OMK 2013

ystiang

好吧，我悲剧了，今年的OMK悲剧了，最后一年的我参加今年的OMK悲剧了。

Section A

1. A large cube is divided into 99 small cube with integer side. Given that 98 of them has side 1 cm, what is the volume of large cube?

2. A 9-digits number contains 9 different digit each, from 1 to 9. Find the GCD of the number.

3. Find remainder when 5^(55555) divided by 10000.

4. A trapezium has perpendicular diagonals, it's known that the longest perpendicular diagonals is 15cm and the height is 12cm. Find the area of trapezium.

5. The digits 2013 can be rearranged into arithmetic progression. Find the possible numbers of positive 4-digits integer with that pattern.
(Note, an arithmetic progression may have the common difference 0)

6. Find the smallest prime factors of 8051.

^印象中，和正式题目可能会有少许出入。

我的答案：
1. (99)^3
好吧，我承认我误解题目。

2. 9
hmmm,1+2+3+4+5+6+7+8+9=45, 9|45

3. 625

4. 2013
完全乱来 TvT

5. 2
0123和3210

6. 83
8051=8100-49=(90)^2 - 7^2 最简单的一题，就是没看得出difference of square也可以慢慢猜~

Section B
1. (a) A triangle ABC has their midpoint A_1, B_1, C_1 each in BC, CA, AB. Construct another triangle DEF with the side of AA_1, BB_1 and CC_1. Prove that their ratio are same no matter the size of ABC?

(b) Find area of DEF if the length of AB, BC, CA is 13cm, 14cm, 15cm respectively.

2. Prove that there exist positive integer in
(a1!)(a2!)(a3!)......(a2013!) = b!

3. Prove that all terms in the sequence is positive integers:
a_(n+1) = [5(a_n+a_n2+an_3+...+a_n)]/n ; a_1 = 1 and a_2 = 143

^对Section B的记忆较模糊，有错勿怪。 为啥全部都是Prove...TvT 严重悲剧化。

呼叫传说中的"hamilan"以及"dunwantotellu"出山~

BatuItu

omk 什么来？？
第一题是 3cm？

GCD？

第三题怎样做？

JamesTea

第三题： Find remainder when 5^(55555) divided by 10000

我知道我的方法很不 generalize，所以如果有更好的方法请指点

Suppose
5^1 = 5 (mod 10000)
5^2 = 25 (mod 10000)
5^3 = 125 (mod 10000)
5^4 = 625 (mod 10000)
5^5 = 3125 (mod 10000)
5^6 = 5625 (mod 10000)
5^7 = 8125 (mod 10000)
5^8 = 625 (mod 10000) = 5^4
5^9 = 3125 (mod 10000) = 5^5
5^10 = 5625 (mod 10000) = 5^6
5^11 = 8125 (mod 10000) = 5^7
...

Observe from 5^4 (mod 10000) every 4n for n=1,2,3,... the remainder recycle, hence, we have
5^(4+4n) = 625 (mod 10000)
5^(5+4n) = 3125 (mod 10000)
5^(6+4n) = 5625 (mod 10000)
5^(7+4n) = 8125 (mod 10000)
for n=1,2,3,...

55555 = 7 + 4(13887)
hence
5^(55555) = 5^(7+4(13887)) = 8125 (mod 10000)

JamesTea

BatuItu 发表于 14-7-2013 12:18 AM
omk 什么来？？
第一题是 3cm？

第一题应该是 27cm^3.. 因为问题要的是 volume

GCD = Greatest Common Divisor

BatuItu

JamesTea 发表于 17-7-2013 03:48 AM
第一题应该是 27cm^3.. 因为问题要的是 volume

GCD = Greatest Common Divisor

对是27,

第二题， 987654321-123456789？

JamesTea

BatuItu 发表于 17-7-2013 06:24 PM
对是27,

第二题， 987654321-123456789？

其实我不是很了解第二题的意思。因为如果要找 GCD，应该是要有两个或以上的号码才可以找。

例如 12345 和 12435。GCD = 15... 因为两个号码可以共同被除的最大号码是 15。

mathlim

这是 2013 年 OMK 高级组的题目。
很多题目都有错误，无法作答。
现在我把原本的题目放上来。

OLIMPIAD MATEMATIK KEBANGSAAN (OMK) 2013
KATEGORI SULONG

Section A
1. A large cube is divided into 99 small cubes with integer side. Given that 98 of them have side 1 cm, what is the volume of large cube?

