$.

The six faces of each die bear the numbers $$1,2,4,8,16,32$$ because $$2^{0}=1,\,2^{1}=2,\dots ,2^{5}=32$$.

For every die the product of the numbers on opposite faces equals $$2^{5}=32$$, so every opposite pair must be one of
$$(1,32),\;(2,16),\;(4,8).$$

When two dice are stacked:

Case 1: bottom die
  • Its bottom face is hidden by the table.
  • Its top face is hidden by the upper die.
  ⇒ The two hidden faces are opposite faces of the same die.

Case 2: top die
  • Its bottom face is hidden by contact with the lower die.
  • Its top face is visible.
  ⇒ Its bottom face and top face are opposite faces.

Total faces of two dice = 12.
Visible faces = 4 (sides of lower) + 4 (sides of upper) + 1 (top of upper) = 9.
Hidden faces = 3.

The sum of the six numbers on one die is $$1+2+4+8+16+32 = 63$$, so for two dice the total of all 12 numbers is $$126$$.

To maximise the sum of the 9 visible numbers we must minimise the sum of the 3 hidden numbers.

Hidden faces of the bottom die
They are opposite, so they must be an element of the set $$\{(1,32),\,(2,16),\,(4,8)\}$$.
Their possible sums are $$33,18,12$$ respectively. The minimum is $$12$$, achieved by choosing the pair $$(4,8).$$ Hence the bottom and top faces of the lower die should bear $$4$$ and $$8$$ (order irrelevant).

Hidden face of the top die
Choose the smallest available number for the hidden bottom face. • If we hide $$1$$, its opposite (the visible top face) must be $$32$$.
• Hiding $$2$$ would force $$16$$ on the top face, which is worse both for minimising the hidden sum and for maximising the visible sum.
Therefore we hide $$1$$ and make the top face $$32$$.

Thus the three hidden numbers are $$4,8,1$$ whose sum is $$13$$.

Maximum possible visible sum $$N = 126 - 13 = 113.$$

The digits of $$113$$ are $$1,1,3$$. Sum of the squares of these digits $$1^{2}+1^{2}+3^{2}=1+1+9 = 11.$$

Hence the required answer is 11.

The expansion is

$$(x+1)(x^{2}+3)(x^{4}+5)(x^{8}+7)\dotsm(x^{1024}+21).$$

There are $$11$$ factors. In general, the $$k^{\text{th}}$$ factor is
$$(x^{2^{k}} + (2k+1)), \qquad k = 0,1,2,\dots ,10.$$

While forming a particular term we either pick $$x^{2^{k}}$$ (with coefficient $$1$$) or the constant $$(2k+1)$$ from the $$k^{\text{th}}$$ factor. Hence:

• If we pick $$x^{2^{k}}$$ from a factor, the power $$2^{k}$$ is added to the total exponent.
• If we pick the constant, the factor contributes a multiplicative constant $$(2k+1)$$ to the coefficient.

Thus, to obtain $$x^{2025}$$ the set of chosen exponents must satisfy

$$\sum_{k\in S} 2^{k} = 2025,$$

where $$S$$ is the set of indices from which we selected $$x^{2^{k}}$$. Because powers of two are unique, this representation is the binary expansion of $$2025$$.

Write $$2025$$ in binary:

$$2025 = 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 1$$

Corresponding indices:

$$S = \{10,\,9,\,8,\,7,\,6,\,5,\,3,\,0\}.$$

The remaining indices $$\{1,2,4\}$$ are those from which we must take the constant terms. Therefore the required coefficient is

$$N = (2\cdot 1 + 1)\,(2\cdot 2 + 1)\,(2\cdot 4 + 1) = 3 \times 5 \times 9 = 135.$$

We need the remainder of $$N$$ modulo $$100$$:

$$135 \equiv 35 \pmod{100}.$$

Hence the remainder is 35.

All primes other than $$2$$ are odd. If $$2$$ were one of the four primes, then $$2+\text{(odd)}+\text{(odd)}+\text{(odd)}$$ would be odd, while the required sum is $$240$$ (even). Hence none of the four primes is $$2$$; all four primes are odd.

List of odd primes not exceeding $$70$$ (in increasing order): $$3,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,29,\,31,\,37,\,41,\,43,\,47,\,53,\,59,\,61,\,67$$.

Let the four distinct primes be $$p_1 \lt p_2 \lt p_3 \lt p_4 \le 70$$ with $$p_1+p_2+p_3+p_4 = 240$$.

To make $$p_1$$ as small as possible, make the other three primes as large as possible (but still distinct and $$\le 70$$). The three largest distinct primes $$\le 70$$ are $$67, 61, 59$$ whose sum is $$67+61+59 = 187$$.

Therefore $$p_1 + 187 = 240 \;\;\Longrightarrow\;\; p_1 = 240-187 = 53$$.

Check: $$53,\,59,\,61,\,67$$ are all primes $$\le 70$$ and $$53+59+61+67 = 240$$. Thus the choice $$p_1 = 53$$ is attainable, and any smaller prime fails (because even using the three largest permissible primes does not reach $$240$$).

Hence the smallest of the four numbers is 53.

Let $$n$$ be a positive integer not exceeding $$100$$.
For every such $$n$$ consider the set of divisors that are demanded by the question:

All positive integers $$i$$ with $$i^{3}\le n \quad\Longleftrightarrow\quad 1\le i\le\sqrt[3]{n}\,.$$

Denote $$k=\left\lfloor\sqrt[3]{n}\right\rfloor$$ (the greatest integer whose cube is at most $$n$$). Then $$n$$ must be divisible by each of $$1,2,\dots ,k$$, i.e. by their least common multiple

$$L_k=\operatorname{lcm}(1,2,\dots ,k).$$

Because $$n\le100$$, $$k$$ can only be $$1,2,3,4$$ (since $$5^{3}=125>100$$). We examine each possible $$k$$.

Range of $$n$$: $$1^{3}\le n\le2^{3}-1\;\Longrightarrow\;1\le n\le7$$.
$$L_1=\operatorname{lcm}(1)=1$$, so every $$n$$ in this range works. Count = $$7$$.

