IOQM Mock Paper 4: 30 Questions and Solutions

Since $10\equiv1\pmod9$, an integer has the same remainder as its digit sum. Here $1+2+\cdots+9=45$, divisible by 9, so the remainder is 0.

Answer / conclusion: 00

Review the idea: Divisibility

We have $36=14+2(xy+yz+zx)$. Hence $xy+yz+zx=11$. The triple (1,2,3) confirms consistency.

Answer / conclusion: 11

Review the idea: Algebraic identities

There are six positions, with three indistinguishable As, two indistinguishable Ns and one B. The number of arrangements is $6!/(3!2!)=60$.

Answer / conclusion: 60

Review the idea: Permutations and arrangements

Writing the smaller angle as t gives $t+(t+20)=120$. Thus $2t=100$, so the smaller angle is $50^\circ$.

Answer / conclusion: 50

Review the idea: Angles

The squares of residues 0 through 9 end in 0,1,4,9,6,5,6,9,4,1. Only residues 4 and 6 work. Each occurs ten times in the stated range, giving 20 integers.

Answer / conclusion: 20

Review the idea: Remainders

The other side is $\sqrt{100-36}=8$. Therefore the perimeter is $2(6+8)=28$.

Answer / conclusion: 28

Review the idea: Pythagoras and stewart

The first eight terms are $1,1,2,3,5,8,13,21$, found by adding the previous two each time. Thus $a_8=21$.

Answer / conclusion: 21

Review the idea: Second order recurrences

The possible sums range from 2 to 12, so the prime sums are 2, 3, 5, 7 and 11. For each sum, the first die determines the second uniquely. Their possible first-die values are, respectively, $\{1\}$, $\{1,2\}$, $\{1,2,3,4\}$, $\{1,2,3,4,5,6\}$, and $\{5,6\}$. Thus the counts are 1, 2, 4, 6 and 2. These sums give disjoint sets of ordered outcomes, so the total is $1+2+4+6+2=15$.

Answer / conclusion: 15

Review the idea: Counting

$\gcd(84,126)=\gcd(84,42)=42$, because 84 is twice 42.

Answer / conclusion: 42

Review the idea: Greatest common divisor

Triangulation gives $(n-2)180=1440$. Thus $n-2=8$, and $n=10$.

Answer / conclusion: 10

Review the idea: Angles

Write $a=6x$, $b=6y$. Since $\gcd(a,b)=6$, we have $\gcd(x,y)=1$. The identity $\gcd(a,b)\operatorname{lcm}(a,b)=ab$ gives $$6\cdot180=36xy,\qquad xy=30.$$ This identity follows prime by prime: the minimum and maximum exponents add to the sum of the two original exponents.

The unordered positive factor pairs of 30 are $(1,30),(2,15),(3,10),(5,6)$, all coprime. Their sums are 31, 17, 13 and 11. Therefore the least $a+b=6(x+y)$ is $6\cdot11=66$, attained by $(a,b)=(30,36)$.

Answer / conclusion: 66

Review the idea: Greatest common divisor

By Vieta and the cubic identity, $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=64-12=52$.

Answer / conclusion: 52

Review the idea: Vietas formulas

Because the incentre $I$ lies on the internal angle bisectors, $\angle IBC=\frac12\angle ABC$ and $\angle ICB=\frac12\angle ACB$. Write the latter two original angles as $B,C$. The angle sum in $ABC$ gives $B+C=180^\circ-60^\circ=120^\circ$.

Now use the angle sum in triangle $BIC$: $$\angle BIC=180^\circ-\frac{B+C}{2}=180^\circ-60^\circ=120^\circ.$$ Hence $\angle BIC-90^\circ=30^\circ$, and the requested integer is 30.

Answer / conclusion: 30

Review the idea: Parallel lines and angle bisectors

There are six available positions for the three Bs, with all other positions filled by A. Hence the count is $\binom63=20$.

Answer / conclusion: 20

Review the idea: Combinations and binomial coefficients

The number is $111(1000^3+1000^2+1000+1)$. Since $111=3\cdot37$, the whole number is divisible by 37. The remainder is 0.

Answer / conclusion: 00

Review the idea: Number bases and digit problems

We obtain $3x+10+5x+10=180$, so $8x=160$ and $x=20$. The angles 70 and 110 degrees are valid.

