O-Level Additional Math

[CoffeeCat]

OK Lor:

Hi Sir,


Need your help for some maths competition questions:

1. When 3^1981 + 2 is divided by 11, the remainder is ...
2. The least positive integer which has remainders 1, 1 and 5 when divided by 3, 5 and 7 respectively, is (A)166 (B)151 (C)145 (D)131 (E)none of these.
3. If x and y are integers such that (x-y)² + 2y² = 27, then the only number x can be are (A)3,5 (B)-6,4 (C)0,4,6 (D)0,-4,4,-6,6 (E)0,-2,2,-4, 4,-6,6.

Thanks.

Hope guanhui won't be angry at me...

2) If the number has the same remainder 1 upon division by 3 and 5, it will leave the same remainder 1 upon division by 15.
the number is of the form 15x + 1 = (14 + 1)x + 1
and 15x will leave a remainder of 4 upon division by 7.
therefore x = 4, or 11, 18, etc
Least integer = 15*4 + 1 = 61

3)notice that the left hand side are all positive so
we are looking at 0, 1, 4, 16, 25 for (x-y)^2
only 25 and 9 fits
and y=1 or -1
x-y= 5 or -5.
or
y= 3 or -3
x-y = 3 or -3
you will get x=6, 4, -6, -4, 0

1) 3^1981 + 2
There are many ways to do this.
for starters use the pattern approach by looking at the remainder of powers of 3 upon division by 11.
3 9 5 4 1 3 9 5 4 1, ...
therefore 3^1981 will leave a remainder of 3.

Since you are doing competition maths, you will eventually come across this method of doing such questions using fermat's little theorem.

[Guan Hui]

haha y would I be angry with you=)

this looks like maths olympia question which give me headaches a month or 2 ago haha

[OK Lor]

Hi Sir,


Thanks to all, excellent .

A few more headache questions :? :
1. Two vertical poles, 20m and 80m high, stand apart on a horizontal plane. The height, in metres, of the point of intersection of the lines joining the top of each pole to the foot of the other is .....

2. Some unit cubes are assembled to form a larger cube and then some of the faces of these larger cubes are painted. After the paint dries, the larger cubes is disassembled into the unit cubes, and it is found that 45 of these have no paint on any of their faces. How many faces of the larger cube were painted?

3. In how many different ways can a careless office boy place four letters in four envelopes so that no one gets the right letter?

Thanks.

[Guan Hui]

here you go I hope you can see this


http://www.postimage.org/image.php?v=aVdEdpi

http://www.postimage.org/image.php?v=PqtXhEA

http://www.postimage.org/image.php?v=gx1SzoJ

[OK Lor]

Hi Sir,


Thanks a lot, but for Q3 need some time to digest.

[Guan Hui]

q3 workings always start with 1x1x because i already dictate where the 1st and 2nd letter has gone to.

I am sure there is a better solution then mine=/

[OK Lor]

Hi Sir,


Thanks.

Please help to resolve the fraction (1-11x-4x^2+3x^3-2x^4)/(6-4x+3x^2-2x^3) into partial fraction.

Thanks.

[CoffeeCat]

OK Lor:

Hi Sir,


Thanks.

Please help to resolve the fraction (1-11x-4x^2+3x^3-2x^4)/(6-4x+3x^2-2x^3) into partial fraction.

Thanks.

Hey, to resolve to partial fraction, one will need to factorise the bottom denominator part.

6 - 4x + 3x^2 - 2x^3

This is not easy to factorise at first sight, this belongs to the case where grouping the 1st and 3rd term together, and the 2nd and 4th term together will show sth.
6 + 3x^2 = 3( x^2 + 2)
-4x - 2x^3 = -2x (x^2 + 2)

Together 6 - 4x + 3x^2 - 2x^3 = (3 -2x ) (x^2 + 2)

I will let tutor guanhui continue from here.

[OK Lor]

Hi Coffeecat,


Thanks, got it. Cheers. 卧虎藏龙 here

[Guan Hui]

haha i got it to this step yest too but still thinking on what to do to the numeratorXD