Range of $$n$$: $$2^{3}=8\le n\le3^{3}-1=26$$.
$$L_2=\operatorname{lcm}(1,2)=2$$, so $$n$$ must be even.
Even numbers between $$8$$ and $$26$$: $$8,10,12,14,16,18,20,22,24,26$$  ⇒  Count = $$10$$.

Range of $$n$$: $$3^{3}=27\le n\le4^{3}-1=63$$.
$$L_3=\operatorname{lcm}(1,2,3)=6$$, so $$n$$ must be a multiple of $$6$$.
Multiples of $$6$$ in $$[27,63]$$: $$30,36,42,48,54,60$$  ⇒  Count = $$6$$.

Range of $$n$$: $$4^{3}=64\le n\le100$$ (upper bound of the problem).
$$L_4=\operatorname{lcm}(1,2,3,4)=12$$, so $$n$$ must be a multiple of $$12$$.
Multiples of $$12$$ in $$[64,100]$$: $$72,84,96$$  ⇒  Count = $$3$$.

Adding the counts from all cases:
$$7+10+6+3 = 26.$$

Hence, the number of positive integers $$n\le100$$ that satisfy the given condition is 26.

Label the rectangle

$$ \begin{array}{ccc} A & B & C\\ D & E & F \end{array} $$

The seven adjacencies are
$$A\!-\!B,\;B\!-\!C,\;D\!-\!E,\;E\!-\!F,\;A\!-\!D,\;B\!-\!E,\;C\!-\!F.$$

Coprimality fails only for the pairs
• any two even numbers ( $$2,4,6$$ ) (gcd$$\ge2$$ )
• the pair $$3$$ and $$6$$ (gcd$$=3$$ ).

Therefore

1. No two even numbers may be adjacent.
2. The numbers $$3$$ and $$6$$ may not be adjacent.

Step 1 : choose the three cells for the even numbers

The grid is bipartite: colour the six cells black (A,C,E) and white (B,D,F). Edges exist only between opposite colours, so a set of cells with no edges inside it must consist wholly of one colour. Hence the only independent triples are

$$\{A,C,E\}\quad\text{and}\quad\{B,D,F\}.$$

Thus there are exactly $$2$$ ways to place the three even numbers $$2,4,6$$ so that they are pairwise non-adjacent.

Step 2 : locate the number 6

Case I Evens on the black cells $$\{A,C,E\}$$
• If $$6$$ is put at $$A$$, its neighbours are $$B,D$$, so $$3$$ must avoid $$B,D$$ and must occupy $$F$$.
• If $$6$$ is put at $$C$$, its neighbours are $$B,F$$, so $$3$$ must occupy $$D$$.
• If $$6$$ is put at $$E$$, its neighbours are $$B,D,F$$; then $$3$$ would be adjacent to $$6$$ wherever it is placed—impossible.
Hence $$6$$ can be placed in $$2$$ of these $$3$$ cells.

Case II Evens on the white cells $$\{B,D,F\}$$ (the argument is symmetric)
• $$6$$ cannot be at $$B$$ (adjacent to $$A,C,E$$).
• $$6$$ may be at $$D$$ (forces $$3$$ to $$C$$) or at $$F$$ (forces $$3$$ to $$A$$).
Again $$6$$ has $$2$$ admissible positions.

Step 3 : count the fillings for each admissible position of 6

After fixing the cell containing $$6$$ and the forced position of $$3$$ :

• The remaining two even numbers $$2,4$$ may be arranged in the two remaining even cells in $$2!$$ ways.
• The remaining two odd numbers $$1,5$$ may be arranged in the two remaining odd cells in $$2!$$ ways.

Thus each admissible placement of $$6$$ yields $$2!\times 2! = 4$$ fillings.

Step 4 : total count

Each colour-pattern (black or white for the evens) allows $$2$$ positions for $$6$$, and each such position gives $$4$$ fillings:

$$2\;(\text{colour patterns}) \times 2\;(\text{positions of }6) \times 4 = 16.$$

Hence the required number of ways is $$16$$.

Answer: 16

The circular disc has area $$1000$$, so

$$\pi r^{2}=1000 \; \Longrightarrow \; r=\sqrt{\frac{1000}{\pi}}.$$

The diameter of the disc is therefore

$$d=2r = 2\sqrt{\frac{1000}{\pi}}.$$

For the largest possible square to be fully contained in the disc, its four vertices must lie on the circle. In that position the diagonal of the square equals the diameter of the circle.

If $$n$$ is the side length of the square, its diagonal is $$n\sqrt{2}$$. Setting this equal to the diameter gives

$$n\sqrt{2}=2\sqrt{\frac{1000}{\pi}} \quad\Longrightarrow\quad n=\frac{2}{\sqrt{2}}\sqrt{\frac{1000}{\pi}} =\sqrt{\frac{2000}{\pi}}.$$

Numerically,

$$\sqrt{\frac{2000}{\pi}}\approx\sqrt{636.62}\approx 25.24.$$

Because $$n$$ must be an integer no larger than this value, the greatest possible integer $$n$$ is $$25$$.

Answer: 25

The number with the repeated pattern $$\overline{ababab}$$ can be written algebraically.

Hundred-thousands, thousands and tens places contain the digit $$a$$, while ten-thousands, hundreds and units places contain $$b$$:
$$\overline{ababab}=a\,(100000+1000+10)+b\,(10000+100+1)=10101(10a+b).$$

Hence the given equation becomes
$$5(2m+1)(2m+3)(2m+5)=10101(10a+b).$$

1. The right side must be divisible by $$5$$, so the last digit of $$\overline{ababab}$$ equals $$b\in\{0,5\}$$.

2. The left side is $$5\times(\text{odd})$$, therefore its last digit is $$5$$ (never $$0$$).
 ⇒ $$b=5$$.

Thus $$10a+b=10a+5=5(2a+1)$$ and
$$\overline{ababab}=10101\bigl(5(2a+1)\bigr)=50505(2a+1).$$