2. Let M be the set of all nine-digit positive integers that contain each digit from 1 to 9 once. Find the highest common factor of all elements of M.

3. What is the remainder when 5^5555 is divided by 10000?

4. Given a trapezium with perpendicular diagonals and height 12. The length of one of its diagonals is 15. Find the area of the trapezium.

5. The digits of 2013 can be rearranged to form an arithmetic progression. Determine the number of four-digit positive integers with this property.
Note: An arithmetic progression might have common difference 0.

6. Determine the smallest prime factor of 8051.

Section B
1. Given a triangle ABC. The midpoints of AB, BC, CA are C₁, A₁, B₁ respectively. Construct another triangle DEF with side lengths equal to the lengths of AA₁, BB₁, CC₁.
   (a) Prove that the ratio of the area of triangle DEF to the area of triangle ABC is a constant, regardless of the choice of triangle ABC.
   (b) Find the area of triangle DEF if AB = 13, BC = 14, CA = 15.

2. Prove that there exist integers a₁, a₂, a₃, …, a₂₀₁₃, b, all greater than 1, such that
    (a₁!)(a₂!)(a₃!)…(a₂₀₁₃!) = b!

3. A sequence x₁, x₂, x₃, … is defined as follows: x₁ = 1, x₂ = 143, and
    x_n+1 = 5(x₁ + x₂ + … + x_n)/n
    for all n ≥ 2. Prove that all terms of the sequence are integers.

mathlim

Section A
1. A large cube is divided into 99 cubes with integer lengths, 98 of them with side 1. Find the volume of the large cube.

一个大立方体被分割成 99 个边长为整数的小立方体，其中 98 个的边长为 1。求大立方体的体积。

观察：
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125
6^3 = 216
7^3 = 343
8^3 = 512
9^3 = 729

因为 125 - 27 = 98，
据此可知大立方体被分割成 98 个边长为 1 及 1 个边长为 3 的小立方体，
而大立方体的边长为 5，体积为 125。

答案：125

mathlim

2. Let M be the set of all nine-digit positive integers that contain each digit from 1 to 9 once. Find the highest common factor of all elements of M.

设 M 是所有由 1 至 9 组成的数字不重复的 9 位数所组成的集合。求集合 M 中所有元素的最大公因数。

因为 1+2+3+4+5+6+7+8+9 = 45 可以被 9 整除，
所以 M 中的所有元素都可以被 9 整除。
现考虑除了 9 以外，是否有其它的公因数？
其中由 123456798 – 123456789 = 9 可知，
123456789 与 123456798 的最大公因数是9，
所以 M 中所有元素的最大公因数是 9。

注：
n(M) = 9! = 362880

mathlim

3. What is the remainder when 5^5555 is divided by 10000?

问 5^5555 除以 10000 所得余数为何？

这类题目，余数一定有循环规律。

5^1 = 5
5^2 = 25
5^3 = 125
5^4 = 625
5^5 = 3125
5^6 = 15625
5^7 = 78125
5^8 = 390625
5^9 = 1953125
5^10 = 9765625
5^11 = 48828125
5^12 = 244140625

因为前四项还未达四位数，
但是从第五项开始，
最后四位数都有循环规律，
即 3125，5625，8125，0625。
n ∈ N
5^(4n+1) 除以 10000 余 3125
5^(4n+2) 除以 10000 余 5625
5^(4n+3) 除以 10000 余 8125
5^(4n+4) 除以 10000 余 0625
5555 = 4×1388 + 3

所以 5^5555 除以 10000 所得余数为 8125。

mathlim

可以发现：
3125，5625，8125，0625，... ...
3125 + 2500 = 5625
5625 + 2500 = 8125
8125 + 2500 = 1 0625

