IOQM Mock Paper 3: 30 Questions and Solutions

Since $60=2^2\cdot3\cdot5$, the allowed divisors have the form $2^a5^b$, with $0\le a\le2$ and $0\le b\le1$. There are $3\cdot2=6$.

Answer / conclusion: 06

Review the idea: Counting and summing divisors

Composition gives $f(f(x))=a^2x+ab+b$. Thus $a^2=4$; positivity yields a=2. Then $3b=9$, so b=3 and $f(1)=5$.

Answer / conclusion: 05

Review the idea: Functions inverses and composition

The first section uses two right steps and one up step, giving 3 choices. The second uses two of each kind, giving 6 choices. Their concatenations are distinct, so the answer is $3\cdot6=18$.

Answer / conclusion: 18

Review the idea: Combinations and binomial coefficients

Let $s$ be the side length and $h=3\sqrt3$ the altitude. In an equilateral triangle the altitude bisects the opposite side: the two right triangles have equal hypotenuses and a common altitude, so their half-bases are equal. Applying Pythagoras in either half gives $$s^2=h^2+(s/2)^2,\qquad \frac34s^2=27.$$ Hence $s^2=36$, and positivity gives $s=6$. The perimeter is $3s=18$.

Answer / conclusion: 18

Review the idea: Pythagoras and stewart

First find powers that leave remainder 1: $2^3=8\equiv1\pmod7$, and $3^6=729=7\cdot104+1\equiv1\pmod7$. Therefore $$2^{30}=(2^3)^{10}\equiv1,\qquad 3^{30}=(3^6)^5\equiv1\pmod7.$$ Their sum is congruent to 2, and $0\le2<7$, so the remainder is 2.

Answer / conclusion: 02

Review the idea: Remainders

There are $\binom72=21$ pairs in total. Exactly 6 have difference 1 and 5 have difference 2. These disjoint excluded cases leave $21-6-5=10$.

Answer / conclusion: 10

Review the idea: Counting

The midpoint of the hypotenuse is equidistant from all three vertices, so the circumradius is 7. The area is $\pi\cdot7^2=49\pi$; hence k=49.

Answer / conclusion: 49

Review the idea: Circles and power of a point

As $(a-b)^2\ge0$, we have $144-4ab\ge0$, so $ab\le36$. Equality occurs at $a=b=6$, which is allowed.

Answer / conclusion: 36

Review the idea: Sum of squares

Use $\binom{32}{16}=32!/(16!)^2$. To count the factors of 2 in $32!$, each of the sixteen even factors supplies one; each of the eight multiples of 4 supplies an additional one; the four multiples of 8, two multiples of 16 and one multiple of 32 supply further factors. This accounts for every factor of 2, giving $16+8+4+2+1=31$.

Similarly, $16!$ contains $8+4+2+1=15$ factors of 2. Dividing by its square subtracts twice this exponent. The required exponent is $31-2\cdot15=1$.

Answer / conclusion: 01

Review the idea: Factorials

A four-digit palindrome has the form $abba$, where $a$ is the first digit and $b$ the second. Thus $1\le a\le9$, $0\le b\le9$, and the digit sum is $2(a+b)$. Divisibility by 9 is equivalent to the digit sum being divisible by 9. Since 2 and 9 are coprime, this is equivalent to $9\mid a+b$.

The range $1\le a+b\le18$ leaves only sums 9 and 18. For sum 9, each $a=1,\ldots,9$ gives one allowable $b=9-a$, so there are nine choices. For sum 18, only $(a,b)=(9,9)$ works. Each pair specifies one palindrome, giving $9+1=10$.

Answer / conclusion: 10

Review the idea: Number bases and digit problems

The hypotenuse is $b=\sqrt{12^2+12^2}=12\sqrt2$, and the triangle’s area is $12\cdot12/2=72$. If $H$ is its perpendicular height to the hypotenuse, then $bH/2=72$, so $H=6\sqrt2$.

