IOQM Mock Paper 5: 30 Questions and Solutions

With five selections, one of the two parity classes contains at least three. Four selections such as 1,2,3,4 contain only two of each parity, so five is the smallest guaranteed number.

Answer / conclusion: 05

Review the idea: Pigeonhole principle

The product ab=3 is nonzero. Thus $1/a+1/b=(a+b)/(ab)=6/3=2$.

Answer / conclusion: 02

Review the idea: Vietas formulas

Label the trapezoid $ABCD$ with $AB\parallel CD$, $AB=7$, $CD=15$. Let $M,N$ be the midpoints of $AD,BC$, respectively, and let $K$ be the midpoint of diagonal $AC$.

In triangle $ADC$, the midpoint theorem gives $MK\parallel DC$ and $MK=15/2$. In triangle $ABC$, it gives $KN\parallel AB$ and $KN=7/2$. Since the two bases are parallel, both lines through $K$ are the same line; hence $M,K,N$ are collinear. In a convex trapezoid the diagonal is inside, so $K$ lies between the two side midpoints. Thus $$MN=MK+KN=\frac{15}{2}+\frac72=11.$$

Answer / conclusion: 11

Review the idea: The midpoint theorem

$5!=120=2^3\cdot3\cdot5$. The divisor exponents have 4,2,2 choices respectively, so the total is $4\cdot2\cdot2=16$.

Answer / conclusion: 16

Review the idea: Counting and summing divisors

The sum is $(2^2-1^2)+(3^2-2^2)+\cdots+(10^2-9^2)=10^2-1^2=99$.

Answer / conclusion: 99

Review the idea: Sequences and sums

If 0 appeared, increasing order would place it first, which is not allowed in a three-digit number. Thus choose any three digits from 1 through 9 and sort them. This gives $\binom93=84$.

Answer / conclusion: 84

Review the idea: Counting with bijections

Since C is on the major arc, angle ACB intercepts the minor arc of measure $120^\circ$. The inscribed-angle theorem gives $\angle ACB=60^\circ$.

Answer / conclusion: 60

Review the idea: Circles and power of a point

There are ten pairs $(-1+2),(-3+4),\ldots,(-19+20)$, each equal to 1. The total is 10.

Answer / conclusion: 10

Review the idea: Sequences and sums

Write the desired positive multiple of 6 as $n=6k$, where $k$ is a positive integer. Since $6\equiv-1\pmod7$, the remainder condition becomes $$-k\equiv4\pmod7,\qquad k\equiv-4\equiv3\pmod7.$$ The least positive such $k$ is 3, so $n=6\cdot3=18$. Indeed, $18=2\cdot7+4$.

Answer / conclusion: 18

Review the idea: Remainders

The common factor one-half times the base cancels from the area ratio. Thus the smaller area is $60\cdot2/3=40$.

Answer / conclusion: 40

Review the idea: Triangle area ratios

We have $\sqrt{x+9}=\sqrt x+1$, so $x+9=x+2\sqrt x+1$. Hence $\sqrt x=4$, giving x=16. Substitution gives 5-4=1, so no extraneous root was introduced.

Answer / conclusion: 16

Review the idea: Powers radicals logarithms

We get $10^4\equiv4$ and $10^8\equiv16\equiv-1\pmod{17}$, hence $10^{16}\equiv1$. Since $100=6\cdot16+4$, the remainder is $10^4\equiv4$.

Answer / conclusion: 04

Review the idea: Remainders

First count positive solutions of $x+y+z=10$ without upper bounds. Place ten identical stars in a row and put two dividers in two of the nine gaps between them. The three nonempty groups have sizes $x,y,z$, and every positive solution is represented uniquely. Thus there are $\binom92=36$ solutions.

If $x\ge7$, replace it by $x-6\ge1$. The three positive variables now sum to 4, and two dividers in the three gaps give $\binom32=3$ solutions. There are three choices of a violating coordinate. Two violations cannot overlap, since their sum with the third positive coordinate would be at least $7+7+1=15>10$. Subtracting these forbidden cases leaves $36-3\cdot3=27$ ordered dice outcomes.

Answer / conclusion: 27

Review the idea: Inclusion exclusion · Counting integer solutions

Applying the median identity gives $m^2=(2\cdot4^2+2\cdot6^2-8^2)/4=(32+72-64)/4=10$. The strict triangle inequality 4+6>8 confirms a nondegenerate triangle.

Answer / conclusion: 10

Review the idea: Pythagoras and stewart

There are $2^5-2=30$ choices of a nonempty proper subset. Each split appears once for each of its two complementary groups, so the count is $30/2=15$.

