#### Number Theory

**Q1** Two positive integers \(m\) and \(n\) exist such that \(2^n + 3^m\) is divisible by 5. Prove that \(2^m + 3^n\) is also divisible by 5.

**Q2** Find all integers \(x\), \(y\) and \(z\) such that

\[x^2 + y^2 + z^2 = 2(yz + 1)\]

\[x + y + z = 4018\]

**Q3** Find all solutions in positive integers to the equation

\[m! + 5 = n^3\]

**Q4** Find all non-negative integers \(n\) such that \(2^{200} + 2^n - 2^{103} + 16\) is a perfect square.

**Q5** Let \(n\) be a positive integer. Prove that there exists a multiple of \(n\), the sum of whose digits is equal to \(n\).

#### Geometry

**Q1** Let \(M\) be any point on the line passing through vertices \(B\) and \(C\) of triangle \(ABC\). A circle passing through A is tangent to side \(BC\) of triangle \(ABC\) at point \(M\) and intersects sides \(AB\) and \(AC\) at points \(D\) and \(E\) respectively.

Prove that \(DE\) is parallel to \(BC\) if and only if \(M\) is an internal or external bisector \(\angle BAC\).

**Q2** Points \(A\), \(B\), \(C\), \(D\) and \(E\) lie in that order on a circle and have the property the lines \(AB\) and \(ED\) are parallel.

Prove that \(\angle ABC = 90^{\circ}\) if and only if \(AC^2 = BD^2 + CE^2\)

**Q3** Triangle \(ABC\) has centroid \(G\). Let \(D\) be the midpoint of \(AC\). The line through \(G\) parallel to \(BC\) intersects \(AB\) at \(E\).

Prove that \(\angle AEC = \angle DGC\) if and only if \(BD^2 + 3CD^2 = AB^2\).

**Q4** Let \(ABC\) be a non-equilateral triangle with incentre \(I\) and centroid \(G\).

Prove that \(BC + AC = 2AB\) if and only if \(IG\) is parallel to \(AB\).

**Q5** Two circles of different radius having centres \(B\) and \(C\) are externally tangent at \(A\). A common tangent (not through \(A\)), touches the first circle at \(D\) and the second one at \(E\). The line through \(A\) which is perpendicular to \(DE\), and the perpendicular bisector of \(BC\) intersect at \(F\).

Prove that \(BC = 2AF\).

#### Algebra

**Q1** Positive real numbers \(a\), \(b\) and \(c\) satisfy \(a + b + c = 1\). Prove that

\[(1 - a)^2 + (1 - b)^2 + (1 - c)^2 > 6 \sqrt {abc}\]

**Q2** Find all functions \(f : \mathbb {R}^{+} \rightarrow \mathbb {R}^{+}\) such that

\[f(xy) f \left(\dfrac {1}{x} + \dfrac {1}{y} \right) = f(x) + y\]

for all \(x, y \in \mathbb{R}^{+}\).

**Q3** Consider the quartic polynomial \(p(x) = x^4 - 4x^3 + Bx^2 - Cx + D\) where \(B\), \(C\) and \(D\) are positive real numbers.

*(a)* Prove that if \(B \geq 6\), then the equation \(p(x) = 0\) cannot have 4 real roots.

*(b)* Prove that if \(36D \geq B^2\), then the equation \(p(x) = 0\) cannot have 4 distinct positive real roots.

**Q4** For each pair of real numbers \(a\), \(b\), let

\[M(a, b) = \max{(3a^2 + 2b, 3b^2 + 2a)}\]

What is the minimum possible value that \(M(a, b)\) can achieve as \(a\), \(b\) are allowed to range over all real numbers?

**Q5** Let \(x\) be any real number. Prove that \(x\) is an integer if and only if

\[\lfloor {x} \rfloor + \lfloor {2x} \rfloor + \lfloor {3x} \rfloor + \ldots + \lfloor {nx} \rfloor = \dfrac {n (\lfloor {x} \rfloor + \lfloor {nx} \rfloor)}{2}\]

holds for all positive integers \(n\).

(Here \(\lfloor {a} \rfloor\) denotes the greatest integer not exceeding the real number \(a\).)

#### Combinatorics

**Q1** Finn likes nine digit positive integers which do not contain the same digit more than once, and which do not contain the digit zero at all.

What is the sum of all the 9 digit numbers Finn likes?

(Express your answer as an actual number, not an uncomputed formula.)

**Q2** Let \(n\) be a positive integer. Krishna is studying the subsets of \(S = {1, 2, \ldots, n}\). Find a simple formula for:

*(a)* The number of subsets of S which contain an odd number of elements.

*(b)* The number of subsets of S the sum of whose elements is an odd number.