Let the square side be $s$. The opposite side of the square is parallel to the hypotenuse and lies at perpendicular distance $s$ from it. The smaller triangle above that side has height $H-s$ and is similar to the original triangle, since their corresponding angles agree. Its base length is therefore $$b\frac{H-s}{H}=b-\frac bH s=12\sqrt2-2s.$$ This base is also a side of the square, so $s=12\sqrt2-2s$. Hence $3s=12\sqrt2$, $s=4\sqrt2$, and the square’s area is $s^2=32$.

Answer / conclusion: 32

Review the idea: Similar triangles

If n is even, $P(n)\equiv P(0)\equiv1\pmod2$. If n is odd, $P(n)\equiv P(1)\equiv1\pmod2$. Thus P(n) is always odd and cannot equal 0. There are no integer roots.

Answer / conclusion: 00

Review the idea: Polynomial functions

There are $2^6=64$ total subsets. Adding or removing one fixed element is a reversible pairing that changes parity of size. Exactly half, namely 32, have even size.

Answer / conclusion: 32

Review the idea: Counting with bijections

The angle-bisector theorem gives $BD/DC=AB/AC=2/3$. Since BD+DC=20, we have $BD=20\cdot2/5=8$.

Answer / conclusion: 08

Review the idea: Parallel lines and angle bisectors

Exactly three divisors requires prime factorisation $p^2$. The primes with $p^2\le50$ are 2, 3, 5 and 7, giving the four integers 4, 9, 25 and 49.

Answer / conclusion: 04

Review the idea: Counting and summing divisors

Telescoping yields $a_{10}=a_1+1+2+\cdots+9=1+9\cdot10/2=46$.

Answer / conclusion: 46

Review the idea: Sequences and sums

The factor pairs are $(1,24),(2,12),(3,8),(4,6)$. Their perimeters are 50, 28, 22 and 20. Thus the least is 20, attained by the 4-by-6 rectangle.

Answer / conclusion: 20

Review the idea: Triangle inequalities

The count is $4!-\binom413!+\binom422!-\binom431!+\binom440!=24-24+12-4+1=9$. Each placement with any correct letters is cancelled by inclusion–exclusion.

Answer / conclusion: 09

Review the idea: Derangements

Subtract a suitable integer multiple of the smaller expression from the larger: $$(2^{18}-1)-2^6(2^{12}-1)=2^6-1=63.$$ Subtracting an integer multiple preserves the common divisors in both directions. Thus the required greatest common divisor is $\gcd(2^{12}-1,63)$. But $$2^{12}-1=(2^6-1)(2^6+1)=63\cdot65,$$ so this greatest common divisor is 63.

Answer / conclusion: 63

Review the idea: Greatest common divisor

Triangle inequality gives $|x-1|+|x-10|\ge9$, and $|x-4|\ge0$. At x=4 the total is $3+0+6=9$, so the minimum is 9.

Answer / conclusion: 09

Review the idea: Absolute value inequalities

Three consecutive integers always include a multiple of 3. If n is odd, n-1 and n+1 are consecutive even integers; one is a multiple of 4, so their product is divisible by 8. All 50 odd values qualify. If n is even, both neighbours are odd, so the entire factor of 8 must come from n itself. There are $\lfloor100/8\rfloor=12$ such even values. The total is 62.

Answer / conclusion: 62

Review the idea: Divisibility

Count by the number of red edges. With zero red edges there are blue triangles. With one red edge, either of the triangles not containing that edge is blue. Interchanging the colours shows that five or six red edges also fail.

With two red edges, if they share a vertex, the other three vertices form a blue triangle. If they are disjoint, every triangle contains one of these red edges and two blue edges, so the colouring is valid. Fix one vertex: its red partner can be any of the other three, and the remaining two vertices must form the second red edge. This gives 3 colourings.

