# International Mathematical Olympiad '62, First Day

The goal of this set of notes is to improve our problem solving and proof writing skills. You are encouraged to submit a solution to any of these problems, and join in the discussion in #imo-discussion on Saturday 12 at 8:30 pm IST ,0800 PDT. For more details, see IMO Problems Discussion Group.

It has been a lot of time since the Last note in the series. Examinations really took over and I couldn't concentrate here... Any ways happy Problem Solving!. For all the Brilliant community here who are scared of IMO Questions try Q1 and Q2. Post any ideas you may have, doubts you come across or solutions you get!

1. (POL) Find the smallest natural number $n$ with the following properties:

(a) In decimal representation it ends with $6$.

(b) If we move this digit to the front of the number, we get a number $4$ times larger.

2. (HUN) Find all real numbers x for which $\sqrt{3-x}-\sqrt{x +1}> \dfrac{1}{2}$

3. (CZS) A cube ABCDA'B'C'D' is given. The point $X$ is moving at a constant speed along the square $ABCD$ in the direction from $A$ to $B$. The point $Y$ is moving with the same constant speed along the square BCC'B' in the direction from $B$ to $C$. Initially, $X$ and $Y$ start out from $A$ and B' respectively. Find the locus of all the midpoints of $XY$.

###### This is part of the set International Mathematical Olympiads

Note by Sualeh Asif
4 years, 11 months ago

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Solution For Q1)

Let the smaller of the two numbers be $n$ and the larger be $m$.

We are given that $n$ ends in $6$ and so the first digit of $m$ is 6. This means that the first digit of $n$ is $\lfloor \frac{6}{4} \rfloor =1$, and so we know that $n$ starts with a $1$.

Since we now know that $n$ starts with a $1$, we know that $m$ starts with a $61$ and so we know that the first two digits of $n$ is $\lfloor \frac{61}{4} \rfloor=15$, so we know that $n$ starts with $15$.

Continuing this same process we get that $n$ starts with $\lfloor\frac{615}{4} \rfloor=153$

$n$ starts with $\lfloor \frac{6153}{4} \rfloor =1538$

$n$ starts with $\lfloor \frac{61538}{4} \rfloor =15384$. In this last case, $\frac{61538}{4}$ is an integer, so our process is over, and we get that

$n=\boxed{153846}$

- 4 years, 11 months ago

Brilliant!

- 4 years, 11 months ago

However, I solved it from the right side by multiplying 4....

- 4 years, 11 months ago

Friends .. this is the first time im attempting these type of probs .... i enjoyed doing them .... ASIF, can u tell me where i can access more such problems??

by th way, i tried Q3 .. i drew a cube and named the 4 vertices in the bottom as ABCD and A' is directly abov A, B' above B, C' above C, and D' above D ... (hope u got the cube i visualized)

Let centre of face ABA'B' be K and centre of face BCB'C' be L ... i got the locus to be the line KL.

AM I RIGHT?? IF I AM WRONG .. HOW SHUD I DO (I tuk the help of CO-ORDINATE GEOMETRY)

- 4 years, 11 months ago

Well two things: The answer is very near but you are missing part of the locus. Try to find if other points exist. (They do!)

These are the problems from the International Mathematical Olympiad. You can either get a copy of the Compendium or use artofproblemsolving.com

- 4 years, 11 months ago

hmm .. i wil try ...

- 4 years, 11 months ago

im not getting any new points ... perhaps cud u pls provide a solution brother ...

- 4 years, 11 months ago

3-x>0,and x+1>0 so -1<x<3 is domain,now before squaring √(3-x)-√(x+1)>0 because it is >1/2 So √(3-x)>√(x+1) i.e. after squaring 3-x>x+1,so x<1 So we now know that -1<x<1 (We can also replace x=cos(2x)=2cos²(x)-1,but no) So now we can square 2√(3-x)-2√(x+1)>1
2√(3-x)>1+2√(x+1) so square it 4(3-x)>1+4√(x+1)+4(x+1) So 4√(x+1)<7-8x So before squaring 7-8x>0,i.e. x<7/8 So now squre it so 16(x+1)<49-112x+64x² i.e. 64x²-128x+33>0 So x<(64-√(64²-64*33))/64 I.e.x<(8-√31)/8 Or x>(8+√31)/8>7/8 But x<7/8 so -1<x<(8-√31)/8 is our answer for 2nd problem

- 4 years, 11 months ago

@Nihar Mahajan @Surya Prakash @Xuming Liang Here is the next note in the series!

- 4 years, 11 months ago

You missed me. -_-

- 4 years, 11 months ago

Thats a big blunder! (Though you were at camp at the time)

- 4 years, 11 months ago

Q2.Hint:Take root(x+2) to the right and then the situation is like f(x)>g(x) which can be solved by drawing graphs of f(x) & g(x).

- 4 years, 11 months ago

Are you sure the Q2 is a IMO question? I think that was a question in 2014 Pre-RMO in India's East Region.

- 4 years, 11 months ago

Solution for Q1 x=10p+6, (k+1 is number of digits of x) We know that 6×10^k+p=4x i.e. 6×10^k+p=40p+24 i.e. 13p=2×10^k-8 So 2×10^k-8=0 (mod 13) I.e. 2×(10^k-4)=0 (mod 13) So because (13,2)=1 so 10^k=4(mod 13) i.e. (-3)^k=-9(mod 13) So putting k=1,2,3,... We get k=5 is smallest such number So now we know p=(2×10^5-8)/13 I.e. p=199992/13=15384 So x=10p+6=153846 is smallest such number.Really 4*153846=615384

- 4 years, 11 months ago