All About Math Olympiad Training & Questions

[MathIzzzFun]

Absolutely Bo Chap:

Thanks a lot, MathIzzzFun. Very nice solutions! Your skills are very impressive! Really appreciated your help.


Presumably, you have represented Singapore in the IMO before? Or you have at least some similar experience?

...maths competitions were not prevalent during my \"era\" .. just pure passion for maths

cheers.

[MathIzzzFun]

http://i47.tinypic.com/2ic9x6s.png\">


cheers.

[Absolutely Bo Chap]

Nice one, MathIzzzFun!


I also have lots of interests in Mathematics, and also competitive maths and brainteaser in particular. They tend to stretch the mind.

After reading your solutions, I would have want to ask you to be my teacher so that I could learn from the Master if I am still in school.

Just out of curiousity, are you also a Maths Olympiad trainer, just like some of other people in this forum? (Even if you are not, you are definitely more than qualified to be one.) Or even a National Team trainer?

[MathIzzzFun]

Absolutely Bo Chap:

Nice one, MathIzzzFun!


I also have lots of interests in Mathematics, and also competitive maths and brainteaser in particular. They tend to stretch the mind.

After reading your solutions, I would have want to ask you to be my teacher so that I could learn from the Master if I am still in school.

Just out of curiousity, are you also a Maths Olympiad trainer, just like some of other people in this forum? (Even if you are not, you are definitely more than qualified to be one.) Or even a National Team trainer?

thanks... 帽子太大了

..I do not conduct formal Maths Olympiad training but I incorporate MO techniques/strategies and MO-styled questions in my tuition sessions..

cheers.

[pinky88]

Hi all, pls help on the following 2 qns:


1) If x & y are positive integers which satisfy x² - 8x - 1001y² = 0, what is
the smallest possible value of x + y?

2a) If y = |x| + |x + 1| + |x - 1|, find the minimum value of y.

b) Find the area, in square units, enclosed by the figure whose boundary
points satisfy |x|+ |y|=4

TIA.

[Maths Olympiad Tutor]

1) If x & y are positive integers which satisfy x² - 8x - 1001y² = 0, what is

the smallest possible value of x + y?

From the question, we know that x² - 8x = 1001y²

So, let us substitute positive intigers into Y and find 1001y²

When Y = 1, 1001y² = 1001
When Y = 2, 1001y² = 4004
When Y = 3, 1001y² = 9009
When Y = 4, 1001y² = 16016

So on and so forth.

Next, we substitute values into x and try to find an x value that is equivalent to one of the 1001y² values. This process involves intelligent guessing

When x is 35, x² - 8x = 945 - not exact
When x is 36, x² - 8x = 1008 - not exact
When x is 67, x² - 8x = 3953 - not exact
When x is 68, x² - 8x = 4080 - not exact
When x is 98, x² - 8x = 8820 - not exact
When x is 99, x² - 8x = 9009 - exactly corresponding to y = 3

Hence, the smallest value is 3 + 99 = 102

2a) If y = |x| + |x + 1| + |x - 1|, find the minimum value of y.

To solve this question, we have to reduce one of the mods to 0.

So either x = 0, therefore |x| = 0
or x = -1, |x + 1| = 0
or x = 1, |x - 1| = 0

Any other substitution of X will produce a larger y.

When x is -1 or 1, y = 3
When x is 0, y is 2

Therefore the minimum value of y is 2.

b) Find the area, in square units, enclosed by the figure whose boundary
points satisfy |x|+ |y|=4

For this question, the minimum and maximum values of both X and Y are -4 and 4 respectively, because anything more or anything less will not satisfy the equation due to the mod signs.

I have plotted the graph on graphmatica, with the values for X and Y.

Apologise for the grainy image, but you should get the idea.

http://i47.tinypic.com/miie6d.png\">

Enjoy!

[pinky88]

Thank u v much, Maths Olympiad Tutor for yr clear explanation:)

[Absolutely Bo Chap]

1) If x & y are positive integers which satisfy x² - 8x - 1001y² = 0, what is

the smallest possible value of x + y?

There is another way to solve this problem.

x² - 8x = x * ( x - 8 ) = 1001y²

Without loss of generality, we may suppose that x and y are coprime, that is, they do not have any common factors.

But we know that 1001 = 7 * 11 * 13, so there are only 4 ways to split 1001 into 2 different factors, namely,

a. 77 * 13
b. 91 * 11
c. 143 * 7
d. 1001 * 1

To get the corresponding value for y, it will be near the square root of the quotient of the 2 factors. The reason is that x * ( x - 8 ) is equal to ( x - 4 )² less 16.

a. about sqrt( 77 / 13 ) = 2 to 3
b. about sqrt( 91 / 11 ) = 2 to 3
c. about sqrt( 143 / 7 ) = 4 to 5
d. about sqrt( 1001 / 1 ) = 31 to 32

So we only need to check the following:

a. y = 2, 3 against x = 69 (=77 - 8), 85 (= 77 +
b. y = 2, 3 against x = 83 (= 91 - 8), 99 (91 +
c. y = 4, 5 against x = 135 (= 143 - 8), 151 (= 143 +
d. y = 31, 32 against x = 993 (= 1001 - 8), 1009 (= 1001 +

Out of the above 8 possibilities, only x = 99 against y = 3 produce an answer.

There are other possibilities when x has has a factor of 2, 4, or 8. I did not look in the details, since that would generate a higher value of x + y even if there are solutions. However, looking at the discriminant suggests that there are no positive integer solutions on x and y.

Hence, this may be the only solution to this Diophantine equation, so there is only 1 possible value of x + y if x and y are to be positive integers.

[pinky88]

Thx for d alternative solution, Absolutely Bo Chap.

[GDragon]

I am joining the math olympiad 2013 this year and can some people give me some tips on how to get into the next round? :xedfingers: