All About Math Olympiad Training & Questions
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pinky88:
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).Hi, need help again on the flwg qns, thx...
http://i44.tinypic.com/24g0g36.png\">
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. -
pinky88:
4a) You need to read up more on modulo operations. In summary,Hi, need help again on the flwg qns, thx...
http://i44.tinypic.com/24g0g36.png\">
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. -
Thank u v much for ur help, MathsOlympiadTrainer.
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pinky88:
Thank u v much for ur help, MathsOlympiadTrainer.
You are welcome! :lol: :lol: -
pinky88:
A square number must ends with digit 00, 1, 4, 5, 6, 9.
http://i44.tinypic.com/24g0g36.png\">
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:
Sorry there was a mistake.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.
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. -
No worries, Maths Hub. My gal spotted the mistake while she was doing d sum. Thx anyway:)
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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? -
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: -
My ds’ teacher informed him about the 2nd round on Sat but did not mention the time- anyone has any idea?