**Q3** Let \(n\) be a positive integer. A class has \(2n\) students in it, and no two people have the same height. Principal Calvin wishes to arrange these \(2n\) students in two rows for a class photo. He wants two rows of \(n\) students, one row directly in front of the other, so that each front row student is shorter than the corresponding back row student.

In how many ways can Calvin organise this?

**Q4** Satvik attempts all six questions on an exam. Each question is given an integral mark out of 10. It turns out that Satvik never scores more in a later question than in any earlier question.

How many different possible sequences of six marks could Satvik achieve subject to the above question.

**Q5** Sharky would like to decorate the surface of a cube of side length 4 with some \(1 \times 3\) rectangular strips of paper. Any such decoration must be neat. This means that if each face of the cube is seen as being composed of 16 unit squares in the natural way, and if each rectangular strip is seen as being composed of three unit squares in the natural way, then each unit square of any such rectangular strip must exactly cover a unit square on the surface of the cube. It is permitted to bend such a strip over the edge of the cube. But it is not permitted to have a rectangular strip overlapping another such strip.

*(a)* What is the maximum number of strips that Sharky can use?

*(b)* Suppose that Sharky only wants to cover three mutually adjacent faces. What is the maximum number of strips that Sharky can use?

#### Harder Mixed

**Q1** Find all right angled triangles whose side lengths are positive integers and whose area is equal to its perimeter.

**Q2** Let \(O\) be the circumcentre of acute triangle \(ABC\). Points \(P\) and \(Q\) are located on sides \(AC\) and \(AB\), respectively such that triangle \(ABC\) and \(BPQ\) are similar.

Prove that \(O\) is the orthocentre of triangle \(BPQ\).

**Q3** Find all function \(f : \mathbb {R} \rightarrow \mathbb {R}\) which satisfy

\[f(x) f(y) = f(x + y) + xy\]

for all \(x, y \in \mathbb {R}\).

**Q4** Around a circular table are seated \(2n\) people, where \(n\) is a positive integer. After a break, they again sit around the circular table in a different order.

Prove that there are at least two people such that the number of participants sitting between them before and after the break is the same.

**Q5** From the set of natural numbers \(1, 2, \ldots, n\), four consecutive even integers are removed. The remaining integers have an average value of \(51 \frac {9}{16}\).

Determine all possible sets of four consecutive even integers whose removal creates this situation.

**Q6** Three schools have 200 students each. Every student has at least one friend in each school. (Friendships are mutual of course.) It is known that there exists a set \(E\) of 300 students among the 600 students such that for any school \(S\) an any two students \(x, y \in E\) which are not in the school \(S\), the number of friends in \(S\) of \(x\) and \(y\) are different.

Show that one can find a student in each school such that they are all friends of each other.

**Q7** Let \(A_1\) be the centre of the square inscribed in an acute triangle \(ABC\) with two vertices of the square on side \(BC\). Thus one of the two remaining vertices of the square on side \(AB\) and the other is on \(AC\). Points \(B_1, C_1\) are defined in a similar way for inscribed squares with two vertices on sides \(AC\) and\(AB\), respectively.

Prove that lines \(AA_1, BB_1, CC_1\) are concurrent.

**Q8** Determine all ordered triples \((x, y, z)\) of positive rational numbers for which all three of

\[x + \dfrac {1}{y}, y + \dfrac {1}{z}, z + \dfrac {1}{x}\]

are integers.

**Q9** A cube of side \(30\) contains \(999\) points in its interior, no four of which are coplanar. Amongst the \(1007\) points which include the 999 points as well as the eight vertices of the cube, prove that there exists a tetrahedron of volume no more than 9 units.

**Q10** A circle passes though points \(B\) and \(C\) of triangle \(ABC\) and intersect the lines \(AB\) and \(AC\) in \(D\) and \(E\) respectively. The projections of the points \(B\) and \(E\) onto \(CD\) are denoted by \(B_1\) and \(E_1\), respectively. The projections of the points \(D\) and \(C\) onto \(BE\) are denoted by \(D_1\) and \(C_1\), respectively.

Prove that the points \(B_1, D_1, E_1, C_1\) are concyclic.

**Q11** Let \(n > 1\) be a given positive integer and let \(f(x) = a_0 + a_1 x + \ldots + a_m x^m\), with \(m \geq 2\), be a polynomial with integer coefficients such that

*(a)* \(a_2, a_3, \ldots, a_m\) are divisible by all prime factors of \(n\),

*(b)* \(a_1\) and \(n\) are relatively prime.

Prove that there exists a positive integer \(c\), such that \(f(c)\) is divisible by \(n^{2014}\)

**Q12** Find a function \(f : \mathbb {R} \rightarrow \mathbb {R}\) satisfying

\[f(xy)(f(x) - f(y)) = (x - y)f(x) f(y)\]

for all \(x, y \in \mathbb {R}\) as well as \(f(2001) = 0\) and \(f(2002) = 2002\).