Divide the original equation by $$5$$:

$$\bigl(2m+1\bigr)\bigl(2m+3\bigr)\bigl(2m+5\bigr)=10101(2a+1). \quad -(1)$$

Let $$n=2m+3$$ (the middle odd number). Then $$n$$ is odd and

Left side $$=(n-2)\,n\,(n+2)=n(n^2-4)=n^3-4n. \quad -(2)$$

Rewrite (1) with this notation:

$$n^3-4n=10101(2a+1). \quad -(3)$$

Because $$a$$ is a non-zero digit different from $$5$$, the possible values of $$2a+1$$ are

$$3,5,7,9,13,15,17,19. \quad -(4)$$

Compute $$10101(2a+1)$$ for each case (use mental multiplication or short work):

$$\begin{aligned} 2a+1=3 &\Rightarrow 30303\\ 2a+1=5 &\Rightarrow 50505\\ 2a+1=7 &\Rightarrow 70707\\ 2a+1=9 &\Rightarrow 90909\\ 2a+1=13&\Rightarrow 131313\\ 2a+1=15&\Rightarrow 151515\\ 2a+1=17&\Rightarrow 171717\\ 2a+1=19&\Rightarrow 191919 \end{aligned}$$

Now solve $$n^3-4n=R$$ for each right-hand value $$R$$ above. Because $$n^3$$ dominates, take the integer cube root of $$R$$ as a first guess:

$$\sqrt[3]{50505}\approx37.$$ Test $$n=37$$ in (2): $$37^3-4(37)=50653-148=50505,$$ which matches $$R$$ exactly.

Therefore $$n=37=2m+3\;\Longrightarrow\;2m=34\;\Longrightarrow\;m=17.$

Since $$2a+1=5\;\Longrightarrow\;a=2$$ and we already had $$b=5$$, the triple $$(m,a,b)=(17,2,5)$$ satisfies every condition.

Checking the other values of $$R$$ shows no integer $$n$$ solves $$n^3-4n=R$$ (the two nearest cubes always differ from $$R$$). Thus the solution found is unique.

Finally, the required sum is
$$m+a+b = 17+2+5 = 24.$$

Answer: 24

Let $$m,n \in \mathbb{Z}^{+},\; m,n \le 20000$$ and suppose

$$m^{2}+n^{4}=2^{k}\qquad (k\in \mathbb{Z}_{\ge 0}).$$

Denote the highest power of $$2$$ dividing an integer $$x$$ by $$v_{2}(x)$$. Write

$$m = 2^{u}\,m_{1},\qquad n = 2^{v}\,n_{1},\quad\text{where } m_{1},n_{1}\text{ are odd},\; u,v\ge 0.$$

Then

$$m^{2}=2^{\,2u}\,m_{1}^{2},\qquad n^{4}=2^{\,4v}\,n_{1}^{4}.$$

Factor the common power of $$2$$ out of the sum:

$$m^{2}+n^{4}=2^{t}\Bigl(m_{1}^{2}\,2^{\,2u-t}+n_{1}^{4}\,2^{\,4v-t}\Bigr),$$

where $$t=\min \{2u,\,4v\}=v_{2}(m^{2}+n^{4})$$. For the bracketed term to be an integer, one of the exponents $$2u-t,\;4v-t$$ is zero.

$$m^{2}+n^{4}=2^{\,4v}\bigl(m_{1}^{2}+n_{1}^{4}\bigr)=2^{k}.$$

Hence $$m_{1}^{2}+n_{1}^{4}=2^{\,k-4v}.$$

Since both $$m_{1},n_{1}$$ are odd, $$m_{1}^{2}\equiv n_{1}^{4}\equiv 1 \pmod{8}$$, so their sum is $$\equiv 2 \pmod{8}$$. A power of $$2$$ that is $$2 \pmod{8}$$ is exactly $$2^{1}=2$$. Therefore

$$m_{1}^{2}+n_{1}^{4}=2,\qquad\Rightarrow\qquad m_{1}=1,\; n_{1}=1.$$

Combining with $$u=2v$$ gives

$$n = 2^{v},\qquad m = 2^{\,2v},\qquad v\in \mathbb{Z}_{\ge 0}.$$

Now apply the upper bound $$m,n \le 20000$$:

$$2^{v}\le 20000,\quad 2^{\,2v}\le 20000.$$

The second inequality is stricter. Compute successive powers:

$$\begin{aligned} v=0:&\;2^{0}=1,\;2^{0}=1 \\ v=1:&\;2^{1}=2,\;2^{2}=4 \\ v=2:&\;2^{2}=4,\;2^{4}=16 \\ v=3:&\;2^{3}=8,\;2^{6}=64 \\ v=4:&\;2^{4}=16,\;2^{8}=256 \\ v=5:&\;2^{5}=32,\;2^{10}=1024 \\ v=6:&\;2^{6}=64,\;2^{12}=4096 \\ v=7:&\;2^{7}=128,\;2^{14}=16384 \\ v=8:&\;2^{8}=256,\;2^{16}=65536\;(\gt 20000) \\ \end{aligned}$$

Thus admissible values are $$v=0,1,2,3,4,5,6,7$$  — eight choices in all.

Each $$v$$ gives the pair $$\bigl(m,n\bigr)=\bigl(2^{\,2v},\,2^{v}\bigr).$$

Therefore the required number of ordered pairs is 08.

Let us place square $$ABCD$$ on a coordinate plane so that
$$A(0,0),\; B(1,0),\; C(1,1),\; D(0,1).$$
Thus $$AB=1$$ lies on the $$x$$-axis.

Because $$AB$$ is a chord of the required circle $$S$$, its centre $$O(x_0,y_0)$$ must lie on the perpendicular bisector of $$AB$$, i.e. on the vertical line $$x=\tfrac12$$.
Write $$O\equiv\left(\tfrac12,\;h\right)$$ where $$h$$ is to be found.

1. Radius in terms of $$h$$:
$$r^2 = OA^2 = \left(\tfrac12\right)^2 + h^2 = 0.25 + h^2.$$ $$-(1)$$

2. Power of point $$C(1,1)$$ with respect to $$S$$.
The tangent length from $$C$$ is given as $$CK=2$$, therefore
$$\text{Power}(C) = CK^2 = 4.$$

Using the power-of-a-point theorem
$$\text{Power}(C)=OC^2-r^2.$$
Compute $$OC^2$$:
$$OC^2=\left(1-\tfrac12\right)^2+(1-h)^2 = 0.25 + (1-h)^2.$$

Hence
$$\left[0.25 + (1-h)^2\right] - \left[0.25 + h^2\right] = 4.$$
The $$0.25$$ terms cancel, giving
$$(1-h)^2 - h^2 = 4.$$ Expand and simplify:
$$1 - 2h + h^2 - h^2 = 4 \;\Longrightarrow\; 1 - 2h = 4 \;\Longrightarrow\; h = -\dfrac32.$$ So the centre lies $$1.5$$ units below the side $$AB$$.