Answer / conclusion: 20

Review the idea: Cyclic and tangential quadrilaterals

We have $100=1\cdot3^4+0\cdot3^3+2\cdot3^2+0\cdot3+1$, so its representation is $10201_3$. The digit sum is $1+0+2+0+1=4$.

Answer / conclusion: 04

Review the idea: Number bases and digit problems

Put $x=2a+1,y=2b+1,z=2c+1$. Then $a,b,c\ge0$ and $a+b+c=4$. Stars and bars gives $\binom62=15$, and the substitution is reversible.

Answer / conclusion: 15

Review the idea: Counting integer solutions

Tangents from the same exterior point have equal lengths, so PA=PB. Thus the base angles PAB and PBA are equal and sum to $180^\circ-50^\circ=130^\circ$. Each is $65^\circ$.

Answer / conclusion: 65

Review the idea: Circles and power of a point

As $n\equiv-12\pmod{n+12}$, the condition is equivalent to $n+12\mid156$. Since n is positive, the divisor must exceed 12. The eligible divisors are 13,26,39,52,78,156, giving six values of n.

Answer / conclusion: 06

Review the idea: Factorisation integer solutions

Rotating a circular order changes none of its neighbour pairs, so put 1 first. Reversing the order also leaves the sum of the four absolute differences unchanged. The six orders of the remaining three numbers therefore form three reverse-pairs, with representatives $$(1,4,7,10),\quad(1,4,10,7),\quad(1,7,4,10).$$ Their sums are, respectively, $$3+3+3+9=18,\quad3+6+3+6=18,\quad6+3+6+9=24.$$ Every circular order belongs to one of these cases. Thus the greatest value is 24, attained by the order $(1,7,4,10)$.

Answer / conclusion: 24

Review the idea: Absolute value inequalities

On $[0,1)$, the expression takes values 0,1,2,3, on intervals cut by 1/3,1/2,2/3. Adding 1 to x raises the sum by $1+2+3=6$. Thus on $[1,2)$ it takes 6,7,8,9, and on $[2,3)$ it takes 12,13,14,15. The three lists are disjoint, giving 12 distinct values.

Answer / conclusion: 12

Review the idea: Floor and ceiling functions

For externally tangent circles above the same line, the centre distance is u+v and the vertical difference is u-v. Pythagoras therefore gives horizontal separation $\sqrt{(u+v)^2-(u-v)^2}=2\sqrt{uv}$. The large-circle separation is $2\sqrt{9\cdot4}=12$. The small circle lies between their tangency points on the line, so $2\sqrt{9r}+2\sqrt{4r}=12$. Hence $10\sqrt r=12$, giving $r=36/25$ and $25r=36$.

Answer / conclusion: 36

Review the idea: Circles and power of a point

There are $5!=120$ permutations. Each forbidden adjacency alone occurs in $2\cdot4!=48$ permutations. Both occur in $2\cdot3!=12$, using the blocks 123 or 321. Thus the count is $120-48-48+12=36$.

Answer / conclusion: 36

Review the idea: Inclusion exclusion

Modulo 3, the equation forces both x and y to be divisible by 3. Write x=3u,y=3v; then $3(u^2+v^2)=z^2$, forcing z=3w. Division by 9 gives $u^2+v^2=3w^2$, another positive solution with smaller sum. A positive solution of least sum therefore cannot exist. There are no positive solutions at all, so the requested count is 0.

Answer / conclusion: 00

Review the idea: Proof methods

Complete a difference of squares: $$4t^4+1=(2t^2+1)^2-(2t)^2=(2t^2-2t+1)(2t^2+2t+1).$$ At $t=5$, the factors are $50-10+1=41$ and $50+10+1=61$.

A composite positive integer has a prime factor no larger than its square root: in a product of two factors, they cannot both exceed the square root. For 41, test primes 2, 3 and 5; none divides it. For 61, test 2, 3, 5 and 7; none divides it (in particular, 61 lies between $7\cdot8=56$ and $7\cdot9=63$). Thus both factors are prime, and the largest is 61.

Answer / conclusion: 61

Review the idea: Algebraic identities

First fix the colour at the first labelled position. Let $E_n$ count valid colourings of a row of $n$ positions whose last colour equals the first, and let $D_n$ count those whose last colour differs. Adjacent positions must have different colours. Initially $(E_1,D_1)=(1,0)$.