Answer / conclusion: 15

Review the idea: Dividing objects into fixed size groups

Write $6n+1=k^2$, taking $k>0$. From $1\le n\le100$, we get $7\le k^2\le601$, and therefore $3\le k\le24$. Squaring residues $0,1,2,3,4,5$ modulo 6 gives, respectively, $0,1,4,3,4,1$. Thus $k^2\equiv1\pmod6$ exactly when $k\equiv1$ or 5, equivalently when $k$ is odd and not divisible by 3.

The possibilities in the stated range are $5,7,11,13,17,19,23$. Each gives an integer $n=(k^2-1)/6$ between 1 and 100, and distinct positive $k$ give distinct $n$. Hence there are 7.

Answer / conclusion: 07

Review the idea: Remainders

Let $t=|x|\ge0$. Then $x^2=t^2$, so the equation becomes $$t^2-5t+6=(t-2)(t-3)=0.$$ Both roots $t=2,3$ are positive, and each gives two distinct real values of $x$: $x=\pm2$ or $x=\pm3$. All four satisfy the original equation, and no other $t$ is possible. Thus there are 4 real solutions.

Answer / conclusion: 04

Review the idea: Absolute value inequalities

The diagonal has length $\sqrt{6^2+8^2}=10$. Its midpoint is equidistant from all four vertices, so the circumradius is 5 and its square is 25.

Answer / conclusion: 25

Review the idea: Circles and power of a point

Write an increasing three-element arithmetic progression as $\{a,a+d,a+2d\}$, with positive integers $a,d$. The largest entry is at most 12, so $1\le a\le12-2d$. This range is nonempty exactly for $d=1,2,3,4,5$. The corresponding numbers of choices for $a$ are 10, 8, 6, 4 and 2.

Each subset uniquely determines its smallest element $a$ and its positive common difference $d$, so no subset is counted twice. The total is $10+8+6+4+2=30$.

Answer / conclusion: 30

Review the idea: Counting

At n=1 the value is 3, prime; at n=2 it is 0, not prime. For n at least 3, a prime product $(n-2)(n+2)$ requires the smaller positive factor to equal 1. Thus n=3, giving 5. There are exactly two values of n.

Answer / conclusion: 02

Review the idea: Prime numbers

The prime 2 fails since $2^2+1=5$ is odd. For an odd prime p, Fermat gives $2^{p-1}\equiv1\pmod p$, so $2^p+1\equiv3\pmod p$. Hence p must divide 3, giving p=3. It works because $2^3+1=9$.

Answer / conclusion: 03

Review the idea: Number theory theorems

Add 1 modulo 11 to every element of a four-element subset, replacing 11 by 0. Distinct elements remain distinct, and subtracting 1 modulo 11 reverses the operation. It is therefore a bijection on all four-element subsets. Since four elements each increase by 1 modulo 11, their sum increases by 4 modulo 11.

Consequently, the number of subsets with sum residue $r$ equals the number with residue $r+4$. Starting at 0 and repeatedly adding 4 modulo 11 visits $$0,4,8,1,5,9,2,6,10,3,7,$$ all eleven residues. Thus all sum-residue classes have equal size. There are $\binom{11}{4}=330$ subsets in total, so exactly $330/11=30$ have sum divisible by 11.

Answer / conclusion: 30

Review the idea: Counting with bijections

Choose coordinates $B=(0,0)$, $C=(8,0)$, and $A=(x,h)$, taking $h>0$. Initially this permits the perpendicular foot $D=(x,0)$ anywhere on line $BC$. The given side lengths yield $$x^2+h^2=25,\qquad(x-8)^2+h^2=49.$$ Subtracting the first equation from the second gives $-16x+64=24$, so $x=5/2$, which lies between 0 and 8.

Substitution now gives $$h^2=25-\left(\frac52\right)^2=\frac{75}{4}.$$ Therefore $4h^2=75$.

Answer / conclusion: 75

Review the idea: Pythagoras and stewart

Ignore the diagonal restriction first. Choose three occupied rows and three occupied columns in $\binom43^2$ ways. Match the chosen rows to the chosen columns in $3!$ ways, placing one rook at each matched cell. This gives exactly the nonattacking placements, so there are $\binom43^2 3!=96$.

There are four forbidden events, one for occupancy of each main-diagonal cell. If a specified diagonal rook is present, its row and column are fixed and unavailable to the other rooks. In the remaining $3\times3$ board, choose two rows and two columns and match them: $\binom32^2 2!=18$ placements. Two specified diagonal rooks leave a $2\times2$ board with four choices for the last rook. Three specified diagonal rooks determine one placement. Four cannot occur because there are only three rooks.