With three red edges, there are $\binom63=20$ choices. Four choices form a red triangle, one for each choice of three vertices. Another four choices consist of one vertex joined in red to the other three; choosing that central vertex gives four, and the other three vertices then form a blue triangle. These two cases are disjoint. They are also all the bad cases: a blue triangle leaves exactly its three complementary edges red, which are precisely the three edges from the remaining vertex. Hence $20-4-4=12$ work.

By interchanging the two colours, four red edges give the same count as two red edges, namely 3. The total is $3+12+3=18$.

Answer / conclusion: 18

Review the idea: Counting

Taking x=y=0 gives f(0)=0. With y=1, $f(n+1)=2f(n)+6-f(n-1)$. Starting from 0 and 3, the next values are 12, 27, 48 and 75. Therefore f(5)=75. Consistency is witnessed by $f(n)=3n^2$, which satisfies the identity.

Answer / conclusion: 75

Review the idea: Solving functional equations

Choose $P=(0,0)$, $A=(-2,0)$, $B=(8,0)$, $C=(0,-4)$, and $D=(0,4)$. A circle’s centre lies on the perpendicular bisector of each chord. Since $PC=PD$, the perpendicular bisector of $CD$ is the horizontal line $AB$. The midpoint of $AB$ has horizontal coordinate 3, so its perpendicular bisector is $x=3$. Therefore the centre is $O=(3,0)$, and the radius is $OA=5$.

A square with half-side $t\ge0$ has vertices $(\pm t,\pm t)$. The two left vertices, with horizontal coordinate $-t$, are farthest from $O$; their squared distance is $$(3+t)^2+t^2=2t^2+6t+9.$$ The disk contains the entire segment joining any two of its points. Thus, if all four square vertices belong to the disk, so do its sides and its interior, which can be filled by joining points on opposite sides.

For side 3, $t=3/2$, and the largest squared distance is $45/2<25$, so the square fits. For side 4, $t=2$, and that squared distance is $29>25$, so it does not fit. For nonnegative $t$, both $t^2$ and $t$ increase when $t$ increases; hence $2t^2+6t+9$ strictly increases. No larger integer side can fit either. The greatest integer side length is 3.

Answer / conclusion: 03

Review the idea: Circles and power of a point

Set $a=\lfloor\sqrt n\rfloor$ and $b=\lfloor\sqrt{100-n}\rfloor$, where the floor is the greatest integer not exceeding its argument. Then $$a^2\le n\le(a+1)^2-1,\qquad b^2\le100-n\le(b+1)^2-1.$$ We need $a+b=13$, while adding the lower bounds gives $a^2+b^2\le100$. We can bound $a$ without solving a quadratic inequality. If $a$ is 0, 1 or 2, then $b=13-a$ is at least 11, so $b^2>100$, impossible. If $a=3$, then $b=10$ and $a^2+b^2=109>100$. Thus $a\ge4$. Interchanging $a$ and $b$ gives $b\ge4$, so $a\le9$. The six values $a=4,5,6,7,8,9$ give $a^2+b^2=97,89,85,85,89,97$, respectively, all at most 100. These are therefore the only possible ordered pairs before the interval checks.

For each pair $(a,b)$, intersect the permitted interval $[a^2,(a+1)^2-1]$ with $[100-(b+1)^2+1,100-b^2]$. It suffices to calculate the first three pairs.

Replacing $n$ by $100-n$ swaps $a,b$ and is reversible. Thus the three reversed pairs contribute the same counts. The pairs are distinct because their sum is odd, and each $n$ determines a unique pair. The total is $2(4+11+12)=54$.

Answer / conclusion: 54

Review the idea: Floor and ceiling functions

We have $3^{10}=59049\equiv49\pmod{100}$, so $3^{20}\equiv49^2\equiv1\pmod{100}$. Also $3^4\equiv1\pmod{20}$, giving $3^9\equiv3\pmod{20}$. Therefore the desired power is congruent to $3^3=27\pmod{100}$.

