IOQM Previous Papers Sept 6 2026

The centre is in row 5, column 5, and moving four squares left and then four squares up preserves the colour, so the top-left corner is also white. Rows $$1, 3, 5, 7, 9$$ therefore contain 5 white squares each, while rows $$2, 4, 6, 8$$ contain 4 each. Hence the total is $$5 \times 5 + 4 \times 4 = 41$$.

Write $$M = 256q + r$$ with $$0 \le r < 256$$. Both $$M$$ and $$256q$$ are divisible by 4, so $$r$$ is divisible by 4, and $$r \ne 0$$ because $$M$$ is not divisible by 256. The possible remainders are therefore $$4, 8, 12, \ldots, 252$$, and each one occurs, giving $$252/4 = 63$$ values.

Every non-zero integer has square at least 1, but the 49 numbers cannot all be $$1$$ or $$-1$$, since a sum of an odd number of odd integers is odd and so cannot be 0. At least one number therefore has absolute value at least 2, giving $$\sum_{i=1}^{49} x_i^2 \ge 48 \times 1 + 2^2 = 52$$. This bound is attained by 25 copies of 1, 23 copies of $$-1$$ and one $$-2$$, since $$25 - 23 - 2 = 0$$ and $$25 + 23 + 4 = 52$$.

To maximise the sum, use the six largest digits $$4, 5, 6, 7, 8, 9$$, and place the three largest of them in the tens places, because swapping a larger digit from a units place into a tens place increases the sum by $$9(v - u) > 0$$. This gives $$N = 10(9 + 8 + 7) + (6 + 5 + 4) = 255$$, attained for instance by $$96 + 85 + 74$$. The digit sum of 255 is $$2 + 5 + 5 = 12$$.

The angles $$\angle BAC$$ and $$\angle BNC$$ stand on the same chord $$BC$$ with their vertices on the same side of it, so $$\angle BNC = \angle BAC = 80^\circ$$. Since $$N$$ lies on the perpendicular bisector of $$BC$$, we have $$NB = NC$$, so triangle $$BNC$$ is isosceles. Hence $$\angle NBC = \frac{180^\circ - 80^\circ}{2} = 50^\circ$$.

With $$a$$ the tens digit and $$b$$ the units digit, the condition $$10a + b = 26 + ab$$ rearranges to $$(a - 1)(10 - b) = 16$$. Since $$1 \le a - 1 \le 8$$ and $$1 \le 10 - b \le 10$$, the only factor pairs are $$(2, 8)$$, $$(4, 4)$$ and $$(8, 2)$$, giving $$(a, b) = (3, 2), (5, 6), (9, 8)$$. The three numbers are 32, 56 and 98, so the count is 3.

At most one digit 8 can appear because $$8 + 8 > 14$$, so at most one digit 1 can appear, since every 1 needs its own preceding 8. But all permitted digits except 1 are even while the digit sum 14 is even, so no digit 1 occurs at all. With no 8, the counts $$r$$ of digit 2 and $$s$$ of digit 4 satisfy $$r + 2s = 7$$, giving $$1 + 6 + \binom{5}{2} + 4 = 21$$ numbers. With one 8 the rest sum to 6, so the digits are $$8, 2, 2, 2$$ with 4 arrangements or $$8, 2, 4$$ with $$3! = 6$$ arrangements, and the total is $$21 + 4 + 6 = 31$$.

Put $$CD = DA = s$$, so $$\angle DAB = 180^\circ - 120^\circ = 60^\circ$$, and divide $$AB = 3s$$ into three equal parts at $$E$$ and $$F$$. Triangle $$ADE$$ is equilateral, and $$DCFE$$ is a parallelogram, so $$CF = s = FB$$ and triangle $$CFB$$ is isosceles with apex angle $$120^\circ$$. Hence $$\angle B = 30^\circ$$ and $$\angle C = 150^\circ$$, so the four angles are $$60^\circ, 30^\circ, 150^\circ, 120^\circ$$ and $$x/y = 150/30 = 5$$.

A number with 225 digits has digit sum at most $$225 \times 9 = 2025$$, so $$N$$ needs at least 226 digits. The smallest such number has leading digit 1 followed by 225 nines, because $$1 + 225 \times 9 = 2026$$. Adding 1 carries through all the nines, giving $$N + 1 = 2 \times 10^{225}$$, whose leading digit is 2.