3125×5 = 3125×(4+1) = 3125×4 + 3125 = 12500 + 3125 = 15625

15625×5 = 15625×(4+1) = 15625×4 + 15625 = 62500 + 15625 = 78125

78125×5 = 78125×(4+1) = 78125×4 + 78125 = 312500 + 78125 = 390625

390625×5 = 390625×(4+1) = 390625×4 + 390625 = 1562500 + 390625 = 1953125

1953125×5 = 1953125×(4+1) = 1953125×4 + 1953125 = 7812500 + 1953125 = 9765625

2500×4 = 10000

再想一想，还有很多的道理... ...

mathlim

BatuItu 发表于 14-7-2013 12:18 AM
omk 什么来？？
第一题是 3cm？

Olimpiad Matematik Kebangsaan

mathlim

4. Given a trapezium with perpendicular diagonals and height 12. The length of one of its diagonals is 15. Find the area of the trapezium.

已知梯形的对角线互相垂直且高为2，其中一对角线的长为15。求此梯形的面积。

BD = 15, AE = BF = 12

DF = √(15^2 – 12^2) = 9

∠AEC = ∠DFB = 90°, ∠CAE = ∠BDF

△AEC ∽ △DFB

CA/AE = BD/DF

CA/12 = 15/9

CA = 20

所以梯形的面积为 15×20/2 = 150。

mathlim

5. The digits of 2013 can be rearranged to form an arithmetic progression. Determine the number of four-digit positive integers with this property.
   Note: An arithmetic progression might have common difference 0.

2013 的数字可以重新组成一个等差数列。求满足这个性质的四位数的个数。
注：等差数列的公差可以为 0。

公差为 0 的等差数列有：
1,1,1,1; 2,2,2,2; 3,3,3,3; 4,4,4,4; 5,5,5,5; 6,6,6,6; 7,7,7,7; 8,8,8,8; 9,9,9,9

公差为 1 的等差数列有：
0,1,2,3; 1,2,3,4; 2,3,4,5; 3,4,5,6; 4,5,6,7; 5,6,7,8; 6,7,8,9

公差为 2 的等差数列有：
0,2,4,6; 1,3,5,7; 2,4,6,8; 3,5,7,9

公差为 3 的等差数列有：
0,3,6,9

其中 4 个数字都相同的有 9 个，即 1111，2222，3333，4444，5555，6666，7777，8888，9999

其中含有 数字 0 的有：0,1,2,3; 0,2,4,6; 0,3,6,9
每一个可以组成 18 个不同的数，如 0,1,2,3 可组成：
1023，1032，1203，1230，1302，1320，
2013，2031，2103，2130，2301，2310，
3012，3021，3102，3120，3201，3210
所以共有 3×18 = 54 个

数字不一样且不含 0 的有：
1,2,3,4; 2,3,4,5; 3,4,5,6; 4,5,6,7; 5,6,7,8; 6,7,8,9; 1,3,5,7; 2,4,6,8; 3,5,7,9
每一个可以组成 24 个不同的数，如 1,2,3,4 可组成：
1234，1243，1324，1342，1423，1432，
2134，2143，2314，2341，2413，2431，
3124，3142，3214，3241，3412，3421，
4123，4132，4213，4231，4312，4321
所以共有 9×24 = 216 个

答案：9 + 54 + 216 = 279

mathlim

6. Determine the smallest prime factor of 8051.

求 8051 的最小质因数。

8051 = 8100 - 49 = 90^2 - 7^2 = (90 + 7)(90 - 7) = 97×83
所以 8051 的最小质因数为 83。

mathlim

今年的成绩公布了！
http://persama.org.my/index.php

mathlim

Section B
1. Given a triangle ABC. The midpoints of AB, BC, CA are C₁, A₁, B₁ respectively.
    Construct another triangle DEF with side lengths equal to the lengths of AA₁, BB₁, CC₁.
    (a) Prove that the ratio of the area of triangle DEF to the area of triangle ABC is a constant, regardless of the choice of triangle ABC.
    (b) Find the area of triangle DEF if AB = 13, BC = 14, CA = 15.

