IOQM Mock Paper 2: 30 Questions and Solutions

Since $18=2\cdot3^2$, a square multiple needs at least one further factor of 2. Taking $k=2$ gives 36, a square, so the least k is 2.

Answer / conclusion: 02

Review the idea: Unique prime factorisation

The identity gives $35=125-15xy$, so $xy=6$. The pair $(2,3)$ confirms that the conditions are attainable.

Answer / conclusion: 06

Review the idea: Algebraic identities

The exterior angles sum to $360^\circ$. With n equal angles, $24n=360$, so $n=15$.

Answer / conclusion: 15

Review the idea: Angles

There are $6\cdot5=30$ ordered selections. Each team is counted twice, giving $30/2=15$ teams.

Answer / conclusion: 15

Review the idea: Combinations and binomial coefficients

Explicit inverse pairs modulo 11 are $(2,6),(3,4),(5,9),(7,8)$; each pair has product 1 modulo 11. The remaining product is $1\cdot10$, so the remainder is 10.

Answer / conclusion: 10

Review the idea: Number theory theorems

The diagonals of a rhombus bisect each other at right angles. To recall why: a rhombus is a parallelogram, so its diagonals bisect each other. The two triangles on either side of one half-diagonal have equal rhombus sides, equal halves of the other diagonal, and a common side. They are congruent, so their adjacent angles at the intersection are equal; those angles sum to $180^\circ$, and therefore each is $90^\circ$.

A side is thus the hypotenuse of a right triangle with legs $10/2=5$ and $24/2=12$. Its length is $\sqrt{5^2+12^2}=13$. All four sides are equal, giving perimeter $4\cdot13=52$.

Answer / conclusion: 52

Review the idea: Quadrilaterals and their diagonals

The axis points are $(\pm3,0),(0,\pm3)$, giving 4. Each of the two magnitude pairs $(1,2),(2,1)$ allows four sign choices, giving 8 more. The total is 12.

Answer / conclusion: 12

Review the idea: Counting integer solutions

Let the digits be $a,b,c$, with $a\ge1$. Then $(a-1)+b+c=1$, whose three possibilities place the 1 in one of three positions. They give 200, 110 and 101.

Answer / conclusion: 03

Review the idea: Number bases and digit problems

We have $a_{n+1}-a_n=(n+1)(n+2)-n(n+1)=2(n+1)$. Thus $n+1=9$ and $n=8$.

Answer / conclusion: 08

Review the idea: Sequences and sums

Let $d$ be the diagonal of the rectangular face with side lengths 4 and 5. By Pythagoras, $d^2=4^2+5^2=41$. Let $L$ be the space diagonal joining opposite vertices of the box. The edge of length 6 is perpendicular to the whole bottom plane, and hence to its diagonal $d$. It forms a right triangle with legs $d,6$ and hypotenuse $L$. Therefore $$L^2=d^2+6^2=41+36=77.$$ A space diagonal uses all three perpendicular edge lengths and is longer than every face diagonal, so this is the requested square.

Answer / conclusion: 77

Review the idea: Pythagoras and stewart

The alternating sum is $7-a$. As $0\le a\le9$, this lies from -2 to 7, whose only multiple of 11 is 0. Thus $a=7$; indeed $374=34\cdot11$.

Answer / conclusion: 07

Review the idea: Divisibility

There are $5!/(2!2!)=30$ total arrangements. With AA treated as one object, the objects AA, B, B, C have $4!/2!=12$ arrangements. The required count is $30-12=18$.

Answer / conclusion: 18

Review the idea: Permutations and arrangements

Since $x^2=(x^2-1)+1$, replacing $x^2$ by 1 does not change the remainder on division by $x^2-1$: the removed part is a multiple of the divisor. Repeating this replacement gives remainder 1 for every even power and remainder $x$ for every odd power.

Thus $x^{20}$ contributes 1, $2x^{11}$ contributes $2x$, $3x^2$ contributes 3, and the constant 4 remains 4. The remainder is $1+2x+3+4=2x+8$. It already has degree below 2, as required. Hence $a=2$, $b=8$, and $b-a=6$.

Answer / conclusion: 06

Review the idea: Polynomial division

The cosine rule gives $BC^2=7^2+8^2-2\cdot7\cdot8\cdot(1/2)=49+64-56=57$.

Answer / conclusion: 57

Review the idea: Trigonometry in geometry

Write $72=2^3 3^2$ and $d=2^a3^b$. Opposite parity requires $a=0$ or $a=3$; for $a=1,2$, both summands are even. There are two choices for a and three for b, so the answer is 6.

Answer / conclusion: 06

Review the idea: Counting and summing divisors

The first three powers have residues 3, 9 and 1, because $27\equiv1\pmod{13}$. Neither of the first two is 1, so the least exponent is 3.

