a*****g
发帖数: 1320 | 1 Thanks a lot!!!!! Will do now! |
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a*****g
发帖数: 1320 | 2 找过了， 只有答案，没有解题步骤，还是不行得求助。 |
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s*****n
发帖数: 742 | 3 友情建议
如果能按照每日一题的形式贴出来，估计有更多的好事者掺乎。 |
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h**6
发帖数: 4160 | 4 没错，以后出题标日期，而不只是1，2，3。还有，要把题目贴上来，不要留链接。 |
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s*****n
发帖数: 742 | 5 最后帮助一下lazy dad
1. 4 possibilities: 0-0, 1-0, 0-1, 1-1
so the answer will be 3/4
2. say have 3i red marbles, total 5i marbles.
3i(3i-1) / 5i(5i-1) = 3/10 => i = 1, so total 2 white marbles.
3. only chance for Manu to win is 1/2 * 1/2 * 1/2= 1/8
so the answer is 1 - 1/8 = 7/8
4. each number has equal 1/6 chance. so the answer should be:
1+2+3+4+5+6 = 21
5。count the area, 1 - (6+4+2)/60 = 4/5 |
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g***y
发帖数: 2942 | 6 说实话这些mathcounts的题都挺简单的，假如都要当家长的辅导，而且家长自己也搞不
定的话，还是不要勉强孩子了。本来在美国又不一定熟悉非得多好，干嘛非要跟自己的
弱项过不去哪。 |
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p**m
发帖数: 3876 | 8 How do you like the MATHCOUNTS program at your kid's school? |
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a*****g
发帖数: 1320 | 9 2 星期前刚开始，每周三早晨上1 小时课， 我刚出差10天回来，只看了他发下来的这
些东东， 还不了解具体的情况。 我同事的孩子在别的学校也有参加，都获得了很好的
名次。 其中一个9年级的自己已经自学了微积分，成功考上了这里最好的贵族学校。 |
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a*****g
发帖数: 1320 | 10 WHAT IS THE PROBABILITY THAT THE LAST THREE DIGITS OF A RANDOMLY SELECTED
PHONE NUMBER ARE ALL PRIME? EXPRESS YOUR ANSWER AS A COMMON FRACTION.
Please help to give the answer. Thanks, |
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Y****a
发帖数: 243 | 11 闲着无聊，来玩玩
3=
b,
expend (x-5)^2+7(x-5)+3=0, you will get the value of b and c
x
expend (x-3)(x-4)(x-5)=0 , you will get the value of b c & d |
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Y****a
发帖数: 243 | 12 3) What is the sum of the reciprocals of the solutions of x3-3x2-13x+15=0?
Express your answer as a common fraction
answer is 3 ?
4) What is the sum of the squares of the solutions of x3-15x2+66x-80=0?
159 ? |
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Y****a
发帖数: 243 | 13 5) The solutions of X3-63x2+cx-1728=0 form a geometric sequence. What is the
value of c?
the solutions are 3,12,48
c=3*12+3*48+12*48=756 |
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T********d
发帖数: 621 | 15 1
b = 5x7 = 35, c = 5^2 x 3 = 25x3 = 75
2.
(x-3)(x-4)(x-5) = 0 ->
b = -12
c = 47
d=-60
3.
if x=a, b , c are the solution
then: (1/a) + (1/b) + (1/c)
= (ab+ac+bc)/(abc)
(ab+ac+bc) = -13
abc = -15
Therefore, solution is 13/15
4.
If x = a,b, c are the solution
Thena^2 + b^2 + c^2 = (a+b+c)^2 - 2 (ab+bc+ac) = 15^2 - 2(66) = 93
5.
if x = a, b, c are the solution
Then abc = 1728
a+b+c = 63
ab + bc + ac = C
and b/c = c/b -> b^2 = ac
-> abc = b^3 =1728 -> b = 12
-> ac = 144 and a+c = 51
C = ab+bc+ac = ... 阅读全帖 |
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a*****g
发帖数: 1320 | 16 Super many thanks you guys great help!! |
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m**********g
发帖数: 114 | 17 好难啊
x2+7x=3=
(b,
and x
ordered
13x+15=0? |
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a*****g
发帖数: 1320 | 18 1.How many of the letters in MATHCOUNTS have a horizontal line of symmetry?