Only the squares $$0, 1, 4, 9, 16$$ can occur, and negative integers give the same squares. Listing each unordered pair once with the smaller square first and keeping sums at most 20 gives $$0, 1, 4, 9, 16$$ from 0, then $$2, 5, 10, 17$$ from 1, then $$8, 13, 20$$ from 4, and 18 from 9. All these sums are distinct, so the count is $$5 + 4 + 3 + 1 = 13$$.

Comparing degrees gives $$2d = d^2$$, so $$d = 2$$ and $$P(x) = ax^2 + bx + c$$ with $$a \ne 0$$. Matching the $$x^4$$ coefficients gives $$a = a^3$$, so $$a = 1$$ or $$a = -1$$; the $$x^3$$ coefficient then forces $$b = 0$$, and the $$x^2$$ coefficient forces $$c = 0$$. Both $$P(x) = x^2$$ and $$P(x) = -x^2$$ satisfy the identity, so there are exactly 2 such polynomials.

A row has 7 squares, so the largest even number of checkers it can contain is 6, and with seven rows the total cannot exceed $$7 \times 6 = 42$$. This bound is attained by leaving the seven main-diagonal squares empty and putting a checker in every other square, since each row and each column then holds exactly 6 checkers. Hence the answer is 42.

Put $$AC = BC = s$$, $$CD = t$$ and $$BD = x = s - t$$, so the given area gives $$\tfrac{1}{2}st = 81$$, that is $$st = 162$$. Drawing $$DF \perp AB$$, triangle $$BDF$$ is a $$45^\circ$$ right triangle so $$BD^2 = 2DF^2$$, while $$\angle DAB = 30^\circ$$ gives $$AD = 2DF$$, hence $$AD^2 = 2BD^2 = 2x^2$$. Pythagoras in triangle $$ACD$$ now gives $$2x^2 = s^2 + t^2 = (s - t)^2 + 2st = x^2 + 324$$, so $$x^2 = 324$$ and $$BD = 18$$.

Writing $$A = 100a + 10b + c$$ and $$B = 100c + 10b + a$$ gives $$|A - B| = 99|a - c| = 3^2 \times 11 \times |a - c|$$. The digits are distinct and nonzero, so $$1 \le |a - c| \le 8$$ and every prime factor of $$|a - c|$$ is at most 7. The factor 11 always occurs and no larger prime can appear, so the largest possible prime factor is 11.

Since $$cd^2 \ge 1$$ we need $$a^2 + b^2 \ge 2027$$, and for a fixed sum $$a + b \le 46$$ the largest possible value of $$a^2 + b^2$$ is $$45^2 + 1 = 2026$$, so $$a + b \ge 47$$. If the total were at most 50 then $$c + d \le 3$$; with $$c = d = 1$$ we would need $$a^2 + b^2 = 2027 \equiv 3 \pmod 4$$, which is impossible, and with $$c + d = 3$$ the value $$a^2 + b^2$$ is even, forcing $$a + b$$ even and contradicting $$a + b = 47$$. Hence the minimum is 51, attained by $$(a, b, c, d) = (45, 2, 3, 1)$$ because $$45^2 + 2^2 - 3 \times 1^2 = 2026$$.

Cross-multiplying and cancelling gives $$a_{n+1}a_{n+3} = a_n a_{n+2}$$, so all products $$a_n a_{n+2}$$ are equal and no term can be zero. Cancelling $$a_{n+2}$$ in $$a_n a_{n+2} = a_{n+2} a_{n+4}$$ gives $$a_{n+4} = a_n$$, so the sequence has period 4, and the given indices force $$a_3 = 6$$, $$a_2 = 2$$, $$a_1 = 1$$, hence $$a_4 = 3$$ and the repeating block $$1, 2, 6, 3$$. Each block contributes $$1 + 4 + 36 + 9 = 50$$, and since $$2026 = 4 \times 506 + 2$$ we get $$N = 506 \times 50 + 1 + 4 = 25305$$, whose digit sum is 15.

If $$m, t, M$$ are the smallest, middle and largest of the three values, the sum of the three distances equals $$2(M - m)$$, so $$M - m = 4$$ and $$(m, M)$$ is one of $$(1, 5), (2, 6), (3, 7), (4, 8)$$. For each pair, a strictly intermediate middle value gives $$3 \times 3! = 18$$ triples and a middle value equal to an endpoint gives $$2 \times 3 = 6$$ triples, so 24 in all. The total is $$4 \times 24 = 96$$.