Answer / conclusion: 03

Review the idea: Remainders

The subset must contain either one odd and three even numbers, or three odd and one even number. These disjoint cases give $\binom51\binom43+\binom53\binom41=20+40=60$.

Answer / conclusion: 60

Review the idea: Combinations and binomial coefficients

Label the quadrilateral $ABCD$ in boundary order, and let $E,F,G,H$ be the midpoints of $AB,BC,CD,DA$, respectively. Write $[XYZ]$ for the area of triangle $XYZ$. Each corner triangle cut off by a midpoint segment is similar to the triangle formed by that corner and its two neighbours, with length scale $1/2$. Its area is therefore $(1/2)^2=1/4$ of the larger triangle’s area.

The four corner areas consequently total $$\frac{[DAB]+[ABC]+[BCD]+[CDA]}4.$$ Diagonal $BD$ partitions the quadrilateral into triangles $DAB,BCD$, whose areas sum to 64. Diagonal $AC$ similarly gives $[ABC]+[CDA]=64$. Thus the corners total $(64+64)/4=32$, leaving area $64-32=32$ for the midpoint parallelogram.

Answer / conclusion: 32

Review the idea: The midpoint theorem

The floor is 0 for k=1,2,3; each value 1,2,3,4 occurs four times; and 5 occurs once. The sum is $4(1+2+3+4)+5=45$.

Answer / conclusion: 45

Review the idea: Floor and ceiling functions

First ignore the upper bounds. Represent $(x,y,z)$ by six identical stars split into three groups by two separators; the group sizes are $x,y,z$. Empty groups are allowed, including at the ends. Conversely every such arrangement gives exactly one nonnegative triple summing to 6. Choose the two separator positions among eight places: there are $\binom82=28$ triples.

If $x\ge4$, put $u=x-4\ge0$. Then $u+y+z=2$, represented by two stars and two separators, so there are $\binom42=6$ triples. The same count applies to a violation by $y$ or $z$. Two violations cannot occur together, since they would make the sum at least $4+4=8>6$. Subtracting the three disjoint forbidden cases leaves $28-3\cdot6=10$.

Answer / conclusion: 10

Review the idea: Inclusion exclusion · Counting integer solutions

Since consecutive integers have greatest common divisor 1, their product is divisible by 4 exactly when one of them is divisible by 4: its even member must supply both factors of 2. Thus $n\equiv0$ or $3\pmod4$. Similarly, divisibility by 5 requires $n\equiv0$ or $4\pmod5$. As 4 and 5 are coprime, both conditions together are equivalent to divisibility by 20.

Among residues $0,1,\ldots,19$, the condition modulo 5 gives $0,4,5,9,10,14,15,19$. Exactly $0,4,15,19$ also satisfy the condition modulo 4. Each of these four residues occurs five times from 1 to 100, so the answer is $4\cdot5=20$.

Answer / conclusion: 20

Review the idea: Number theory theorems

If the first two letters are AA, the next letter must be B. Completing legally gives AABBCC or AABCBC. If the first two letters are AB, the third is A or C. A third A leaves one B and two Cs, giving ABABCC or ABACBC. A third C gives ABC, after which the remaining A,B,C are forced in that order. Thus ABCABC is the fifth string, and the branching exhausts all possibilities.

Answer / conclusion: 05

Review the idea: Counting

Order the two distinct roots as $r<s$. Since the polynomial is monic, $P(x)=(x-r)(x-s)$. Therefore $$12=P(1)=(1-r)(1-s)=(r-1)(s-1).$$ Neither root can be 1, and both are positive integers, so both are at least 2. Thus $r-1,s-1$ form a positive unequal factor pair of 12.

The possibilities are $(1,12),(2,6),(3,4)$, giving root pairs $(2,13),(3,7),(4,5)$. Each pair determines one monic polynomial with integer coefficients, and each has the required value at 1. Hence there are 3 polynomials.

Answer / conclusion: 03

Review the idea: Vietas formulas

Let $G$ be the centroid, and let $M$ be the midpoint of $AB$, with the given medians drawn from $A,B$. The centroid divides each median in the ratio $2:1$ from its vertex, so $GA=(2/3)9=6$ and $GB=(2/3)12=8$. Their lines are perpendicular.

Choose perpendicular coordinate axes through $G$, directed towards $A$ and $B$. Then $A=(6,0)$, $B=(0,8)$, and their midpoint is $M=(3,4)$. Pythagoras gives $GM=\sqrt{3^2+4^2}=5$. On the third median $CM$, the centroid property gives $CG:GM=2:1$. Therefore the entire median has length $CM=3GM=15$.

