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

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

$n=\boxed{153846}$

Brandon Monsen - 5 years, 6 months ago

Brilliant!

Sualeh Asif - 5 years, 6 months ago

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

Symon Saroar - 5 years, 6 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)

AWAITING A GOOD REPLY

Ganesh Ayyappan - 5 years, 6 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

I hope you can explain to us your argument so that we help you in it.

hmm .. i wil try ...

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

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

Nikola Djuric - 5 years, 6 months ago

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

You missed me. -_-

Sharky Kesa - 5 years, 5 months ago

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

Sualeh Asif - 5 years, 5 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).

Deepak Kumar - 5 years, 6 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.

Anupam Nayak - 5 years, 6 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

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$