## Comments

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TopNewest## Number Theory

Q1Residues of \(2^n\) and \(3^n\) both have cycles of length \(4\), as shown in the table below. Here is a list of \(n\), with \(2^n, 3^n\) residues \(\text{mod }5\):

\[ \begin{array}{c|c|c} n&2^n&3^n \\ \hline 0&1&1\\ 1&2&3\\ 2&4&4\\ 3&3&2\\ \hline 4&1&1\\ \vdots &\vdots &\vdots \end{array} \] (the fact that \(2^0\equiv 3^0\equiv 2^4\equiv 3^4\equiv 1\pmod{4}\) shows that \(2^n, 3^n\) both cycle with length \(4\).)

Thus:

If \(n\equiv 0\pmod{4}\), then \(m\equiv 2\pmod{4}\)

If \(n\equiv 1\pmod{4}\), then \(m\equiv 1\pmod{4}\)

If \(n\equiv 2\pmod{4} \), then \(m\equiv 0\pmod{4}\)

If \(n\equiv 3\pmod{4}\), then \(m\equiv 3\pmod{4}\)

Because \(m\) and \(n\) are symmetric (we can switch them around and it will still give a true statement), combining the first and third statements, give (in addition to the second and fourth statements individually):

\[n\equiv 0\pmod{4}\iff m\equiv 2\pmod{4}\]

\[n\equiv 2\pmod{4}\iff m\equiv 0\pmod{4}\]

\[n\equiv 1\pmod{4}\iff m\equiv 1\pmod{4}\]

\[n\equiv 3\pmod{4}\iff m\equiv 3\pmod{4}\]

Thus, \(5\mid 2^n+3^m\) if and only if \(5\mid 2^m+3^n\) \(\Box\)

Q2\[x^2+y^2+z^2=2(yz+1)\iff x^2+(y-z)^2=2\]

Since \(x,y,z\in \mathbb{Z}\) then \(x=\pm 1\) and \(y-z=\pm 1\)

Case 1:\(x=1.y-z=1\)Then \(y+z=4017\) and \(y-z=1\) gives \[(x,y,z)=(1, 2009, 2008)\]

Case 2:\(x=-1. y-z=1\)Then \(y+z=4019\) and \(y-z=1\) gives \[(x,y,z)=(-1, 2010, 2009)\]

Case 3:\(x=1. y-z=-1\)Then \(y+z=4017\) and \(y-z=-1\) gives \[(x,y,z)=(1, 2008, 2009)\]

Case 4:\(x=-1. y-z=-1\)Then \(y+z=4019\) and \(y-z=-1\) gives \[(x,y,z)=(-1, 2009, 2010)\]

\(\Box\)

Q3We know that \(n^3\equiv -1, 0, 1\pmod{9}\)

If \(m\ge 9\), then \(m!+5\equiv 5\not\equiv n^3\pmod{9}\)

Thus, \(m\le 8\). We just have to check all integers \(m=1\to 8\) for solutions. It turns out that the only solution out of these is \[(m,n)=(5,5)\] \(\Box\)

Q4\(2^{200}+2^n-2^{103}+16=(2^{100}-2^2)^2+2^n=x^2\) for some positive integer \(x\).

Thus, \[2^n=(x+2^{100}-2^2)(x-2^{100}+2^2)\] Thus, integers \(a,b\) satisfy \(a+b=n\), \(a > b\), and \[2^a=x+2^{100}-2^2\] \[2^b=x-2^{100}+2^2\]

Thus \[2^a-2^b=2^{101}-2^3\]

Thus, \(b=3\), \(a=101\), and \(n=\boxed{104}\) \(\Box\) – Daniel Liu · 2 years, 3 months ago

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– Abdur Rehman Zahid · 2 years, 3 months ago

Daniel Liu in Q2 3rd case,shouldn't i be 2008 instead of 208?Log in to reply

– Daniel Liu · 2 years, 3 months ago

fixed, thanks.Log in to reply

– Sharky Kesa · 2 years, 3 months ago

Also, Cases 2 - 4 are labelled incorrectly.Log in to reply

CombinatoricsQ1.Let us fix the unit digit of our number to be \(1\). Then, we can arrange the remaining digits in the remaining \(8\) places in \(8!\) ways. This way we can continue by fixing the last digit from \(1\) to \(9\).So, its value becomes \(8!(1+2+3+4+5+6+7+8+9)\) which is equal to \(1814400\).