3. Radius and diameter.
From (1):
$$r^2 = 0.25 + \left(-\dfrac32\right)^2 = 0.25 + 2.25 = 2.5 = \dfrac52.$$ Therefore
$$r = \sqrt{\dfrac52},\qquad d = 2r = 2\sqrt{\dfrac52}= \sqrt{10}.$$

4. Verification of the given geometric conditions.
The highest point of the circle is at $$y = h + r = -1.5 + 1.581\ldots \approx 0.08$$, which is below the line $$CD$$ ( $$y=1$$ ). Hence $$CD$$ does not meet the circle, exactly as stated in the problem.

Finally
$$d^{2} = (\sqrt{10})^{2} = 10.$$

Answer: 10

Let $$a,b,c,d$$ be positive integers satisfying
$$17\bigl(abcd+ab+ad+cd+1\bigr)=20\bigl(bcd+b+d\bigr)\quad -(1)$$

Step 1 Introduce the common factor $$S=bcd+b+d$$.
Because $$\gcd(17,20)=1,$$ equation (1) forces
$$17\mid S\;\Longrightarrow\;S=17k,\;k\in\mathbb{N}$$
and consequently
$$20\mid\bigl(abcd+ab+ad+cd+1\bigr).$$

Step 2 Rewrite the left‐hand expression in terms of $$S$$.
Note that
$$abcd+ab+ad+cd+1=a(bcd+b+d)+cd+1=aS+cd+1.$$ Substituting this and $$S=17k$$ in (1):
$$17\bigl(aS+cd+1\bigr)=20S \;\Longrightarrow\; 17aS+17(cd+1)=20S.$$ Divide by $$17$$ and replace $$S$$ by $$17k$$:
$$a(17k)=20k-cd-1 \;\Longrightarrow\; (20-17a)k=cd+1\quad -(2)$$

Step 3 Fix the only possible value of $$a$$.
All quantities are positive, so the left side of (2) must be positive: $$20-17a\gt 0\;\Longrightarrow\;a\le 1.$$ Since $$a$$ is a positive integer, we must have
$$\boxed{a=1}$$

With $$a=1$$, equation (2) becomes
$$3k=cd+1\quad -(3)$$

Step 4 Express $$k$$ and $$S$$ in terms of $$c,d$$.
From (3): $$k=\dfrac{cd+1}{3}$$ (so $$cd\equiv2\pmod 3$$).
Using $$S=17k$$:
$$S=17\bigl(\tfrac{cd+1}{3}\bigr)=\tfrac{17}{3}(cd+1).$$

Step 5 Set up the final condition for $$b,c,d$$.
But $$S=bcd+b+d$$, hence
$$bcd+b+d=\frac{17}{3}(cd+1).$$ Multiply by 3:
$$3bcd+3b+3d=17cd+17.$$ Rearrange:
$$cd(3b-17)+3(b+d)-17=0\quad -(4)$$

Step 6 Solve (4) for the small integer $$b$$.
• If $$3b-17\ge 0$$ (that is $$b\ge 6$$), then the first term in (4) is non-negative and the second term is positive, so the sum cannot be zero. Hence $$b\lt 6$$.
• List the possibilities:

Then $$3b-17=-2$$ and (4) gives
$$-2cd+3d-2=0\;\Longrightarrow\;2cd-3d+2=0$$ $$d(2c-3)=-2.$$ Because $$d,c\gt0$$, the only way is $$2c-3=-1\;(c=1)$$ and $$d=2.$$
Thus $$b=5,c=1,d=2$$ is a solution.

For these values, $$3b-17\le -5$$, so the term $$cd(3b-17)$$ is too negative to be balanced by the positive remainder $$3(b+d)-17$$, and no solution occurs.

Hence the unique positive integer solution is
$$(a,b,c,d)=(1,5,1,2).$$

Step 7 Compute the required sum of squares:
$$a^{2}+b^{2}+c^{2}+d^{2}=1^{2}+5^{2}+1^{2}+2^{2}=1+25+1+4=31.$$

Therefore, the answer is
31.

We first evaluate the left‐hand expression bottom-up.

Start with the innermost term:
$$6+\frac1{7}=\frac{43}{7}$$

Move one level outward each time:

$$5+\frac1{\displaystyle 6+\frac1{7}} =5+\frac{1}{\tfrac{43}{7}} =5+\frac{7}{43} =\frac{222}{43}$$

$$4+\frac1{\displaystyle 5+\frac1{6+\frac1{7}}} =4+\frac{1}{\tfrac{222}{43}} =4+\frac{43}{222} =\frac{931}{222}$$

$$3+\frac1{\displaystyle 4+\frac1{5+\cdots}} =3+\frac{1}{\tfrac{931}{222}} =3+\frac{222}{931} =\frac{3015}{931}$$

$$2+\frac1{\displaystyle 3+\frac1{4+\cdots}} =2+\frac{1}{\tfrac{3015}{931}} =2+\frac{931}{3015} =\frac{6961}{3015}$$

Finally,

$$1-\frac1{\displaystyle 2+\frac1{3+\cdots}} =1-\frac{1}{\tfrac{6961}{3015}} =1-\frac{3015}{6961} =\frac{3946}{6961}$$

Hence

$$1-\frac{1}{2+\frac{1}{3+\cdots}}=\frac{3946}{6961} =\frac1{x_1+\frac1{x_2+\frac1{x_3+\frac1{x_4+\frac1{x_5+\frac1{x_6+\frac1{x_7}}}}}}}$$

Taking reciprocals gives the ordinary continued fraction we need:

$$x_1+\frac1{x_2+\frac1{x_3+\frac1{x_4+\frac1{x_5+\frac1{x_6+\frac1{x_7}}}}}} =\frac{6961}{3946}$$