Answer / conclusion: 15

Review the idea: Concurrency and collinearity

Interchanging the primes changes neither the equation nor their sum, so assume $p\le q$. Then $3p^2\le p^2+pq+q^2=109$, so $p^2\le109/3<49$ and $p\le6$. The possible primes are 2, 3 and 5.

Regard the equation as $q^2+pq+p^2-109=0$. Completing the square after multiplying by 4 gives $$(2q+p)^2=436-3p^2.$$ Since $q$ is an integer, the right side must be a perfect square. For $p=2,3,5$ it is, respectively, 424, 409 and 361. The first two lie strictly between $20^2=400$ and $21^2=441$, whereas $361=19^2$. Thus $p=5$, and positivity gives $2q+5=19$, so $q=7$. Both are prime and $25+35+49=109$. Hence $p+q=12$.

Answer / conclusion: 12

Review the idea: Prime numbers

Order the distinct side lengths as $a<b<c$; this counts each noncongruent scalene triangle once. The triangle inequality gives $c<a+b=18-c$, so $c<9$ and hence $c\le8$. If $c\le6$, the largest possible distinct sides would be 4, 5 and 6, whose sum is only 15. Thus $c\ge7$.

For $c=7$, we need $a+b=11$ and $a<b<7$, yielding only $(a,b)=(5,6)$. For $c=8$, we need $a+b=10$ with $a<b<8$, giving $(3,7)$ or $(4,6)$. Each pair has sum larger than $c$, so all three triangles exist. They are $(5,6,7),(3,7,8),(4,6,8)$, giving answer 3.

Answer / conclusion: 03

Review the idea: Triangle inequalities

There are $\binom63=20$ colourings of six labelled positions. Define a colouring’s least rotational period $d$ as the smallest positive shift of positions that leaves it unchanged. It divides 6: write $6=qd+r$, $0\le r<d$. A shift by 6 and a shift by $qd$ both preserve the colouring, so their difference, a shift by $r$, does too. Minimality forces $r=0$. Thus $d$ is 1, 2, 3 or 6.

Period 1 would make every bead the same colour. Period 3 would repeat one three-bead block twice, giving an even number of black beads; both are impossible. Exactly two labelled colourings have period 2: the alternating patterns black-white-black-white-black-white and its shift. They form one rotation class.

Every remaining colouring has period 6, meaning its six rotations are distinct. They form $(20-2)/6=3$ classes. Including the alternating class gives $1+3=4$. Reflections have not been identified.

Answer / conclusion: 04

Review the idea: Circular permutations

Put $d=\gcd(x,y)$, and write $x=da$, $y=db$, where $a,b\ge1$ and $\gcd(a,b)=1$. The equation becomes $$d(a+b+1)=30.$$ Set $m=a+b=30/d-1$. Since $m\ge2$, the possible divisors $d$ of 30 are $1,2,3,5,6,10$; larger divisors give $m<2$.

For fixed $m$, choose $1\le a<m$; then $b=m-a$ is forced. Also $\gcd(a,m-a)=\gcd(a,m)$, because subtracting one number from the other does not change the common divisors. Thus we count the following eligible values.

Each choice determines one ordered pair $(x,y)=(da,d(m-a))$, and its greatest common divisor recovers $d$, so none is counted twice. The total is $28+6+6+4+2+1=47$.

Answer / conclusion: 47

Review the idea: Greatest common divisor

Cauchy gives $(x+y+z)^2\le3\cdot49=147$, so the integer sum is at most 12. But each integer has the same parity as its square, so the sum must be odd. Hence it is at most 11. The triple $(2,3,6)$ has squared sum 49 and sum 11, proving attainability.

Answer / conclusion: 11

Review the idea: Cauchy schwarz inequality

Let the square side be $a>0$. Choose $A=(0,0)$, $B=(a,0)$, $D=(0,a)$, and $P=(x,y)$. Since $P$ is strictly inside, $0<x,y<a$. The three distance conditions give $$x^2+y^2=5,\quad(x-a)^2+y^2=20,\quad x^2+(y-a)^2=10.$$ Subtracting the first equation from each of the other two yields $a^2-2ax=15$ and $a^2-2ay=5$.

Put $s=a^2$, the square’s area. Then $x=(s-15)/(2a)$ and $y=(s-5)/(2a)$. Substitute into $x^2+y^2=5$ and multiply by $4a^2=4s$: $$(s-15)^2+(s-5)^2=20s.$$ Expanding gives $s^2-30s+125=0$, or $(s-5)(s-25)=0$. Since $x>0$, we must have $s>15$; hence $s=25$. This is attainable with $a=5$, $x=1$, $y=2$, which lies inside the square and gives squared distances 5, 20 and 10. The area is 25.

Answer / conclusion: 25

Review the idea: Pythagoras and stewart