2.Billy takes two marbles, without replacement, from a bag that contains
only six yellow marbles and three blue marbles. What is the probability that
he gets one marble of each color? Express your answer as a common fraction. |
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B*Q
发帖数: 25729 | 19 1. HCO
2. 1/2 (6/9*3/8 + 3/9*6/8) |
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a*****g
发帖数: 1320 | 20 Thanks a lot! The answers are correct! |
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a*****g
发帖数: 1320 | 21 In base b, 321-123=154. What is the value of b? |
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a*****g
发帖数: 1320 | 24 You mean B+1 -3=4? Thanks a lot! |
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G******t
发帖数: 1782 | 25 A b-based number with digits d2,d1,d0
has value d2*b^2+d1*b+d0
So 321-123=154 becomes an equation like this:
( 3*b^2 + 2*b + 1 ) - (1*b^2 + 2*b + 3 )
= 2*b^2 -2
= (1*b^2 + 5*b + 4)
b^2 - 5*b - 6 = 0
(b + 1)( b -6) = 0
b = -1 or b = 6
since the number contains up to 5, so, b >= 6
thus, b = 6 |
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G******t
发帖数: 1782 | 26 I agree, a tutor would help ... |
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f******n
发帖数: 198 | 28 The middle digit is 2 for both operands, so the middle digit in the result
can be either 0 or (b - 1), which gives b = 6. |
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a*****g
发帖数: 1320 | 30 1) If $1000 is invested at 6% annual interest compounded semiannually , how
much interest has been earned at the end of one year?
2) Four children (A,B,C AND D) are arranged in a line. If B and C cannot be
next to each other, in how many ways can the four children be arranged? |
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n**p
发帖数: 1150 | 31
how
1060.9
be
P4,4-2*P3,3 = 12 |
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t**x
发帖数: 20965 | 32 牛比support.
如果微软的support有你这么快速准确就好了。 |
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a*****g
发帖数: 1320 | 33 非常感谢，请问是否可以将第一题的解题步骤写出，再次谢谢。 |
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a*****g
发帖数: 1320 | 36 1.At Marsh Mel Low Middle School 34 students play soccer, 36 students play
football and 29 students play basketball. Of these students 15 play both
soccer and football, 18 play both basketball and football and 13 play both
soccer and basketball. What is the smallest possible number of students
thatplay all three sports?
2. If x and y are two triangular numbers less than 100 that when added,
produce a sum that is a square number, what is the largest possible value of
x or y?
3. Suppose that six b... 阅读全帖 |
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B******1
发帖数: 9094 | 37 比如：中学的 mathscounts 和 science olympiad 俱乐部多半是亚裔。TJ 高中的
2012 新生， 大于60％是亚裔。前两天本版一个 O8 总统与中学数学代表队的合影是一
水儿亚裔。反之，中学橄榄球队，篮球队，甚至啦啦队，很少见到亚裔。
问问你家孩子，在学校是不是有同学说过：XX math is so good since (s)he is
Asian! 更有少数亚裔学生视 A－ 为耻辱。这就跟 DC 版是吃货版一样：Your
reputation proceeds you.
AMO 并不评估学生，而是评估学校，鼓励差校把差生教得更好，缩小族裔学生之间的差
距。这和歧视有本质差别。这就是事实。种族平等是终极目标。种族差异举世共睹，现
实还差得很远。 |
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e***l
发帖数: 710 | 38 0-9里面有4个素数，概率是0.4。最后结果是0.4的三次方=8/125 |
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d********a
发帖数: 55 | 41 应该是6/250吧。
分子是4个质数挑3个进行排列，分母是后三位数字的可能性10的三次方。 |
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x******i
发帖数: 3022 | 44 出这种难度的题，是对考生的侮辱。
3=0. What are the value of b and c? Express your answer as an ordered pair (
b,
x |
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h******g
发帖数: 11250 | 45
6 two digit positive integers are perfect square (4^2 to 9^)
2 two digit positive integers are perfect cube (3^3 and 4^3)
so the probability is 8/90 (totally 90 two digit positive integers)
THE
3/5
3 odd numbers out of 1-5
NUMBER
to be divisible by 4, the first two digits (count from right) has to be
divisible by 4, so only xxx32, xxx72 meet this requirement, therefore. the
fraction is
2*P(3,3)/P(5,5)=2*3!/5!=1/10
need to think about it |
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h******g
发帖数: 11250 | 48
0
there is a rule, if the sum of all digits of a number is 9, this number is
divisible by 9, therefore, it is not a prime.
the proof is following
suppose there is a n-digit number XnX(n-1)....X2X1, where Xi is a integer
between 0 and 9, then
XnX(n-1)....X2X1=sum(Xi*10^(i-1))=sum(Xi*(10^(i-1)-1))+sum(Xi), i=1,2,...,n
the first term is divisible by 9, if the second is divisible as well, the
number has to be divisible |
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h******g
发帖数: 11250 | 49 对任意一个数字 XnX(n-1)....X1，其中Xi为0-9的整数
XnX(n-1)...X2X1=XnX(n-1)...X3*100+X2X1
第一项可被4整除，因为是100的倍数
所以只要第二项可比4整除，整个数字就可以被4整除
剩下的就排列组合而已 |
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k*b
发帖数: 15 | 50 8^2=4^3, can not double count. so 7/90 then. |
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