Practise your Math Skills

Number Theory

Q1 Two positive integers mm and nn exist such that 2n+3m2^n + 3^m is divisible by 5. Prove that 2m+3n2^m + 3^n is also divisible by 5.

Q2 Find all integers xx, yy and zz such that

x2+y2+z2=2(yz+1)x^2 + y^2 + z^2 = 2(yz + 1)

x+y+z=4018x + y + z = 4018

Q3 Find all solutions in positive integers to the equation

m!+5=n3m! + 5 = n^3

Q4 Find all non-negative integers nn such that 2200+2n2103+162^{200} + 2^n - 2^{103} + 16 is a perfect square.

Q5 Let nn be a positive integer. Prove that there exists a multiple of nn, the sum of whose digits is equal to nn.


Geometry

Q1 Let MM be any point on the line passing through vertices BB and CC of triangle ABCABC. A circle passing through A is tangent to side BCBC of triangle ABCABC at point MM and intersects sides ABAB and ACAC at points DD and EE respectively.

Prove that DEDE is parallel to BCBC if and only if MM is an internal or external bisector BAC\angle BAC.

Q2 Points AA, BB, CC, DD and EE lie in that order on a circle and have the property the lines ABAB and EDED are parallel.

Prove that ABC=90\angle ABC = 90^{\circ} if and only if AC2=BD2+CE2AC^2 = BD^2 + CE^2

Q3 Triangle ABCABC has centroid GG. Let DD be the midpoint of ACAC. The line through GG parallel to BCBC intersects ABAB at EE.

Prove that AEC=DGC\angle AEC = \angle DGC if and only if BD2+3CD2=AB2BD^2 + 3CD^2 = AB^2.

Q4 Let ABCABC be a non-equilateral triangle with incentre II and centroid GG.

Prove that BC+AC=2ABBC + AC = 2AB if and only if IGIG is parallel to ABAB.

Q5 Two circles of different radius having centres BB and CC are externally tangent at AA. A common tangent (not through AA), touches the first circle at DD and the second one at EE. The line through AA which is perpendicular to DEDE, and the perpendicular bisector of BCBC intersect at FF.

Prove that BC=2AFBC = 2AF.


Algebra

Q1 Positive real numbers aa, bb and cc satisfy a+b+c=1a + b + c = 1. Prove that

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

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

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

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

Q3 Consider the quartic polynomial p(x)=x44x3+Bx2Cx+Dp(x) = x^4 - 4x^3 + Bx^2 - Cx + D where BB, CC and DD are positive real numbers.

(a) Prove that if B6B \geq 6, then the equation p(x)=0p(x) = 0 cannot have 4 real roots.

(b) Prove that if 36DB236D \geq B^2, then the equation p(x)=0p(x) = 0 cannot have 4 distinct positive real roots.

Q4 For each pair of real numbers aa, bb, let

M(a,b)=max(3a2+2b,3b2+2a)M(a, b) = \max{(3a^2 + 2b, 3b^2 + 2a)}

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

Q5 Let xx be any real number. Prove that xx is an integer if and only if

x+2x+3x++nx=n(x+nx)2\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 nn.

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


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 nn be a positive integer. Krishna is studying the subsets of S=1,2,,nS = {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 nn be a positive integer. A class has 2n2n students in it, and no two people have the same height. Principal Calvin wishes to arrange these 2n2n students in two rows for a class photo. He wants two rows of nn 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 1×31 \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 OO be the circumcentre of acute triangle ABCABC. Points PP and QQ are located on sides ACAC and ABAB, respectively such that triangle ABCABC and BPQBPQ are similar.

Prove that OO is the orthocentre of triangle BPQBPQ.

Q3 Find all function f:RRf : \mathbb {R} \rightarrow \mathbb {R} which satisfy

f(x)f(y)=f(x+y)+xyf(x) f(y) = f(x + y) + xy

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

Q4 Around a circular table are seated 2n2n people, where nn 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,,n1, 2, \ldots, n, four consecutive even integers are removed. The remaining integers have an average value of 5191651 \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 EE of 300 students among the 600 students such that for any school SS an any two students x,yEx, y \in E which are not in the school SS, the number of friends in SS of xx and yy are different.

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

Q7 Let A1A_1 be the centre of the square inscribed in an acute triangle ABCABC with two vertices of the square on side BCBC. Thus one of the two remaining vertices of the square on side ABAB and the other is on ACAC. Points B1,C1B_1, C_1 are defined in a similar way for inscribed squares with two vertices on sides ACAC andABAB, respectively.

