All About Math Olympiad Training & Questions

[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:

[trileets]

May I know how this qn is done?


Find the number of ways to put 4 different coloured marbles into 4 identical empty boxes.

Ans: 15 (my ans of 24 is wrong)



TIA

[MathsOlympiadtrainer]

trileets:

May I know how this qn is done?


Find the number of ways to put 4 different coloured marbles into 4 identical empty boxes.

Ans: 15 (my ans of 24 is wrong)



TIA

4 0 0 0: 1 way
3 1 0 0: 4 ways (4 ways to choose the odd one out)
2 2 0 0: 3 ways (AB, AC, AD, BC, BD, CD total 6 ways in the first box but we have to divide by 2 as AB CD is the same as CD AB and there is double counting)
2 1 1 0: 6 ways (AB, AC, AD, BC, BD, CD total 6 ways)
1 1 1 1: 1 way

[trileets]

:thankyou:

Quekcc for the clear explanation .so it's not putting different marbles in each box.

Sorry blur me.. Missed out mathsolympiadtrainer's post. Thank you very much for your explanation too!

[iRabbit]

Anyone’s kid participating in the GAUSS MO contest this May? DS is registered for Grade 7 but I see that the Sect C qns are beyond P3 syllabus. He won’t be able to ans those as he’s not taught geometry and other advanced topics. What level (relative to local primary std) is grade 7? Thx