Now expand $$\frac{6961}{3946}$$ as a simple continued fraction using the Euclidean algorithm:

$$\frac{6961}{3946}=1+\frac{3015}{3946}=1+\frac1{\tfrac{3946}{3015}} \quad\Rightarrow\; x_1=1$$
$$\frac{3946}{3015}=1+\frac{931}{3015}=1+\frac1{\tfrac{3015}{931}} \quad\Rightarrow\; x_2=1$$
$$\frac{3015}{931}=3+\frac{222}{931}=3+\frac1{\tfrac{931}{222}} \quad\Rightarrow\; x_3=3$$
$$\frac{931}{222}=4+\frac{43}{222}=4+\frac1{\tfrac{222}{43}} \quad\Rightarrow\; x_4=4$$
$$\frac{222}{43}=5+\frac{7}{43}=5+\frac1{\tfrac{43}{7}} \quad\Rightarrow\; x_5=5$$
$$\frac{43}{7}=6+\frac{1}{7}=6+\frac1{\tfrac{7}{1}} \quad\Rightarrow\; x_6=6$$
$$\frac{7}{1}=7 \quad\Rightarrow\; x_7=7$$

Thus
$$x_1=1,\;x_2=1,\;x_3=3,\;x_4=4,\;x_5=5,\;x_6=6,\;x_7=7.$$

The required sum is
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7 =1+1+3+4+5+6+7=27.$$

Answer: 27

The product of the drawn numbers will be a multiple of $$6$$ exactly when the set of cards drawn supplies both a factor $$2$$ (an even number) and a factor $$3$$ (a multiple of $$3$$).
Hence Mainak fails only if, after drawing his cards, at least one of these two prime factors is still missing.

So we ask: “How many cards can be drawn while avoiding at least one of the factors $$2$$ or $$3$$?”  The answer to that question, plus one more card, will give the desired guarantee.

Case 1: Avoid every even number (draw only odd cards).
There are $$50$$ odd numbers between $$1$$ and $$100$$. After drawing all $$50$$ odds the product still lacks a factor $$2$$, so it is not divisible by $$6$$.

Case 2: Avoid every multiple of $$3$$ (draw numbers not divisible by $$3$$).
Count the numbers between $$1$$ and $$100$$ that are not multiples of $$3$$:
There are $$\left\lfloor\frac{100}{3}\right\rfloor = 33$$ multiples of $$3$$, so $$100-33 = 67$$ numbers are not divisible by $$3$$.
If Mainak draws these $$67$$ cards their product has plenty of factors $$2$$ (because $$34$$ of them are even) but has no factor $$3$$, therefore the product is still not a multiple of $$6$$.

Case 2 shows that, even after drawing $$67$$ cards, Mainak could still miss the factor $$3$$, keeping the product free of $$6$$. Hence drawing $$67$$ cards is not sufficient.

However, once he draws any additional card—i.e. the $$68^{\text{th}}$$ card—he is forced to pick from the $$33$$ multiples of $$3$$ that remain in the box (by the pigeon-hole principle).
Because the first $$67$$ cards already contain even numbers, this new multiple of $$3$$ supplies the missing factor $$3$$. The combined set now has both $$2$$ and $$3$$, so its product is necessarily a multiple of $$6$$.

Therefore Mainak must draw at least $$68$$ cards to guarantee that the product of the numbers on the drawn cards is a multiple of $$6$$.

Final answer: 68.

Let $$W(x,y)$$ denote the number of admissible paths from the origin $$(0,0)$$ to the lattice point $$(x,y)$$ when we are

  • allowed moves : one step East (E) $$:(x,y)\rightarrow (x+1,y)$$ or one step North (N) $$:(x,y)\rightarrow (x,y+1)$$
  • forbidden points : those with both coordinates odd, i.e. $$(x,y)$$ with $$x$$ odd and $$y$$ odd.

The usual additive recurrence for rectangular‐grid paths remains valid provided the destination point itself is not forbidden:

If $$(x,y)$$ is not forbidden, then
$$W(x,y)=W(x-1,y)+W(x,y-1)$$   -(1)

If $$(x,y)$$ is forbidden (both $$x,y$$ odd) we put $$W(x,y)=0$$.

Initialisation : $$W(0,0)=1$$. On the coordinate axes at least one coordinate is even, hence no point on either axis is forbidden. Therefore

$$W(x,0)=1\;(0\le x\le 7),\qquad W(0,y)=1\;(0\le y\le 12).$$

Using (1) row by row (or column by column) we fill the $$8\times 13$$ array $$\{0\le x\le 7,\;0\le y\le 12\}.$$ The forbidden points are left blank (value 0).

y \x  0 1 2 3 4 5 6 7
0  1 1 1 1 1 1 1 1
1  1 0 1 0 1 0 1 0
2  1 1 2 2 3 3 4 4
3  1 0 2 0 3 0 4 0
4  1 1 3 3 6 6 10 10
5  1 0 3 0 6 0 10 0
6  1 1 4 4 10 10 20 20
7  1 0 4 0 10 0 20 0
8  1 1 5 5 15 15 35 35
9  1 0 5 0 15 0 35 0
10 1 1 6 6 21 21 56 56
11 1 0 6 0 21 0 56 0
12 1 1 7 7 28 28 84 84

The last entry in the table is $$W(7,12)=84$$, because $$(7,12)$$ itself is admissible (7 is odd, 12 is even).

Hence, the required number of paths is

$$n = 84$$

Write each positive integer in the form $$k\;x^2$$ where
  • $$k$$ is square-free (no prime square divides it),
  • $$x$$ is a positive integer.

If $$m=k\,u^2$$ and $$n=k\,v^2$$ then
$$mn=(k\,u^2)(k\,v^2)=k^2(u\,v)^2=(k\,u\,v)^2,$$
which is a perfect square. Conversely, if $$mn$$ is a perfect square, the parity of every prime’s exponent in $$m$$ and $$n$$ must be the same, so their square-free parts are identical. Hence

mn is a perfect square $$\Longleftrightarrow$$ $$m=k\,u^2,\;n=k\,v^2$$ with the same square-free $$k$$.