Prove that lines AA1,BB1,CC1AA_1, BB_1, CC_1 are concurrent.

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

x+1y,y+1z,z+1xx + \dfrac {1}{y}, y + \dfrac {1}{z}, z + \dfrac {1}{x}

are integers.

Q9 A cube of side 3030 contains 999999 points in its interior, no four of which are coplanar. Amongst the 10071007 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 BB and CC of triangle ABCABC and intersect the lines ABAB and ACAC in DD and EE respectively. The projections of the points BB and EE onto CDCD are denoted by B1B_1 and E1E_1, respectively. The projections of the points DD and CC onto BEBE are denoted by D1D_1 and C1C_1, respectively.

Prove that the points B1,D1,E1,C1B_1, D_1, E_1, C_1 are concyclic.

Q11 Let n>1n > 1 be a given positive integer and let f(x)=a0+a1x++amxmf(x) = a_0 + a_1 x + \ldots + a_m x^m, with m2m \geq 2, be a polynomial with integer coefficients such that

(a) a2,a3,,ama_2, a_3, \ldots, a_m are divisible by all prime factors of nn,

(b) a1a_1 and nn are relatively prime.

Prove that there exists a positive integer cc, such that f(c)f(c) is divisible by n2014n^{2014}

Q12 Find a function f:RRf : \mathbb {R} \rightarrow \mathbb {R} satisfying

f(xy)(f(x)f(y))=(xy)f(x)f(y)f(xy)(f(x) - f(y)) = (x - y)f(x) f(y)

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

Note by Sharky Kesa
4 years, 4 months ago

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Number Theory
Q1

Residues of 2n2^n and 3n3^n both have cycles of length 44, as shown in the table below. Here is a list of nn, with 2n,3n2^n, 3^n residues mod 5\text{mod }5:

n2n3n011123244332411 \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 203024341(mod4)2^0\equiv 3^0\equiv 2^4\equiv 3^4\equiv 1\pmod{4} shows that 2n,3n2^n, 3^n both cycle with length 44.)

Thus:

If n0(mod4)n\equiv 0\pmod{4}, then m2(mod4)m\equiv 2\pmod{4}

If n1(mod4)n\equiv 1\pmod{4}, then m1(mod4)m\equiv 1\pmod{4}

If n2(mod4)n\equiv 2\pmod{4} , then m0(mod4)m\equiv 0\pmod{4}

If n3(mod4)n\equiv 3\pmod{4}, then m3(mod4)m\equiv 3\pmod{4}

Because mm and nn 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):

n0(mod4)    m2(mod4)n\equiv 0\pmod{4}\iff m\equiv 2\pmod{4}

n2(mod4)    m0(mod4)n\equiv 2\pmod{4}\iff m\equiv 0\pmod{4}

n1(mod4)    m1(mod4)n\equiv 1\pmod{4}\iff m\equiv 1\pmod{4}

n3(mod4)    m3(mod4)n\equiv 3\pmod{4}\iff m\equiv 3\pmod{4}

Thus, 52n+3m5\mid 2^n+3^m if and only if 52m+3n5\mid 2^m+3^n \Box


Q2

x2+y2+z2=2(yz+1)    x2+(yz)2=2x^2+y^2+z^2=2(yz+1)\iff x^2+(y-z)^2=2

Since x,y,zZx,y,z\in \mathbb{Z} then x=±1x=\pm 1 and yz=±1y-z=\pm 1

Case 1: x=1.yz=1x=1.y-z=1

Then y+z=4017y+z=4017 and yz=1y-z=1 gives (x,y,z)=(1,2009,2008)(x,y,z)=(1, 2009, 2008)

Case 2: x=1.yz=1x=-1. y-z=1

Then y+z=4019y+z=4019 and yz=1y-z=1 gives (x,y,z)=(1,2010,2009)(x,y,z)=(-1, 2010, 2009)

Case 3: x=1.yz=1x=1. y-z=-1

Then y+z=4017y+z=4017 and yz=1y-z=-1 gives (x,y,z)=(1,2008,2009)(x,y,z)=(1, 2008, 2009)

Case 4: x=1.yz=1x=-1. y-z=-1

Then y+z=4019y+z=4019 and yz=1y-z=-1 gives (x,y,z)=(1,2009,2010)(x,y,z)=(-1, 2009, 2010)

\Box


Q3

We know that n31,0,1(mod9)n^3\equiv -1, 0, 1\pmod{9}

If m9m\ge 9, then m!+55≢n3(mod9)m!+5\equiv 5\not\equiv n^3\pmod{9}

Thus, m8m\le 8. We just have to check all integers m=18m=1\to 8 for solutions. It turns out that the only solution out of these is (m,n)=(5,5)(m,n)=(5,5) \Box


Q4

2200+2n2103+16=(210022)2+2n=x22^{200}+2^n-2^{103}+16=(2^{100}-2^2)^2+2^n=x^2 for some positive integer xx.

