IOQM Mock Paper 1: 30 Questions and Solutions

We have $360=2^3 3^2 5$. A square divisor has exponent 0 or 2 on each of 2 and 3, and exponent 0 on 5. Thus there are $2\cdot2\cdot1=4$ choices.

Answer / conclusion: 04

Review the idea: Counting and summing divisors

Expanding gives $(a+b)^2=a^2+b^2+2ab$. Hence $a^2+b^2=81-28=53$. The given conditions are consistent, for example with $a=7,b=2$.

Answer / conclusion: 53

Review the idea: Algebraic identities

The ten pairs are $\{1,20\},\{2,19\},\ldots,\{10,11\}$. Selecting 11 numbers forces both members of one pair. Selecting only $1,2,\ldots,10$ avoids a sum of 21, so 11 is smallest.

Answer / conclusion: 11

Review the idea: Pigeonhole principle

Triangles $ABD$ and $ADC$ have a common altitude from $A$ to line $BC$. Thus their areas have ratio $2:1$. The required area is $54\cdot2/3=36$.

Answer / conclusion: 36

Review the idea: Triangle area ratios

As $7^2=49$ and $7^4=2401\equiv1\pmod{100}$, we have $7^{20}=(7^4)^5\equiv1\pmod{100}$.

Answer / conclusion: 01

Review the idea: Remainders

Each vertex has seven non-neighbouring vertices, giving $10\cdot7$ endpoint counts. Each diagonal has two endpoints, so there are $70/2=35$ diagonals.

Answer / conclusion: 35

Review the idea: Counting

The residues of $p,p+10,p+14$ cover all three residue classes modulo 3. A prime divisible by 3 equals 3. Since $p+10$ and $p+14$ exceed 3, we must have $p=3$. It works because 13 and 17 are prime.

Answer / conclusion: 03

Review the idea: Prime numbers

We get $x^2+2+x^{-2}=9$, so the answer is 7. Such real numbers exist because $x^2-3x+1=0$ has positive discriminant.

Answer / conclusion: 07

Review the idea: Algebraic identities

Writing $\angle B=t$, the angle sum gives $2t+t+3t=180^\circ$, so $t=30^\circ$.

Answer / conclusion: 30

Review the idea: Angles

If the final digit is 0, the other digits have $3!=6$ arrangements. If it is 2, there are two nonzero choices for the first digit and two orders for the middle digits, giving 4. The total is $6+4=10$.

Answer / conclusion: 10

Review the idea: Permutations and arrangements

The equation implies both $x,y\gt 12$. It rearranges to $(x-12)(y-12)=144$. Since $144=2^4 3^2$ has 15 positive divisors, its unordered factor pairs number $(15+1)/2=8$; the square pair is included once. Every pair gives a valid solution.

Answer / conclusion: 08

Review the idea: Factorisation integer solutions

Let $r$ be the inradius, the radius of the circle tangent to all three sides. Pythagoras gives the hypotenuse $\sqrt{9^2+12^2}=15$. The area is $\Delta=9\cdot12/2=54$, and the semiperimeter (half the perimeter) is $s=(9+12+15)/2=18$.

Join the incentre $I$ to the three vertices. This divides the triangle into three smaller triangles with bases 9, 12 and 15. Each has altitude $r$, since the perpendicular distance from $I$ to each side is a radius. Thus $$\Delta=\frac{9r}{2}+\frac{12r}{2}+\frac{15r}{2}=rs=18r.$$ Therefore $r=54/18=3$.

Answer / conclusion: 03

Review the idea: Triangle area ratios

Write the selected elements in increasing order as $a<b<c$. No two being consecutive means $b\ge a+2$ and $c\ge b+2$. Remove one compulsory gap before the second element and two before the third: the new numbers are $a,b-1,c-2$, and $$1\le a<b-1<c-2\le5.$$

Conversely, take any increasing triple from $1,2,\ldots,5$ and add 0, 1 and 2 to its successive entries. The resulting entries lie between 1 and 7 and differ by at least 2. These two operations undo each other, so the answer is the number of three-element subsets of five elements: $\binom53=10$.

Answer / conclusion: 10

Review the idea: Counting with bijections

For every $n\ge4$, $n!$ contains $4!=24$ as a factor and is divisible by 12. Thus the remainder is that of $1+2+6=9$.

Answer / conclusion: 09

Review the idea: Factorials

We need $(x-y)(x+y)=45$. Both factors are positive odd integers, with the first smaller. Their possibilities are $(1,45),(3,15),(5,9)$, giving $(x,y)=(23,22),(9,6),(7,2)$. Thus there are 3 pairs.

Answer / conclusion: 03

Review the idea: Factorisation integer solutions

Choose two of the three rows and two of the four columns. Each choice gives one distinct rectangle and all such rectangles arise this way. The answer is $\binom32\binom42=3\cdot6=18$.