Thus for every square-free $$k\le 50$$ count ordered pairs $$(u,v)$$ with $$u\lt v,\qquad k\,u^2\le 50,\;k\,v^2\le 50.$$ Put

$$t(k)=\Bigl\lfloor\sqrt{\frac{50}{k}}\Bigr\rfloor$$ (the number of positive integers $$x$$ satisfying $$k\,x^2\le 50$$). For a fixed $$k$$ the choices of $$(u,v)$$ are all $$\binom{t(k)}{2}$$ pairs with $$u\lt v$$.

List every square-free $$k\le 50$$, evaluate $$t(k)$$, keep only those with $$t(k)\ge 2$$:

$$\begin{array}{c|c|c} k & t(k)=\left\lfloor\sqrt{50/k}\right\rfloor & \binom{t(k)}{2} \\ \hline 1 & 7 & 21\\ 2 & 5 & 10\\ 3 & 4 & 6\\ 5 & 3 & 3\\ 6 & 2 & 1\\ 7 & 2 & 1\\ 10 & 2 & 1\\ 11 & 2 & 1 \end{array}$$

All other square-free $$k$$ (13,14,15,17,19,\,$$\dot$$s,47) give $$t(k)=1$$, hence contribute 0 pairs.

Add the contributions:

$$21+10+6+3+1+1+1+1=44.$$ Therefore the number of ordered pairs $$(m,n)$$ with $$1\le m\lt n\le 50$$ for which $$mn$$ is a perfect square equals

44.

Let the required four-digit number be $$\overline{abcd}$$, where the digits $$a,b,c,d$$ take values from $$\{1,2,\dots ,9\}$$ (zero is not allowed).

The given conditions are
$$a+c = bd \qquad -(1)$$
$$b+d = ac \qquad -(2)$$

Because every digit is at most $$9$$, the largest possible value of the left-hand sides (the sums) is $$9+9 = 18$$. Hence from $$-(1)$$ and $$-(2)$$ we immediately get

$$bd \le 18, \qquad ac \le 18 \qquad -(3)$$

Thus only those ordered pairs of digits whose product does not exceed $$18$$ can appear in the pairs $$(b,d)$$ and $$(a,c)$$. This already eliminates the large majority of possibilities.

Fix an ordered pair $$(a,c)$$ that satisfies $$ac \le 18$$. Define
$$S_1 = a+c, \qquad S_2 = ac$$
Equations $$-(1)$$ and $$-(2)$$ can now be read as

$$bd = S_1, \qquad b+d = S_2$$

If we regard $$b$$ and $$d$$ as the two roots of a quadratic, they must satisfy

$$t^{2}-S_2\,t+S_1 = 0 \qquad -(4)$$

For $$b,d$$ to be (positive) integral digits, the quadratic in $$-(4)$$ must have

1. A non-negative discriminant: $$\Delta = S_2^{2}-4S_1 \ge 0$$.
2. A perfect-square discriminant (so that the roots are integral).
3. Both roots lying between $$1$$ and $$9$$ (inclusive).

We now list all ordered pairs $$(a,c)$$ with $$ac \le 18$$ and test them with the above three criteria. Because the list is short, the check can be done by hand in a few minutes; the successful pairs are summarised below.

No other ordered pair $$(a,c)$$ with $$ac \le 18$$ satisfies the discriminant and root conditions; hence the list above is complete.

Collecting all the valid four-digit numbers:

$$\{\,1253,1352,2135,2222,2531,3125,3521,5213,5312\,\}$$

The count of such numbers is $$9$$.

Answer: 09

We are given the functional equation

$$4f(3-x)+3f(x)=x^{2}\qquad\forall x\in\mathbb{R}$$ $$-(1)$$

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

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

Equations $$-(1)$$ and $$-(2)$$ form a linear system in the two unknowns $$f(x)$$ and $$f(3-x)$$:

$$\begin{cases} 3f(x)+4f(3-x)=x^{2} &\qquad -(1)\\[4pt] 4f(x)+3f(3-x)=(3-x)^{2} &\qquad -(2) \end{cases}$$

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

$$\begin{aligned} &(9f(x)+12f(3-x))-(16f(x)+12f(3-x))\\ &=3x^{2}-4(3-x)^{2} \end{aligned}$$

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

Compute the right‐hand side:

$$(3-x)^{2}=x^{2}-6x+9$$
$$4(3-x)^{2}=4x^{2}-24x+36$$
$$3x^{2}-4(3-x)^{2}=3x^{2}-(4x^{2}-24x+36)=x^{2}-24x+36$$

Hence

$$7f(x)=x^{2}-24x+36\quad\Longrightarrow\quad f(x)=\frac{x^{2}-24x+36}{7}$$ $$-(3)$$

Now evaluate at the required points.

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

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

Therefore

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

The value rounded to the nearest integer is $$8$$.

Answer: 08

Label the three girls $$G_1,G_2,G_3$$ and the four stories $$S_1,S_2,S_3,S_4$$.
For every story record, in the same order $$G_1,G_2,G_3$$, who likes it (1 = likes, 0 = does not).
Thus every story is represented by a 3-digit 0-1 string.

The string $$111$$ (liked by all three) is forbidden by the first condition, whereas any of the other seven strings are allowed:

Pairs only (each must appear at least once, second condition):
$$A = 110,\; B = 101,\; C = 011$$

Singletons: $$100,\;010,\;001$$

Liked by none: $$000$$

Because the stories themselves are different, choosing strings for $$S_1,S_2,S_3,S_4$$ is an ordered assignment.
We have to count those assignments in which every one of the three pair-patterns $$A,B,C$$ appears at least once.

Let $$U$$ be the set of all possible assignments.
Since each story has 7 available patterns, $$|U| = 7^4 = 2401$$.

Define the following “bad” events:

$$E_A:$$ no story is of type $$A = 110$$
$$E_B:$$ no story is of type $$B = 101$$
$$E_C:$$ no story is of type $$C = 011$$

The required count $$n$$ equals the number of assignments in $$U$$ that avoid all three bad events. Use the Principle of Inclusion-Exclusion (PIE).

• If $$A$$ is forbidden, each story may take any of the remaining 6 patterns, hence $$|E_A| = 6^4$$, and likewise $$|E_B| = 6^4,\; |E_C| = 6^4$$.

• If both $$A$$ and $$B$$ are forbidden, only 5 patterns are left, so $$|E_A \cap E_B| = 5^4$$. All three pairwise intersections have the same size.

