All About Math Olympiad Training & Questions

[MathsOlympiadtrainer]

pinky88:

Hi, need help again on the flwg qns, thx...




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3. This is a not-obvious qn on Pythagoras. We make use of pythagorean triplets to find them. Some common pythagorean triplets include (3,4,5), (5,12,13), (7,24,25), (8,15,17).

870:464=15:8
615:855=41:57
123:477=41:159
326:614=163:307
536:462=268:231

Since only in the first pair, it is part of the triplets (8,15,17), and none of the other 4 are, therefore only A is the answer.

[MathsOlympiadtrainer]

pinky88:

Hi, need help again on the flwg qns, thx...




http://i44.tinypic.com/24g0g36.png\">

4a) You need to read up more on modulo operations. In summary,
If we take 59 / 29 = 2 R 1. We just need to look at the remainders for powers.

59/29 = 2R1
87/29 = 3R0
115/29 = 3R28 = 4R(-1)

So remainder in 59^59 + 87^87 - 115^115 = 1^59 + 0^87 + (-1)^115 = 1 + 0 -1 = 0

4b) n^6039<2013^2013
((n^3)^2013)<2013^2013
n^3<2013

Since 12*12*12=1728
and 13*13*13=2197

Therefore the largest n will be 12.

[pinky88]

Thank u v much for ur help, MathsOlympiadTrainer.

[MathsOlympiadtrainer]

pinky88:

Thank u v much for ur help, MathsOlympiadTrainer.

You are welcome! :lol: :lol:

[kiasiparent]

pinky88:

http://i44.tinypic.com/24g0g36.png\">

A square number must ends with digit 00, 1, 4, 5, 6, 9.

In addition, if it ends with 5, then it must end with 25. this is because for eg a number (m5)^2
(10m +5)^2 = 100m^2 + 100 m + 25 will end with 25

It can be calculated that (b), (c), (d) and (e) will not work just by looking at their last 2 digits when calculating.

[Maths Hub]

Maths Hub:

pinky88:

Hi,


I need help for the following questions:

1.\tIn a competition consisting of 30 problems. Lydia was given 12 points for each correct solution & 7 points were subtracted from her score for each incorrect solution. Problems not attempted contributed 0 points. How many correct solutions did Lydia have if her score was 209? (Hint : 19 X 11 = 209).

TIA.

Sorry there was a mistake.

1. Assume all are correct.
Total Score will be 30 X 12 = 360
For each wrong question, a total of 12 + 7 = 19 marks is deducted from the total marks.
For each unanswered question, 12 marks are deducted from the total marks.
Now, there is a difference of 360 - 209 = 151 from the total score.
We try to find 151 as a sum of multiples of 19 and 12.
151 = 1 X 19 + 132 (a multiple of 12)
= 2 X 19 + 113 (not a multiple of 12)
= 3 X 19 + 94 (not a multiple of 12)
= 4 X 19 + 75 (not a multiple of 12)
= 5 X 19 + 56 (not a multiple of 12)
= 6 X 19 + 37 (not a multiple of 12)
= 7 X 19 + 18 (not a multiple of 12)

The only case is 1 X 19 + 11 X 12. So there must be 1 wrong, 11 unanswered and thus 30 - 1 - 11 = 18 correct.

Hope that helps.

[pinky88]

No worries, Maths Hub. My gal spotted the mistake while she was doing d sum. Thx anyway:)

[kiasiparent]

Just take 30x12 = 360 (not possible), then subtract 209, then check if it is divisible by 7.

Since 209 is not divisible by 12, she must have at least one question wrong.

29 x 12 = 348, -209 = 139 (very far)

22 x 12 = 264, -209 = 55 (not multiple of 7)
21 x 12 = 252, -209 = 43 (not multiple of 7)
20 x 12 = 240, - 209 = 31 (not multiple of 7)
19 x 12 = 228, - 209 = 19 (not multiple of 7)
18 x 12 = 216, -209 = 7
Anything lower, not possible.

One solution confirmed. Why so complicated?

[MathsOlympiadtrainer]

Congrats to all NMOS 2013 invitation round qualifying students!


http://oas.nushigh.edu.sg/NMOS/NMOS%202013%20Special%20Round%20Results.pdf

:rahrah: :rahrah: :rahrah: :rahrah:

[burp]

My ds’ teacher informed him about the 2nd round on Sat but did not mention the time- anyone has any idea?