The vertex receiving label 1 can be chosen in 8 ways, and at each later step the next label must sit at one of the two ends of the remaining boundary chain, since an interior choice would strand unvisited vertices and force a crossing. That leaves 2 choices for each of the labels $$2, 3, 4, 5, 6, 7$$, and the label 8 is forced, so $$N = 8 \times 2^6 = 512$$. The remainder on division by 100 is 12.

Since $$\angle AEC = \angle BED = 90^\circ$$, the midpoint-of-hypotenuse property gives $$FA = FE = FC$$ and $$GB = GE = GD$$. Writing $$\angle EAC = \alpha$$, the isosceles triangle $$AFE$$ gives $$\angle EFG = 2\alpha$$, while the isosceles triangle $$BGE$$ gives $$\angle EGF = 180^\circ - 2(45^\circ + \alpha) = 90^\circ - 2\alpha$$. The angle sum in triangle $$EFG$$ then yields $$\angle FEG = 180^\circ - 2\alpha - (90^\circ - 2\alpha) = 90^\circ$$, so $$x = 90$$.

If $$M$$ has $$k$$ digits then $$N = 10(M - 3 \times 10^{k-1}) + 3 = 10M - 3(10^k - 1)$$, and $$N = M/4$$ rearranges to $$M = \frac{4(10^k - 1)}{13}$$. Thus $$10^k \equiv 1 \pmod{13}$$, which first happens at $$k = 6$$, giving $$M = \frac{4 \times 999999}{13} = 307692$$, and indeed $$4 \times 76923 = 307692$$. The digit sum is $$3 + 0 + 7 + 6 + 9 + 2 = 27$$.

Subtracting the equations in pairs gives $$(x - y)(x + y - z - 1) = 0$$ together with its two cyclic versions, and if $$x, y, z$$ were all different the second factors would give $$x + y - z = y + z - x = z + x - y = 1$$, forcing $$x = y = z$$, a contradiction. If all three equal $$t$$, then $$2t^2 + t = t^2$$ gives $$t = 0$$ or $$t = -1$$, with moduli 0 and 8. If exactly two are equal, say $$x = y = t$$ and $$z = u \ne t$$, then $$u = 1$$ and $$t^2 = -1$$, and the product is $$(-1 - t)(-2)(1 - t) = 2(1 - t^2) = 4$$. The distinct moduli are $$0, 4, 8$$, whose sum is 12.

Treating the six given numbers as blocks, there are $$6! - 5! = 600$$ arrangements not beginning with 0, but different arrangements can produce the same 8-digit number. Counting each number once through the representation in which the block 10 is the first occurrence of 10 and the block 23 is the first occurrence of 23, the arrangements breaking these rules number $$\frac{5!}{2} = 60$$ and $$\frac{5! - 4!}{2} = 48$$, with $$\frac{4!}{4} = 6$$ restored by inclusion and exclusion. Hence $$N = 600 - 60 - 48 + 6 = 498$$ and the digit sum is $$4 + 9 + 8 = 21$$.

Colour the leftmost vertical edge in 3 ways. Once a square's left edge is fixed, its other three edges must supply the two missing colours, and inclusion and exclusion gives $$3^3 - 2^3 - 2^3 + 1 = 12$$ completions, the same count whatever that left edge is. Each of the five squares introduces three new edges, so no choice is counted twice and $$N = 3 \times 12^5 = 746496$$, leaving remainder 96 on division by 100.

Draw $$CF \perp BD$$. A half-turn about the centre of the rectangle takes $$A$$ to $$C$$ and $$E$$ to $$F$$, so $$CF = AE = 12$$ and $$EF^2 = CE^2 - CF^2 = 193 - 144 = 49$$, giving $$EF = 7$$. Putting $$BE = p$$ and $$ED = q$$, the symmetry gives $$|p - q| = 7$$ while the altitude relation in right triangle $$ABD$$ gives $$pq = AE^2 = 144$$, so $$BD^2 = (p - q)^2 + 4pq = 49 + 576 = 625$$ and $$BD = 25$$. The area is $$2 \times \tfrac{1}{2} \times BD \times AE = 25 \times 12 = 300$$.