• If $$A,B,C$$ are all forbidden, only the 4 patterns $$000,100,010,001$$ remain, giving $$|E_A \cap E_B \cap E_C| = 4^4$$.

Applying PIE,

$$ \begin{aligned} n &= 7^4 - 3\cdot 6^4 + 3\cdot 5^4 - 4^4 \\[4pt] &= 2401 - 3(1296) + 3(625) - 256 \\[4pt] &= 2401 - 3888 + 1875 - 256 \\[4pt] &= 132. \end{aligned} $$

The digits of $$n$$ are 1, 3 and 2. Their squared values are $$1^2 = 1,\; 3^2 = 9,\; 2^2 = 4$$. Hence the required sum is $$1 + 9 + 4 = 14$$.

Answer: 14

Assign masses $$m_A , m_B , m_C$$ to the vertices $$A , B , C$$ respectively and use the Mass-Points rule:

If a point $$P$$ lies on the segment joining two points with masses $$m_1 , m_2$$, then
$$\dfrac{\text{distance from }P\text{ to point with mass }m_1}{\text{distance from }P\text{ to point with mass }m_2}= \dfrac{m_2}{m_1}$$.

1. Along $$B \!-\! P \!-\! E$$ we have $$\dfrac{BP}{PE}= \dfrac{5}{2}$$.
 Mass at $$B$$ is $$m_B$$, mass at $$E$$ is the sum of the endpoint masses on $$CA$$, i.e. $$m_E = m_A + m_C$$.
 Therefore $$\dfrac{m_E}{m_B}= \dfrac{5}{2}\;\; \Rightarrow\;\; m_A + m_C = \tfrac{5}{2}\,m_B$$.  $$(1)$$

2. Along $$C \!-\! P \!-\! F$$ we have $$\dfrac{CP}{PF}= \dfrac{7}{3}$$.
 Mass at $$C$$ is $$m_C$$, mass at $$F$$ is the sum on $$AB$$, i.e. $$m_F = m_A + m_B$$.
 Therefore $$\dfrac{m_F}{m_C}= \dfrac{7}{3}\;\; \Rightarrow\;\; m_A + m_B = \tfrac{7}{3}\,m_C$$.  $$(2)$$

Choose $$m_B = 2t$$ so that equation $$(1)$$ becomes $$m_A + m_C = 5t$$.

Put $$m_C = c$$. Then $$m_A = 5t - c$$ and substitute these in $$(2)$$:
$$(5t - c) + 2t = \tfrac{7}{3}\,c \;\; \Rightarrow\;\; 7t - c = \tfrac{7}{3}\,c$$
$$\Rightarrow\; 7t = \tfrac{10}{3}\,c \;\; \Rightarrow\;\; c = \tfrac{21}{10}\,t.$$

Hence $$m_A = 5t - \tfrac{21}{10}t = \tfrac{29}{10}t,\qquad m_B = 2t = \tfrac{20}{10}t,\qquad m_C = \tfrac{21}{10}t.$$

Multiply by $$10$$ to clear the denominator and obtain integral masses:
$$m_A : m_B : m_C = 29t : 20t : 21t.$$ The common factor $$t$$ is irrelevant, so take
$$m_A = 29,\; m_B = 20,\; m_C = 21.$$

3. Point $$D$$ lies on $$BC$$, hence its mass is the sum of the masses at $$B$$ and $$C$$:
$$m_D = m_B + m_C = 20 + 21 = 41.$$

Along $$A \!-\! P \!-\! D$$ we therefore have
$$\dfrac{AP}{PD}= \dfrac{m_D}{m_A}= \dfrac{41}{29}.$$

Thus $$\dfrac{AP}{PD}= \dfrac{p}{q}= \dfrac{41}{29} \; \Rightarrow\; (p,q)=(41,29).$$

Since $$\gcd(41,29)=1$$, we get $$p+q = 41 + 29 = 70.$$

Answer: 70

We want all natural numbers $$n \le 105$$ that satisfy the congruence

$$2^{\,n} \equiv n^{2} \pmod{7}\,.$$

Step 1 : Period of $$2^{\,n} \pmod{7}$$
Since $$2^{3}=8 \equiv 1 \pmod{7}$$, the powers of $$2$$ repeat every $$3$$. Therefore

$$2^{\,n} \equiv \begin{cases} 1 & \text{if } n \equiv 0 \pmod{3},\\ 2 & \text{if } n \equiv 1 \pmod{3},\\ 4 & \text{if } n \equiv 2 \pmod{3}. \end{cases}$$

Step 2 : Possible values of $$n^{2} \pmod{7}$$
Because there are only seven residues, list the squares:

$$\begin{array}{c|ccccccc} n \pmod{7} & 0 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline n^{2} \pmod{7} & 0 & 1 & 4 & 2 & 2 & 4 & 1 \end{array}$$

Step 3 : Combine the two moduli
The pattern of $$2^{\,n}$$ repeats every $$3$$, and that of $$n^{2}$$ repeats every $$7$$. Hence the combined congruence repeats every $$\operatorname{lcm}(3,7)=21$$. It suffices to test $$n=1,2,\dots ,21$$ and then extend the count up to $$105=5\times 21$$.

Step 4 : Check each residue $$n \pmod{21}$$
Compare $$2^{\,n}\pmod{7}$$ with $$n^{2}\pmod{7}$$:

$$\begin{array}{c|cccccccccccccccccccccc} n & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20&21\\ \hline 2^{\,n}\!\!\pmod{7} & 2&4&1&2&4&1&2&4&1&2&4&1&2&4&1&2&4&1&2&4&1\\ n^{2}\!\!\pmod{7} & 1&4&2&2&4&1&0&1&4&2&2&4&1&0&1&4&2&2&4&1&0 \end{array}$$

The congruence holds precisely for

$$n=2,\,4,\,5,\,6,\,10,\,15.$$

Thus there are $$6$$ solutions in one block of length $$21$$.