If the last colour equals the first, either of the other two colours may be appended, and both differ from the first. Thus each such colouring contributes two to $D_{n+1}$. If the last colour differs from the first, one allowed next colour is the first, and the other is the third colour. Thus each contributes one to $E_{n+1}$ and one to $D_{n+1}$. Therefore $$E_{n+1}=D_n,\qquad D_{n+1}=2E_n+D_n.$$

For $n=2,3,4,5$, these pairs are $(0,2),(2,2),(2,6),(6,10)$. Closing the row into a cycle requires the last colour to differ from the first, giving $D_5=10$ choices for the fixed first colour. There are three possible first colours, so the answer is $3\cdot10=30$.

Answer / conclusion: 30

Review the idea: Counting with recurrences

The semiperimeter is $s=(10+17+21)/2=24$. Heron’s formula gives the area $$\Delta=\sqrt{s(s-10)(s-17)(s-21)}=\sqrt{24\cdot14\cdot7\cdot3}=84.$$ Let $r$ be the inradius. Joining the incentre $I$ to the three vertices gives triangles whose altitudes to the three side lines all equal $r$. Their areas add to $\Delta=r(10+17+21)/2=rs$. Hence $r=84/24=7/2$.

Let the incircle touch $AB,AC,BC$ at $T,U,V$, respectively. Tangent segments from the same external point are equal, so put $AT=AU=t$. Then $BT=BV=10-t$ and $CU=CV=17-t$. Adding along $BC$ gives $$(10-t)+(17-t)=21,\qquad t=3.$$ A radius to a tangency point is perpendicular to the tangent, so $IT\perp AB$. In right triangle $AIT$, Pythagoras now gives $$AI^2=AT^2+IT^2=3^2+\left(\frac72\right)^2=\frac{85}{4}.$$ Therefore $4AI^2=85$.

Answer / conclusion: 85

Review the idea: Cyclic and tangential quadrilaterals · Trigonometry in geometry

Count each triangle at its right-angle vertex; a triangle has only one right angle, so this avoids double counting. The grid has four corners, four side midpoints and one centre.

At a corner, all rays towards other grid points lie in a single $90^\circ$ sector. Two can be perpendicular only when they are the horizontal and vertical boundary rays. Each ray has two possible endpoints, giving $2\cdot2=4$ triangles per corner.

At a side midpoint, the horizontal/vertical pairs give $2\cdot2=4$ triangles: there are two choices along the inward perpendicular and one on each side along the boundary. The two inward $45^\circ$ diagonal rays give one further triangle. For example, from $(1,0)$ these go to $(0,1),(2,1)$. The only remaining rays from $(1,0)$ go to $(0,2)$ and $(2,2)$: each moves one unit horizontally and two units vertically. A $90^\circ$ turn swaps these horizontal and vertical distances, so a perpendicular direction would require twice as much horizontal movement as vertical movement. Any nonzero vertical movement between grid points is at least one unit, so this would require at least two horizontal units. But from $(1,0)$, every grid point is at most one horizontal unit away. Hence neither remaining ray has a perpendicular partner in the grid. Thus there are exactly 5 triangles per side midpoint.

At the centre, there are four axial neighbours and four diagonal neighbours, one on each available ray. In each group of four rays, the perpendicular unordered pairs are the four adjacent pairs around the centre. This gives four axial pairs and four diagonal pairs. No axial ray is perpendicular to a diagonal ray, so the centre contributes 8. The total is $$4\cdot4+4\cdot5+8=44.$$

Answer / conclusion: 44

Review the idea: Counting

Let $a$ be the number of 2s and $b$ the number of 3s. Then $2a+3b=30$, with $a,b\ge0$. Consequently $b$ is even, $0\le b\le10$, and $a=(30-3b)/2$. The sequence has $a+b=15-b/2$ terms.

For fixed $b$, choose which $b$ positions contain 3; all other positions contain 2. There are $\binom{15-b/2}{b}$ choices. For $b=0,2,4,6,8,10$, the counts are $$\binom{15}{0},\binom{14}{2},\binom{13}{4},\binom{12}{6},\binom{11}{8},\binom{10}{10}.$$ Therefore $N=1+91+715+924+165+1=1897$, whose remainder modulo 100 is 97.

Answer / conclusion: 97

Review the idea: Combinations and binomial coefficients