Answer / conclusion: 18

Review the idea: Combinations and binomial coefficients

Let $O$ be the centre, $AB$ the chord, and $M$ the perpendicular foot from $O$ to $AB$. The perpendicular from a circle’s centre bisects a chord: the right triangles $OMA,OMB$ have equal hypotenuses $OA=OB$ and common leg $OM$, so their other legs are equal. Write $AM=MB=h$.

In right triangle $OMA$, the radius is $OA=13$ and $OM=5$. Pythagoras gives $$h^2=13^2-5^2=169-25=144.$$ Since $h>0$, $h=12$, and the whole chord is $AB=2h=24$.

Answer / conclusion: 24

Review the idea: Circles and power of a point

We have $a_n+1=2\cdot3^{n-1}$. Thus $a_4=2\cdot27-1=53$. Directly the terms are 1, 5, 17, 53.

Answer / conclusion: 53

Review the idea: First order linear recurrences

Since $30=2\cdot3\cdot5$, an integer is relatively prime to 30 exactly when it is divisible by none of 2, 3 and 5. In $1,2,\ldots,30$, the eligible values are $$1,7,11,13,17,19,23,29.$$ Adding 30 preserves divisibility by each of 2, 3 and 5. Thus each block $1\text{–}30$, $31\text{–}60$, $61\text{–}90$ contributes eight values, giving 24.

In the remaining block $91\text{–}100$, the even numbers are excluded; 93 and 99 are divisible by 3, and 95 is divisible by 5. Only 91 and 97 qualify. The total is $24+2=26$.

Answer / conclusion: 26

Review the idea: Complete and reduced residue systems

For $2\le x\le8$, the sum is 6, so there is no solution there. For $x\lt 2$, $10-2x=10$ gives $x=0$. For $x\gt 8$, $2x-10=10$ gives $x=10$. Their sum is 10.

Answer / conclusion: 10

Review the idea: Absolute value inequalities

The product $n!$ contains all integers from 1 to $n$. Each multiple of 7 supplies one factor of 7; a multiple of $49=7^2$ supplies another, and higher powers supply further factors in the same way.

Up to 48, the only multiples of 7 are $7,14,21,28,35,42$. None is divisible by 49, so $48!$ contains exactly six factors of 7. Every smaller factorial contains at most these six, and therefore is not divisible by $7^8$. At 49, the new factor $49=7^2$ supplies two more, giving eight. Hence the smallest possible $n$ is 49.

Answer / conclusion: 49

Review the idea: Factorials

Repeatedly remove factors of 2 from each number until an odd number remains; call it the odd part. This process is unique. Group numbers having the same odd part. For example, odd part 3 gives the group $3,6,12,24$. In any one group every smaller number divides every larger one, since their ratio is a power of 2. Hence at most one member of each group may be selected.

The possible odd parts are $1,3,5,\ldots,29$, fifteen values, so the subset has size at most 15. The fifteen numbers $16,17,\ldots,30$ attain the bound: a larger distinct multiple of any selected number is at least twice it, hence at least 32 and outside the set. The greatest size is 15.

Answer / conclusion: 15

Review the idea: Pigeonhole principle

Write $R$ for a right step and $U$ for an up step. Any unrestricted path uses four of each, so choosing the positions of the four right steps gives $\binom84=70$ paths.

Call a path bad if it visits $y<x$. At its first such visit, it reaches $x=y+1$. Its prefix up to that point contains exactly one more $R$ than $U$. Swap $R$ and $U$ in this prefix only. The total number of right steps decreases from 4 to 3, while the number of up steps increases from 4 to 5. The resulting path therefore ends at $(3,5)$.

This operation is reversible. A path ending at $(3,5)$ must have a first point on $y=x+1$, since $y-x$ begins at 0, ends at 2 and changes by 1 per step. Swapping its prefix up to that first point recovers a path to $(4,4)$ whose first bad point has $x=y+1$. Thus bad paths are in one-to-one correspondence with all paths to $(3,5)$, of which there are $\binom83=56$. The required number is $70-56=14$.

Answer / conclusion: 14

Review the idea: Counting with bijections

The semiperimeter is 21. Heron gives $\Delta=\sqrt{21\cdot8\cdot7\cdot6}=84$. The extended sine rule yields $R=13\cdot14\cdot15/(4\cdot84)=65/8$. Therefore $8R=65$.

Answer / conclusion: 65

Review the idea: Trigonometry in geometry

AM–GM gives $xyz\le((x+y+z)/3)^3$. A sum at most 12 would force $xyz\le64$, contrary to 72. Thus the integer sum is at least 13. The triple $(3,4,6)$ has product 72 and sum 13, so the bound is attained.

