奥数问题大家讨论

davidbin

Work backwards:1973->1954->1940->1924->1907->1885->1865->1846->1832->1816 (X)
1971->1953->1935->1917(x)
remove 1973,1971 from the list. I stop here, waiting for smart way!

静雯

第八期

davidbin

静雯 发表于 14-3-2012 12:17
第七期

受楼主提示，给出以下解法
一个数除以3的余数= 这个数的各位数字之和除以3的余数
比如：mod (16,3) =1； mod ((1+6),3) =1
1,2,4,8,16,23,28,38,...
以上的数列，后一个数=前一个数加上其各位数字之和
ｍod（后一个数，3 ）＝　ｍｏｄ（前一个数，３）＋ｍｏｄ（前一个数各位数字之和，３）＝　２ｍｏｄ（前一个数，３）
1,2,4,8,16,23,28,38,...　　　ｍｏｄ　３

１，２，１，２，１，２，．．．．．

ｍｏｄ（１９７１，３）　＝０　Ｘ
ｍｏｄ（１９７４，３）　＝０　Ｘ
ｍｏｄ（１９７３，３）　＝２
因为是第１５１个数，其余数必是１，故１９７３也不可能，答案是１９７２

davidbin

静雯 发表于 21-3-2012 10:10
第八期

猜一猜　３６

静雯

Problem 9
How many 5-digit multiples of 3 have at least one of its digits equal to 3?

Solution
Method 1:  (provided by frekiwang )
Let
a(n) denote the number of n-digit numbers which are mutiples of 3 and include at least a '3';
b(n) ................................................................. are not mutiples of 3 and include at least a '3';
c(n) ................................................................. are mutiples of 3 and include no '3';
d(n) ................................................................. are not mutiples of 3 and include no '3';
By considering adding an addtional number at the end of n-digit number to form a (n+1)-digit number,
we can conclude:
a(n+1)=4a(n)+3b(n)+c(n) (numbers in a(n), you can add 0,3,6,9 at the end; numbers in b(n), you can add either 1,4,7 or 2,5,8 at the end; numbers in c(n), you can add a 3)
b(n+1)=6a(n)+7b(n)+d(n) (similar reasoning)
c(n+1)=3c(n)+3d(n) (similar reasoning)
d(n+1)=6c(n)+6d(n) (similar reasoning)

We know a(1)=1, b(1)=0, c(1)=2, d(1)=6
work recursively, we have
a(2)=6, b(2)=12, c(2)=24, d(2)=48
a(3)=84, b(3)=168, c(3)=216, d(3)=432
a(4)=1056, b(4)=2112, c(4)=1944, d(4)=3888
a(5)=12504, b(5)=25008, c(5)=17496, d(5)=34992

So the answer is 12504. The method above works for b(n),c(n),d(n) groups very well, so it is a more 'general' approach.

There is a 'simpler' way, which is by considering the number of 5-digit numbers which include 3 (90000-9*8*8*8*8=37512), and approximately 1/3 of them are divisibly by 3, so the answer is 37512/3=12504 as well, but you need a paragraph to explain why it is exactly 1/3 but not approximately.

Method 2: (provided by davidbin)
Total 5 digits number from 10000~99999: 99999-10000+1 = 90000,
90000/3 = 30000
So in total there are 30000 number multiples of 3 among 5 digits number
Without single digit of 3 could be
1a1a2a3a4
2a1a2a3a4
4a1a2a3a4
5a1a2a3a4
.....
9a1a2a3a4
a1,a2,a3,a4 belong (0,1,2,4,5,6,7,8,9)
0,1,2,4,5,6,7,8,9 mod 3 , the result can be grouped into 3
R0(0,6,9), R1(1,4,7), R2(2,5,8),each group has 3 numbers
1a1a2a3a4 total possiblity is 9X9X9X3= 2187
get a1,a2,a3 from (0,1,2,4,5,6,7,8,9), 9 choices for each number. For every 1, a1,a2,a3 combination, you always have 3 choices for a4 to ensure the number 1a1a2a3a4 can be devided by 3
(1,2,4~9)a1a2a3a4 total possibility = 8x2187=17496
Result = 30000- 17496 = 12504
Below formula valid for 3 & 9
(90000- 8x9x9x9x9)/X x=3 or 9

静雯

Problem 10.
A goat in a horizontal ground is tied to one end of 14 m long rope. The other end of the rope is attanched to a ring which is free to slide along a fixed 20 m long horizontal rail. Find the maximum possible area that the goat can graze.
Ignore the dimension of the ring and take pi to be 22/7.

The given answer is 1204 square m, but the correct answer should be 1176 square m after discussion.

静雯

Problem 11.
A theme part isssues entrance tickets bearing 5-digit serial numbers from 00000 to 99999. If any adjacent numbers in the serial numbers differ by 5 (for example 12493), customers holding such a ticket could use the ticket to reddem a free drink. Find the number of tickets that have serial numbers with this property.
(这个是2011年Asia Pacific Mathematical Olympiad for Primary Schools第一轮竞赛的其中一道试题.)

Solution: (provided by jjrchome )

先求相邻2个不等于5的。
首位可取10个，第2、3、4、5位只能取9个，共10*9*9*9*9。
所以相差5的数有 100000-10*9*9*9*9=34390个

静雯

Problem 12.
P, Q, R, S and T are equally spaced on a straight rod.      P----Q----R----S----T
If the rod is frist rotated 180 degree about T, then 180 degree about S and finally 180 degree about P, which point's position remains unchanged?

The given answer is R, but the correct answer should be Q.

Solution: (provided by 我要开心)

P Q R S T T S R Q P P Q R S T T S R Q P Answer is Q.

静雯

Problem 13                                                                                            16 April
A code is to be formed by using only digits 0 and 1. The length of the code is the number of digits in the code. Find the number of codes of length 10 such that no two 1s can be together(For example, 0101010010).

davidbin

Assume code : XX XX XX XX XX
先去除 XX= 11
XX XX XX XX XX    XX = 00, 01,10, 共有3X3X3X3X3 种可能
再去除 01 10
   （） XX （） XX  ( )   XX  ( )
01 10 可放在其中任何一个（）位子 ， 共有4X3X3X3 种可能
01 10 XX XX XX
XX 01 10 XX XX
XX XX 01 10 XX
XX XX XX 01 10
其中
01 10 01 10 XX   ( 01 10 XX XX XX and XX XX 01 10 XX 重复计算了3次 01 10 01 10 XX )
01 10 XX 01 10   ( 01 10 XX XX XX and XX XX XX 01 10 重复计算了3次 01 10 XX 01 10 )
XX 01 10 01 10   ( XX 01 10 XX XX and XX XX XX 01 10 重复计算了3次 XX 01 10 01 10 )
各多算了3 次，共9 次
3x3X3X3X3 - (4x3x3x3 -9) = 144