12. 設 [x] 代表不超過 x 的最大整數，且 {x} = x −[x]，例如 [1.1] = 1、{6.9} = 0.9 和 [5] =
5。若x是正數，且 2[x]{x}+ 4 = 3{x}+ 3x，求x所有可能值之和。
Let [x] denote the greatest integer not exceeding x and {x} = x −[x] . For example, [1.1] = 1,
{6.9} = 0.9 and [5] = 5. If x is a positive number such that 2[x]{x}+ 4 = 3{x}+ 3x , find the
sum of all possible values of x.16. 已知n 是正整數，它被7 除時餘3，被11 除時餘5，被13 除時餘6，被101 除時餘
50。求n的最小可能值。
Let n be a positive integer which leaves a remainder of 3 when divided by 7, a remainder of 5
when divided by 11, a remainder of 6 when divided by 13 and a remainder of 50 when divided
by 101. Find the smallest possible value of n.17. 設x、y、z 為正整數，其中 x < y < z 且 x + y + z =155。敏儀計算了x 和y 的最大公因
數、x 和z 的最大公因數及y 和z 的最大公因數，並發現這三個最大公因數當中，其中
一個等於另外兩個之積。求x所有可能值之和。
Let x, y, z be positive integers such that x < y < z and x + y + z =155 . Mandy computes the
H.C.F. of x and y, the H.C.F. of x and z as well as the H.C.F. of y and z. She finds that among
these three H.C.F.s, one is equal to the product of the other two. Find the sum of all possible
values of x.18. 一個袋子中共有100枚硬幣，其中每個都是二元、五元或十元硬幣。若這100個硬幣共
值n元，問n有多少個不同的可能值？
There are 100 coins in a bag, each of denomination 2 dollars, 5 dollars or 10 dollars. If the
total value of the 100 coins is n dollars, how many different possible values of n are there?
培正數學邀請賽
2012-05-06 6:55 am
回答 (2)
2012-05-07 12:09 am
✔ 最佳答案
12. x=[x]+{x}2[x]{x}+4=3{x}+3x=>2[x]{x}+4=3{x}+3[x]+3{x}(2[x]-6){x}=3[x]-4{x}=(3[x]-4)/(2[x]-6)Hence 1> (3[x]-4)/(2[x]-6)≥0Note [x]≠3if [x]>3 then 2[x]-6>3[x]-4≥0=>-2>[x]≥4/3 (no solution)if [x]<3 then 2[x]-6<3[x]-4≤0=>-2<[x]≤4/3Hence [x] can be 1,0 or-1When [x]=1,{x}=1/4 so x= 5/4When [x]=0,{x}=2/3 so x= 2/3When [x]=-1,{x}=7/8 so x=- 1/8Sum= 5/4+2/3-1/8=43/24 16. Note all letters in below are integersLet n=7a+3=11b+5=13c+6=101d+507a+3=11b+5=>7a=11b+27a≡2 (mod 11)…(1)11a≡0 (mod 11)…(2)(2)-(1)=>4a≡-2 (mod 11)=>8a≡-4 (mod 1)…(3)(3)-(1)=>a≡-6 ( mod 11)≡5 (mod 11)So a=11p+5 and n=77p+38…(4) 13c+6=101d+50=>13c=101d+4413c≡44 (mod 101)…(5)101c≡0 (mod 101)…(6)7×(5)=>91c≡308 (mod 101)≡5 (mod 101)…(7)(6)-(7)=>10c≡-5 (mod 101)=>100c≡-50 (mod 101)…(8)(6)-(8)=>c≡50 (mod 101)So c=101m+50 and n=1313m+656 77p+38=1313m+656=>77p=1313m+61877p≡618 (mod 1313)…(9)1313p≡0 (mod 1313)…(10)17×(9)=>1309p≡10506 (mod 1313)≡2 (mod 1313)…(11)(10)-(11)=>4p≡=-2 (mod 1313)=>1312p≡-656 (mod 1313)…(12)(10)-(12)=>p≡656 (mod 1313)n=77(1313q+656)+38=101101q+50550minimum n=50550 17.x<y<z and x+y+z=155y>x=>y≥x+1 and z>y=>z≥y+1≥x+2x+y+z=155=>x+x+1+x+2<155=>x< 152/3=>x≤50If x is even, then one of (y,z) is even and one is odd, say y is evenHCF of y,z must be odd, HCF of x,y is even and HCF of x,z is oddproduct of any 2 HCF' s cannot equal to the third oneHence x must be oddSo x can only be 1,3,5,7…,49.For all these values, we can always find odd values y and z so that they are all relative primeSum of all possible values of x=1+3+5+…+49=25×25=625 2012-05-06 16:09:40 補充：
18. Let the number of 2 dollars, 5 dollars and 10 dollars coins be x,y and z:
0≤x,y,z≤100,x+y+z=100
Then n=2x+5y+10z
2012-05-06 16:10:55 補充：
For some values of (x,y,z ) the calculated n may be equal to that of (X,Y,Z )
that is 2x+5y+10z=2X+5Y+10Z
2(x+y+z )+3+8=2(X+Y+Z )+3Y+8Z
200+3y+8z=200+3Y+8Z
3(y-Y )=8(Z-z) where 3 and 8 are relative primes
2012-05-06 16:11:18 補充：
For all values of y≥8, we can always find Y=y-8 and Z=z+3 to make n the same
So all these y can be removed in the process of finding unique n
For y=7, (x,z) can be (0,93),(1,92),…(93,0) 94 choices
For y=6, (x,z) can be (0,94),(1,93),…(94,0) 95 choices
…
For y=0, (x,z) has 101 choices
2012-05-06 16:11:28 補充：
So the number of unique n=94+95+…+101=780
2012-05-07 2:52 am
可否再幫我 http://hk.knowledge.yahoo.com/question/question?qid=7012050500826
2012-05-06 18:53:43 補充：
仲有 http://hk.knowledge.yahoo.com/question/question?qid=7012050500803
2012-05-06 18:53:43 補充：
仲有 http://hk.knowledge.yahoo.com/question/question?qid=7012050500803
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