已知三角形 ABC。AB，BC，CA 的中点分别为 C₁，A₁，B₁。以 AA₁，BB₁，CC₁ 为边长组成另一个三角形 DEF。
(a) 证明三角形 DEF 的面积与三角形 ABC 的面积之比是一常数。
(b) 若 AB = 13，BC = 14，CA = 15，求三角形 DEF 的面积。

设 AB = c，BC = a，CA = b，AA₁ = x，BB₁ = y，CC₁ = z
△ABC = √[s(s-a)(s-b)(s-c)]，s = (a+b+c)/2
△ABC = (1/4)√[(a+b+c)(a+b-c)(b+c-a)(c+a-b)]
          = (1/4)√(2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4)
△ABA₁ = △ACA₁
(1/4)√[(c+x+a/2)(c+x-a/2)(x+a/2-c)(a/2+c-x)] = (1/4)√[(b+x+a/2)(b+x-a/2)(x+a/2-b)(a/2+b-x)]
x^2 = 2b^2 + 2c^2 - a^2
同理可得
y^2 = 2c^2 + 2a^2 - b^2
z^2 = 2a^2 + 2b^2 - c^2
△DEF = (1/4)√(2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 - x^4 - y^4 - z^4)
          = (3/16)√(2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4)
          = (3/4)△ABC
∴ △DEF/△ABC = 3/4
∴ △DEF = (3/4)√(21×8×7×6) = 63

mathlim

2. Prove that there exist integers a₁, a₂, a₃, …, a₂₀₁₃, b, all greater than 1, such that
   (a₁!)(a₂!)(a₃!)…(a₂₀₁₃!) = b!

证明存在大于一的正整数 a₁, a₂, a₃, …, a₂₀₁₃, b, 使得 (a₁!)(a₂!)(a₃!)…(a₂₀₁₃!) = b!。

设 a₂₀₁₃ = (a₁!)(a₂!)(a₃!)…(a₂₀₁₂!) - 1,
则 b = (a₁!)(a₂!)(a₃!)…(a₂₀₁₂!),
b! = (a₁!)(a₂!)(a₃!)…(a₂₀₁₃!) = [(a₁!)(a₂!)(a₃!)…(a₂₀₁₂!)]! 即得证。
举例：
a₁ = a₂ = a₃ = … = a₂₀₁₂ = 2
a₂₀₁₃ = 2^2012 – 1, b = 2^2012
b! = (2^2012)!

mathlim

3. A sequence x₁, x₂, x₃, … is defined as follows: x₁ = 1, x₂ = 143, and

   x_n+1 = 5(x₁ + x₂ + … + x_n)/n for all n ≥ 2.
   Prove that all terms of the sequence are integers.

一数列 x₁, x₂, x₃, … 定义成：x₁ = 1, x₂ = 143, 及
x_n+1 = 5(x₁ + x₂ + … + x_n)/n 对于所有 n ≥ 2。
证明此数列的每一项都是整数。

x₃ = (5/2)(x₁+x₂)
x₄ = (35/6)(x₁+x₂)

x₅ = (35/3)(x₁+x₂)
x₆ = (21)(x₁+x₂)
x₇ = (35)(x₁+x₂)
x₈ = (55)(x₁+x₂)
x₉ = (165/2)(x₁+x₂)
x₁₀ = (715/6)(x₁+x₂)

x₄ = (7/3)x₃
x₅ = (8/4)x₄
x₆ = (9/5)x₅
x₇ = (10/6)x₆
x₈ = (11/7)x₇
x₉ = (12/8)x₈
x₁₀ = (13/9)x₉

x₁ = 1

x₂ = 143

x₃ = 360 = 6×5×4×3

x₄ = (7/3)×360 = 7×6×5×4

x₅ = (8/4)(7/3)×360 = 8×7×6×5

x₆ = (9/5)(8/4)(7/3)×360 = 9×8×7×6

x₇ = (10/6)(9/5)(8/4)(7/3)×360 = 10×9×8×7

x₈ = (11/7)(10/6)(9/5)(8/4)(7/3)×360 = 11×10×9×8

x₉ = (12/8)(11/7)(10/6)(9/5)(8/4)(7/3)×360 = 12×11×10×9

x₁₀ = (13/9)(12/8)(11/7)(10/6)(9/5)(8/4)(7/3)×360 = 13×12×11×10

x_n = _n+3P₄, n ≥ 3，n ∈ Z

所以此数列的每一项都是整数。

ystiang

话说自己的水平很差

mathlim前辈很厉害...

话说OMK有"参加文凭"吗？