But wait, we can do this for all places of digits and not for only the unit digit. So, for fixing it at the

tensplace, we have to multiply \(1814400\) by \(10\), athundredsplace by \(100\) and so on till the first place.Now, this becomes \[1814400\times (10^{8}+10^{7}+10^{6}+10^{5}+10^{4}+10^{3}+10^{2}+10^{1}+1)\] which is equal to \(201599999798400\) which is our required answer.

BTW Finn, why don't you like the number \(0\). All I can say to you is that Don't Underestimate the Power of Zero

Q2.\((i)\) We know that a set of \(n\) elements have a total number of \(2^{n}\) subsets including the Phi Subset which doesn't have any element. So, we have to select an odd number of elements from \(n\) elements to find out the total number of sets including an odd number of elements. This means:\[\dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}....\dbinom{n}{k}\] where \(n≥k\) and \(n-k\) is either \(0\) or \(1\).

This expression is equal to \(2^{n-1}\) and hence our required answer is \(2^{n-1}\) subsets.

Note:The Phi Subset will not be added to this list as it represents a set which doesn't contain any element. Phi is not an element of the set, it is just a symbol to represent an empty set. – Yash Singhal · 2 years, 3 months agoLog in to reply

Nice set of problems. Which ones did you create yourself?

(Edited after seeing your reply to Siddhartha) – Calvin Lin Staff · 2 years, 3 months ago

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## Created myself:

Number Theory:Q1, Q2, Q3

Geometry:

Q2, Q4, Q5

Algebra:

Q1, Q3, Q5

Combinatorics:

Q2, Q5

Harder Mixed:

Q1, Q3, Q5, Q8, Q9, Q12 – Sharky Kesa · 2 years, 3 months ago

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Number Theory Q5

The number

\(10^{a+b}.(10^{\varphi(t)}+10^{2.\varphi(t)}+...+10^{n.\varphi(t)})\) , where \(n=2^a.5^b.t\) obviously fits the criteria. – Bogdan Simeonov · 2 years, 3 months ago

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– Des O Carroll · 2 years, 3 months ago

please explain why it fits and perhaps you would illustrate when n=7 say thanksLog in to reply

– Bogdan Simeonov · 2 years, 3 months ago

Well, firstly see that this is a sum of n different powers of ten, thus the sum of its digits is n.Also, it obviously is divisible by 2^a and 5^b.Also, \(10^{\varphi(t)}\equiv1 \pmod{t}\), so the inner sum is \(\equiv n \pmod{t}\), but \(t|n\) so the number is divisible by n.Log in to reply

@Sharky Kesa Hey for algebra Q.2 I am getting ans as : f(x)=f(-1)f(0)+1/x for all x. But I am unable to find f(-1) and f(0). My solution is just the substitution of y= -1/x. – Shrihari B · 1 year, 4 months ago

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Q2. of Combinatorics part (ii) answer is 2^(n-1).

Doubt: Is 6,6,5,4,3,2 is a valid sequence for Q4. of Combinatirics? – Herman Pert · 2 years, 3 months ago

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I have a doubt regarding Q4. of the Combinatorics Section Sharky! Can Satvik be awarded \(0\) marks for a question? Also, can he be awarded equal marks in two or more questions. Thanks! – Yash Singhal · 2 years, 3 months ago

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– Sharky Kesa · 2 years, 3 months ago

0 marks is the minimum that can be awarded. I have also stated that he does not get more in a later question than in any earlier question.Log in to reply

Small typo:- You interchange between Finn and Tim in Q1 of Combinatorics.

Also, where do you get these questions from? – Siddhartha Srivastava · 2 years, 4 months ago

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– Sharky Kesa · 2 years, 4 months ago

Some of these question I found in some papers which I found quite interesting. Others I make myself.Log in to reply

a square and an eqilateral triangle have the same perimeter. if the area of triangle is 16√3, what is the area of square – Oladayo Daniel · 1 year, 4 months ago

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Please solve this: (1001-ab)/(a+b) =c where a,b and c are positive integers and a<b<c. Find how many ordered triples satisfy the condition! Please help! – Herman Pert · 2 years, 3 months ago

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– Sharky Kesa · 2 years, 3 months ago

Do not litter my comments section with irrelevant comments. Delete this and post it as a note if you want it answered.Log in to reply

– Herman Pert · 2 years, 3 months ago

But first tell me whether 6,6,5,4,3,2 is a valid sequence for Q4. of Combinatirics section or not. And get chilled!Log in to reply

– Sharky Kesa · 2 years, 3 months ago

I just don't like it when irrelevant comments are posted. Also, yes, it is allowed.Log in to reply

@Sharky Kesa Solutions please. – Satvik Golechha · 2 years, 4 months ago

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– Sharky Kesa · 2 years, 4 months ago

Dude, you have to figure them out yourself. That's the point of the note.Log in to reply