Thus, 2n=(x+210022)(x2100+22)2^n=(x+2^{100}-2^2)(x-2^{100}+2^2) Thus, integers a,ba,b satisfy a+b=na+b=n, a>ba > b, and 2a=x+2100222^a=x+2^{100}-2^2 2b=x2100+222^b=x-2^{100}+2^2

Thus 2a2b=2101232^a-2^b=2^{101}-2^3

Thus, b=3b=3, a=101a=101, and n=104n=\boxed{104} \Box

Daniel Liu - 4 years, 4 months ago

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

Abdur Rehman Zahid - 4 years, 4 months ago

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Also, Cases 2 - 4 are labelled incorrectly.

Sharky Kesa - 4 years, 4 months ago

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fixed, thanks.

Daniel Liu - 4 years, 4 months ago

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

(Edited after seeing your reply to Siddhartha)

Calvin Lin Staff - 4 years, 4 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 - 4 years, 4 months ago

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Combinatorics

Q1. Let us fix the unit digit of our number to be 11. Then, we can arrange the remaining digits in the remaining 88 places in 8!8! ways. This way we can continue by fixing the last digit from 11 to 99.

So, its value becomes 8!(1+2+3+4+5+6+7+8+9)8!(1+2+3+4+5+6+7+8+9) which is equal to 18144001814400.

But wait, we can do this for all places of digits and not for only the unit digit. So, for fixing it at the tens place, we have to multiply 18144001814400 by 1010, at hundreds place by 100100 and so on till the first place.

Now, this becomes 1814400×(108+107+106+105+104+103+102+101+1)1814400\times (10^{8}+10^{7}+10^{6}+10^{5}+10^{4}+10^{3}+10^{2}+10^{1}+1) which is equal to 201599999798400201599999798400 which is our required answer.

BTW Finn, why don't you like the number 00. All I can say to you is that Don't Underestimate the Power of Zero

Q2. (i)(i) We know that a set of nn elements have a total number of 2n2^{n} subsets including the Phi Subset which doesn't have any element. So, we have to select an odd number of elements from nn elements to find out the total number of sets including an odd number of elements. This means:

(n1)+(n3)+(n5)....(nk)\dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}....\dbinom{n}{k} where nkn≥k and nkn-k is either 00 or 11.

This expression is equal to 2n12^{n-1} and hence our required answer is 2n12^{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 - 4 years, 4 months ago

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Number Theory Q5

The number

10a+b.(10φ(t)+102.φ(t)+...+10n.φ(t))10^{a+b}.(10^{\varphi(t)}+10^{2.\varphi(t)}+...+10^{n.\varphi(t)}) , where n=2a.5b.tn=2^a.5^b.t obviously fits the criteria.

Bogdan Simeonov - 4 years, 4 months ago

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please explain why it fits and perhaps you would illustrate when n=7 say thanks

Des O Carroll - 4 years, 4 months ago

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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, 10φ(t)1(modt)10^{\varphi(t)}\equiv1 \pmod{t}, so the inner sum is n(modt)\equiv n \pmod{t}, but tnt|n so the number is divisible by n.

Bogdan Simeonov - 4 years, 4 months ago

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

Also, where do you get these questions from?

Siddhartha Srivastava - 4 years, 4 months ago

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

Sharky Kesa - 4 years, 4 months ago

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I have a doubt regarding Q4. of the Combinatorics Section Sharky! Can Satvik be awarded 00 marks for a question? Also, can he be awarded equal marks in two or more questions. Thanks!

Yash Singhal - 4 years, 4 months ago

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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.

Sharky Kesa - 4 years, 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 - 4 years, 4 months ago

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@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 - 3 years, 5 months ago

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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 - 3 years, 5 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 - 4 years, 4 months ago

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

Sharky Kesa - 4 years, 4 months ago

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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!

Herman Pert - 4 years, 4 months ago

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@Herman Pert I just don't like it when irrelevant comments are posted. Also, yes, it is allowed.

Sharky Kesa - 4 years, 4 months ago

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@Sharky Kesa Solutions please.

Satvik Golechha - 4 years, 4 months ago

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

Sharky Kesa - 4 years, 4 months ago

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