Factor the expression as $$p^4 + p^2 - 2 = (p^2 - 1)(p^2 + 2)$$. For a prime $$p > 3$$ the numbers $$p - 1$$ and $$p + 1$$ are consecutive even numbers with one divisible by 4, so $$8 \mid p^2 - 1$$; also $$p^2 \equiv 1 \pmod 3$$, so both $$p^2 - 1$$ and $$p^2 + 2$$ are divisible by 3 and their product by 9. Hence every member is divisible by $$8 \times 9 = 72$$, and since $$p = 5$$ and $$p = 7$$ give $$648 = 72 \times 9$$ and $$2448 = 72 \times 34$$ with $$\gcd(9, 34) = 1$$, nothing larger works.

Take a point $$A$$ with red neighbours $$B, C, D$$. For any other point $$X$$ the segment $$AX$$ is blue, and applying the first condition to $$B, A, X$$ forces $$BX$$ to be red, so $$B$$ has at least $$1 + (n - 4) = n - 3$$ red neighbours and $$n \le 6$$, while clearly $$n \ge 4$$. Counting endpoints of red segments, $$3n$$ must be even, so $$n$$ is even, leaving $$n = 4$$ and $$n = 6$$. Both occur, by colouring every segment red when $$n = 4$$, and by splitting six points into two groups of three with blue inside groups and red between them, so the sum is $$4 + 6 = 10$$.

Label the consecutive vertices $$A, B, C, D$$, so $$BC^2 + CD^2 = (a + 3) + (a + 2) = BD^2$$ and the converse of Pythagoras gives $$\angle BCD = 90^\circ$$. Drawing $$AH \perp CD$$ and $$AK \perp CB$$ and applying Pythagoras leads to $$\frac{(a + 4)^2}{a + 3} + \frac{a + 2}{4} = 2a + 5$$, which simplifies to $$(3a + 10)(a - 1) = 0$$, so $$a = 1$$. Then $$BK = \tfrac{1}{2}$$ and $$AK = \frac{\sqrt{3}}{2}$$ give $$\angle ABK = 60^\circ$$ and hence $$\angle ABC = 120^\circ$$, while the angles at $$A$$ and $$D$$ are acute base angles of isosceles triangles. The two largest angles are $$120^\circ$$ and $$90^\circ$$, so $$\theta = 30$$.

Writing $$a_1 = x$$, $$a_2 = y$$ and $$b_1 = k$$, repeated use of the recurrences gives $$a_{10} = 21x + 34y$$ and $$b_{10} = 2^9 k = 512k$$, so $$21x + 34y = 512k < 2026$$ and $$k \in \{1, 2, 3\}$$. Since $$y > x$$ we get $$55x < 512k$$, and reducing modulo 34 gives $$x \equiv 26k \pmod{34}$$, which leaves only $$(x, y, k) = (18, 19, 2)$$ and $$(10, 39, 3)$$. The possible values of $$a_1 + b_1$$ are 20 and 13, whose sum is 33.

From the definition, $$2^{62} = 4^{31} = 3n + 1$$, so $$2^{62} \equiv 1 \pmod{n}$$. Also $$n = 1 + 4 + 4^2 + \cdots + 4^{30}$$ is odd, so $$n - 1$$ is even, and since $$4^5 = 1024 = 31 \times 33 + 1$$ we have $$4^{30} \equiv 1 \pmod{31}$$, so 31 divides $$3(n - 1) = 4(4^{30} - 1)$$ and hence divides $$n - 1$$. Therefore $$n - 1 = 62t$$ and $$2^{n-1} = (2^{62})^t \equiv 1 \pmod{n}$$, so the remainder is 1.

Construct the equilateral triangle $$ABF$$ on the same side of $$AB$$ as $$C$$. Since $$\angle DAB = 30^\circ$$, the line $$AD$$ is the perpendicular bisector of $$BF$$, giving $$DB = DF$$, and $$BE$$ bisects $$\angle ABF = 60^\circ$$, giving $$EA = EF$$. The isosceles triangles $$AEF$$ and $$BDF$$ have base angles $$\angle EAF = 60^\circ - 50^\circ = 10^\circ$$ and $$\angle DBF = 70^\circ - 60^\circ = 10^\circ$$ with equal bases, so they are congruent and $$EF = DF$$, while $$\angle EFD = (60^\circ - 10^\circ) + 10^\circ = 60^\circ$$ makes triangle $$EFD$$ equilateral. Hence $$ED = DF = DB$$, so triangle $$BDE$$ is isosceles and $$\angle DEB = \angle DBE = 70^\circ - 30^\circ = 40^\circ$$, giving $$x = 40$$.