Inclusion–exclusion subtracts each single forbidden event, adds intersections of two, then subtracts intersections of three. The required number is $$96-4\cdot18+\binom42\cdot4-\binom43=44.$$

Answer / conclusion: 44

Review the idea: Inclusion exclusion

The hypotenuse is $AB=AD+DB=9+16=25$. Triangles $ACD$ and $ABC$ share angle $A$ and are both right triangles, so they are similar by the angle-angle criterion. Matching corresponding sides gives $$\frac{AC}{AB}=\frac{AD}{AC},\qquad AC^2=AD\cdot AB=9\cdot25=225.$$ Thus $AC=15$. Likewise, using the shared angle at $B$ in $BCD$ and $BAC$, $BC^2=DB\cdot AB=16\cdot25=400$, so $BC=20$.

The area is $\Delta=15\cdot20/2=150$. Let $r$ be the inradius and $I$ the incentre. Joining $I$ to the three vertices partitions the triangle into three triangles whose altitudes to the side lines all equal $r$. Hence $$150=\Delta=\frac{r(15+20+25)}2=30r.$$ Therefore $r=5$.

Answer / conclusion: 05

Review the idea: Similar triangles

Write $[XYZ]$ for the area of triangle XYZ and set $[BGD]=t$. Since BD=DC and both small triangles have the same altitude from G, $[CGD]=t$, so $[BCG]=2t$. The ratio AE:EC=2:1 gives $[ABE]=60$ and $[CBE]=30$. Thus $[CGE]=30-2t$. Triangles AGE and CGE share an altitude to AC, so $[AGE]=2[CGE]=60-4t$. Consequently $[ABG]=[ABE]-[AGE]=4t$. Finally D is the midpoint of BC, so $[ABD]=45$, and $[ABG]+[BGD]=5t=45$. Hence the required area is $t=9$.

Answer / conclusion: 09

Review the idea: Triangle area ratios

Degree 2 means that each vertex is joined to exactly two other vertices. Start at a vertex, follow an edge, and at each new vertex leave by the edge other than the one just used to arrive. As there are finitely many vertices, a vertex must eventually repeat. The first repeat must be the starting vertex: any other previously visited vertex already has its entering and leaving edges, so an additional arrival from a new vertex would give it degree at least 3. Thus the walk closes into a cycle.

Each vertex of that cycle already has both its edges within the cycle, so no further vertex can attach to it. Its entire connected component is therefore that cycle. Loops and repeated edges are forbidden, so each cycle has at least three vertices. Five vertices cannot split into two such cycles, so the whole graph is one cycle through all five labelled vertices.

Fix one labelled starting vertex. The remaining four appear around the cycle in $4!$ orders. Each edge set is counted twice, once in each direction, and no further starting-point overcount remains because the first vertex was fixed. The answer is $4!/2=12$.

Answer / conclusion: 12

Review the idea: Circular permutations

The ordering gives $1\le3/x$, so x is at most 3. It cannot be 1 because the other fractions are positive. If x=3, all fractions are at most 1/3 and equality forces (3,3,3). If x=2, then y,z>2 and $(y-2)(z-2)=4$. Its ordered-by-size factor pairs (1,4) and (2,2) give (2,3,6) and (2,4,4). Thus there are exactly three triples.

Answer / conclusion: 03

Review the idea: Factorisation integer solutions

Let $E_n$ count alternating permutations of n distinct values starting with a rise; reversing all value ranks shows the falling-start count is the same. Set $E_0=E_1=1$. For n>1 the maximum is in an even position j. Choose the j-1 values to its left, then independently arrange both sides alternately: $E_n=\sum_{j\text{ even}}\binom{n-1}{j-1}E_{j-1}E_{n-j}$. This gives $E_2=1,E_3=2,E_4=5,E_5=16$. Therefore $E_6=\binom51E_1E_4+\binom53E_3E_2+\binom55E_5E_0=25+20+16=61$.

Answer / conclusion: 61

Review the idea: Counting with recurrences

First $a^2+b^2+c^2=12^2-2\cdot44=56$. Put $x=a-4,y=b-4,z=c-4$. Then $x+y+z=0$ and $x^2+y^2+z^2=56-8\cdot12+48=8$. Each deviation has absolute value at most 2, so its square is 0,1 or 4. The only way three such squares total 8 is 4+4+0. The zero-sum condition forces deviations 2,-2,0 in some order. Thus a,b,c are 6,2,4 in some order and $abc=48$.

Answer / conclusion: 48

Review the idea: Symmetric polynomials