Answer / conclusion: 27

Review the idea: Number theory theorems

Choose a term $x^{a_i}$ from the $i$-th factor, where $a_i$ is 0, 1 or 2. Multiplying the six choices gives $x^{a_1+\cdots+a_6}$, each with coefficient 1. Thus the desired coefficient counts ordered six-tuples with $$a_1+\cdots+a_6=8,\qquad 0\le a_i\le2.$$

Without upper bounds, represent the tuple by eight identical stars divided into six groups by five separators; empty groups are allowed. Each arrangement is exactly one tuple, so choosing the separator positions gives $\binom{13}{5}=1287$.

If a specified coordinate is at least 3, subtract 3 from it. The remaining sum is 5, counted by five stars and five separators: $\binom{10}{5}=252$. There are six choices of the violating coordinate. If two specified coordinates are at least 3, subtract 3 from each; the remaining sum is 2, giving $\binom75=21$. There are $\binom62=15$ such pairs. Three violations are impossible since they would have sum at least 9.

Subtract the single violations, then add back the double violations because those were subtracted twice. The answer is $$1287-6\cdot252+15\cdot21=90.$$

Answer / conclusion: 90

Review the idea: Binomial expansions and generating functions · Counting integer solutions

Let $k\ge2$ be the number of terms and $a\ge1$ the first. Then $$30=a+(a+1)+\cdots+(a+k-1)=ka+\frac{k(k-1)}2,$$ so $$a=\frac{60/k-k+1}{2}.$$ The smallest possible sum of $k$ positive consecutive terms is $1+\cdots+k=k(k+1)/2$. This is at most 30 only if $k\le7$.

For $k=2,3,4,5,6,7$, the formula gives, respectively, $$a=\frac{29}{2},\ 9,\ 6,\ 4,\ \frac52,\ \frac97.$$ Only $k=3,4,5$ give positive integers. They produce $9+10+11$, $6+7+8+9$, and $4+5+6+7+8$, each equal to 30. These exhaust every possible length, so the count is 3.

Answer / conclusion: 03

Review the idea: Factorisation integer solutions

A row with $r_i$ marks uses $\binom{r_i}{2}$ pairs of columns. If a column pair is used in two rows, those four marks form a forbidden rectangle. There are only $\binom42=6$ column pairs, so $\sum_{i=1}^4\binom{r_i}{2}\le6$.

If an allowed array had at least 10 marks, erasing extras would give an allowed array with exactly 10. Consider the row counts of such an array. If two counts satisfy $a\ge b+2$, transfer one unit from the larger count to the smaller. The pair sum changes by $$-(a-1)+b\le-1.$$ Therefore, among all nonnegative row counts summing to 10, the smallest pair sum occurs when no counts differ by more than 1. The counts are then $3,3,2,2$, giving $3+3+1+1=8>6$. This contradiction proves that at most 9 marks are possible.

To attain 9, use marked-column sets $\{1,2,3\},\{1,4\},\{2,4\},\{3,4\}$ in rows 1 through 4. Their column pairs are 12, 13, 23; 14; 24; and 34. None repeats, so no forbidden rectangle occurs. The mark count is $3+2+2+2=9$, proving the answer.

Answer / conclusion: 09

Review the idea: Pigeonhole principle

Let $a,b,c$ count the octagon edges along the three successive arcs between the chosen triangle vertices, following one fixed direction around the circle. All three are positive integers and $a+b+c=8$. Each arc-edge has measure $360^\circ/8=45^\circ$. By the inscribed-angle theorem, the opposite triangle angle is half its intercepted arc, hence $22.5^\circ$ times the corresponding gap.

All three angles are acute exactly when each gap is less than 4, that is, at most 3. The only three positive integers at most 3 summing to 8 are 2, 3 and 3. Choose the labelled starting vertex in eight ways, then choose which of the three successive gaps is 2 in three ways. Each actual triangle has been counted three times, once from each of its vertices; the direction around the circle was fixed throughout. Thus the answer is $8\cdot3/3=8$.

Answer / conclusion: 08

Review the idea: Circles and power of a point