Answer / conclusion: 13

Review the idea: Arithmetic geometric and harmonic means

Since $(r-1)(r+1)$ is divisible by a prime only if one factor is, the choices are $r\equiv\pm1\pmod7$ and $r\equiv\pm1\pmod{13}$. The Chinese remainder theorem gives exactly one residue modulo 91 for each of the four choices. They are 1, 27, 64 and 90.

Answer / conclusion: 04

Review the idea: Number theory theorems

Write the equation as $y^2-3xy+x^2-1=0$. If $x=1$, this reduces to $y(y-3)=0$, so positivity forces $y=3$.

For $x>1$, regard it as a quadratic in $y$. Its two roots sum to $3x$ and multiply to $x^2-1$. Since one root is $y$, the other is $$y'=3x-y=\frac{x^2-1}{y}.$$ The first expression is an integer; the second, together with $y\ge x$, gives $$0<y'\le\frac{x^2-1}{x}<x.$$ As $y'$ is also a root, $(y',x)$ satisfies the same equation. Its larger coordinate is smaller than the larger coordinate of $(x,y)$. Repeatedly descending cannot continue forever through positive integers, so it reaches $(1,3)$.

Given the smaller pair $(u,v)=(y',x)$, the previous pair is uniquely $(v,3v-u)$. Hence reversing the descent produces the single chain $$(1,3),(3,8),(8,21),(21,55),\ldots.$$ The replacement preserves the equation by the same root relation. Also, if $y\ge x>0$, then $3y-x\ge2y>y$, so the entries and pair sums strictly increase along the chain. The first sums are 4, 11, 29 and 76. Exactly the first three are at most 50, giving 3 pairs.

Answer / conclusion: 03

Review the idea: Factorisation integer solutions

Let $P=AD\cap BE$, $Q=BE\cap CF$, and $R=CF\cap AD$. Write $[XYZ]$ for the area of triangle $XYZ$, and put $t=[ABP]$.

Triangles $ABE,CBE$ have bases $AE,CE$ on the same line and the same altitude from $B$, so $[ABE]:[CBE]=AE:CE=1:2$. Using $BE$ as their common base shows that the perpendicular distances of $A,C$ from line $BE$ are also in ratio $1:2$. Triangles $ABP,CBP$ have common base $BP$ on this line and those same respective altitudes. Consequently $[CBP]=2t$.

Now $BD:BC=2:3$. Triangles $BDP,BCP$ share their altitude from $P$ to line $BC$, so $$[BDP]=\frac23[BCP]=\frac{4t}{3}.$$ Likewise $[ABD]=\frac23[ABC]=14$. Since $P$ lies on segment $AD$, triangle $ABD$ is split into $ABP$ and $BDP$. Thus $$t+\frac{4t}{3}=14,\qquad t=6.$$

Repeat this reasoning after cyclically relabelling $A,B,C$; the three given side ratios are identical. It gives $[BCQ]=[CAR]=6$. The three outer triangles $ABP,BCQ,CAR$ and the central triangle $PQR$ have disjoint interiors and together fill $ABC$, as their drawn boundaries show. Hence $$[PQR]=21-6-6-6=3.$$

Answer / conclusion: 03

Review the idea: Triangle area ratios

Let $F_n$ count binary strings in a row of length $n$ with no adjacent 1s. Set $F_0=1$, for the empty string, and $F_1=2$. For $n\ge2$, a string starting with 0 may be followed by any valid string of length $n-1$. If it starts with 1, the next digit must be 0, followed by any valid string of length $n-2$. These disjoint cases give $F_n=F_{n-1}+F_{n-2}$.

Successive values $F_0,\ldots,F_8$ are $$1,2,3,5,8,13,21,34,55.$$ A valid row string fails the circular condition exactly when both endpoints are 1. Their neighbours must then be 0, leaving four middle positions subject only to the ordinary row rule. There are $F_4=8$ such strings. Thus the labelled circular fillings number $55-8=47$.

Answer / conclusion: 47

Review the idea: Counting with recurrences

Suppose the expression equals $k^2$, with $k>0$. Completing the square gives $$k^2=(n+10)^2+75,\qquad (k-n-10)(k+n+10)=75.$$ Since $n>0$, we have $k>n+10$. Put $u=k-n-10$, $v=k+n+10$. Then $u,v$ are positive, $u<v$, and $uv=75$, so both are odd. Conversely, a positive odd factor pair gives integers $$n=\frac{v-u}{2}-10,\qquad k=\frac{u+v}{2}.$$

The positive factor pairs with $u<v$ are $(1,75),(3,25),(5,15)$. They give $n=27,1,-5$, respectively. Only 27 and 1 are positive. They work: the expression is $1444=38^2$ for $n=27$, and $196=14^2$ for $n=1$. Thus there are 2 possible positive integers.

Answer / conclusion: 02

Review the idea: Factorisation integer solutions