Step 5 : Extend to $$n \le 105$$
The interval $$1 \le n \le 105$$ contains exactly $$5$$ full blocks of length $$21$$:

$$105 = 5 \times 21.$$

Therefore the total number of acceptable $$n$$ is

$$6 \times 5 = 30.$$

Answer: 30

Let $$g=\gcd(a,b)$$. Write $$a=gx,\;b=gy$$ where $$x,y\in\mathbb{N}$$ and $$\gcd(x,y)=1$$.

Then
$$ab=g^2xy,$$
$$\operatorname{lcm}(a,b)=gxy.$$

The given relation becomes
$$g^2xy \;=\;406+11(gxy)+7g.$$

Bring every term to the left and factor out $$g$$:
$$g^2xy-11gxy-7g-406=0$$
$$\Longrightarrow\;g\bigl[xy(g-11)-7\bigr]=406.$$

Set
$$k=xy(g-11)-7.$$
Then $$gk=406.$$ Because $$a,b$$ are positive, $$xy\gt0$$, so $$k\gt0$$ and consequently $$g-11\gt0\;(\text{i.e. }g\gt11).$$

All positive divisors of $$406=2\cdot7\cdot29$$ are $$1,2,7,14,29,58,203,406.$$ The divisors exceeding $$11$$ are $$14,29,58,203,406.$$

Then $$k=\dfrac{406}{14}=29,$$ and

$$xy(g-11)=xy\cdot3=k+7=36\;\Longrightarrow\;xy=12.$$

Since $$x,y$$ are coprime and $$xy=12$$, the admissible pairs are $$(1,12),\,(3,4),\,(4,3),\,(12,1).$$ For each pair $$a+b=g(x+y)=14(x+y).$$ The smallest sum arises from $$(x,y)=(3,4)\text{ or }(4,3):$$ $$a+b=14(3+4)=98.$$

Now $$k=\dfrac{406}{29}=14$$ and $$xy(g-11)=xy\cdot18=21\;\bigl(k+7\bigr),$$ giving $$xy=\dfrac{21}{18}$$ - not an integer. Hence no solution.

Then $$k=7$$ and $$xy(g-11)=xy\cdot47=14,$$ which again yields a non-integer $$xy$$. No solution.

Here $$k=2$$ and $$xy(g-11)=xy\cdot192=9,$$ impossible.

Here $$k=1$$ and $$xy(g-11)=xy\cdot395=8,$$ impossible.

The only feasible case is $$g=14$$ with $$(x,y)=(3,4)\text{ or }(4,3).$$ Taking $$a=42,\;b=56$$ (or vice-versa) indeed satisfies

$$ab=42\cdot56=2352,$$ $$406+11\cdot\operatorname{lcm}(42,56)+7\cdot\gcd(42,56)=406+11\cdot168+7\cdot14=2352.$$

Therefore the minimum possible value of $$a+b$$ is $$\mathbf{98}$$.

Label the vertices of the regular dodecagon $$0,1,2,\dots ,11$$ anticlockwise.
The three vertices of every equilateral triangle are obtained by jumping four steps, and the four vertices of every square are obtained by jumping three steps:

Equilateral triangles (step $$4$$):
$$\{0,4,8\},\; \{1,5,9\},\; \{2,6,10\},\; \{3,7,11\}$$

Squares (step $$3$$):
$$\{0,3,6,9\},\; \{1,4,7,10\},\; \{2,5,8,11\}$$

Introduce the $$4\times3$$ array whose rows correspond to the four triangles and whose columns correspond to the three squares:

$$ \begin{array}{ccc} 0 & 4 & 8\\ 9 & 1 & 5\\ 6 & 10 & 2\\ 3 & 7 & 11 \end{array} $$

Row $$r$$ consists of the vertices of triangle $$r$$ and column $$c$$ consists of the vertices of square $$c$$.
The conditions of the problem translate to

1. every row of the array is not monochromatic (avoids a monochromatic triangle);
2. every column of the array is not monochromatic (avoids a monochromatic square).

Count colourings row-wise first.

Step 1: impose the triangle condition (rows)
For one row of length $$3$$ there are $$2^3=8$$ possible colourings. Excluding the two monochromatic ones (BBB, RRR) leaves $$6$$ admissible patterns.
Because rows are independent at this stage, $$|U| = 6^4 = 1296$$ colourings satisfy all the triangle conditions. Call this set $$U$$.

Step 2: exclude colourings that break at least one square condition (columns)
Let $$C_0,C_1,C_2$$ be the events “column $$0,1,2$$ is monochromatic”, respectively. We shall use the Principle of Inclusion-Exclusion inside the universe $$U$$.

• Single events. Fix column $$0$$ to be BBBB (the case RRRR is symmetric). In each row at least one of the remaining two entries must be red, so the allowed pairs are (BR, RB, RR) - three choices. Hence $$|C_0| = 2\cdot 3^4 = 162,$$ and similarly $$|C_1|=|C_2|=162$$.

• Intersections of two events. Suppose columns $$0,1$$ are monochromatic.   (i) If they carry the same colour (BB or RR), the third entry in every row is forced to be the opposite colour - exactly one choice per row. This gives $$2$$ colourings.
  (ii) If the columns carry different colours (BR or RB), the third entry in a row may be B or R (two choices). This gives $$2\cdot2^4=32$$ colourings.
Thus $$|C_0\cap C_1| = 2 + 32 = 34,$$ and by symmetry $$|C_0\cap C_2|=|C_1\cap C_2|=34$$.

• Intersection of three events. All three columns are monochromatic. There are $$2^3=8$$ colour triples; the two triples (BBB) and (RRR) violate the row condition, the remaining six comply. Hence $$|C_0\cap C_1\cap C_2| = 6.$$

Apply inclusion-exclusion:

$$ \begin{aligned} |C_0\cup C_1\cup C_2| &= \bigl(162+162+162\bigr) - \bigl(34+34+34\bigr) + 6\\ &= 486 - 102 + 6\\ &= 390. \end{aligned} $$

Step 3: final count
Colourings obeying both the triangle and square restrictions: $$ N = |U| - |C_0\cup C_1\cup C_2| = 1296 - 390 = 906. $$

Write $$N = 100p + q$$ with $$0\lt p,q\lt 100$$. Here $$N = 100\cdot 9 + 6$$, so $$p = 9$$ and $$q = 6$$.

Therefore, $$p + q = 9 + 6 = 15$$.

Answer: 15