Making Light: A small puzzle【Puzzle迷◆！时隔三年！为了见你终于做到了！_华晨宇吧_百度贴吧
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【Puzzle迷◆！时隔三年！为了见你终于做到了！收藏
2013年的那个夏天，我知道了你，穿着黑白t，棕色的七分裤，红色的鞋子，黑框眼镜拿着你的琴当然还有架子吃着你爱的橘子讲真那时候真的没想到后来会那么喜欢你，66进20的时候我都快忘记了你，然后我姐说那个傻傻的男的唱了吗(ps: 求不打)我说还没，后来你要一首the kill惊呆了(那段视频看了好多遍呢)但是事情介绍一堆橘子真心有点傻然后记住了你华晨宇20进10somebody that I used to know怎么办从不听英文歌的我爱上了呢?男生学院那个时候每日每夜的看，老心疼花了呢10进9我惊艳了，哥哥的歌真心的很多人翻唱，很多翻唱不好听，花花唱到心坎了不知道这首歌收了多少人的心
小时候和邻家小妹一起玩的游戏
三年一个轮回，我还在
原谅我的错别字，有点激动，上一次的帖子还是三年前的
大家可以补充哦
忍不住了，约起来好吗？
我也记得我高三暑假疯狂的补视频的日子海选就看上了你，但是却觉得错过了很多。三年第一次赴约北京，华晨宇，你太美好。
9进8We are young真正让我难过的是那个时候的短片事情介绍，有时候会怕，晚上会担心有怪兽…但唱歌的时候大主唱的架子一下就出来了，很嗨呢！但是那个时候你应该很不舒服吧，很难受的对不对？8进7亲爱的小孩收割很多姐姐粉这其中就包括了我的两个姐姐，她们说花花就这样，不帅，不高（原谅我们看脸）但就是没办法不喜欢他，看似孤僻的谁都不在乎的人其实比任何一个人都在乎对他好的人，他保护我们的方式很不一般，我想这也是我爱他爱了那么久的原因
说说最喜欢那个时候他说的一个什么字
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为兴趣而生，贴吧更懂你。或Number Puzzles and Sequences
Number Puzzles and Sequences
Number Puzzles and Sequences
The day before yesterday I was 25 and the next year I will be 28. This is true only one day in a year.
What day is my birthday?
He was born on December 31st and spoke about it on January 1st.
Easy Deduction
A teacher thinks of two consecutive numbers between 1 and 10. The first student knows one number and the second student knows the second number. The following exchange takes place:
First: I do not know your number.
Second: Neither do I know your number.
First: Now I know.
What are the 4 solutions of this easy number puzzle?
None of the students can have numbers 1 or 10, since they would guess the other one's number with no problems. I will describe solutions at one end of the interval of numbers 1-10 (the same can be done on the other end).
Information that the second student does not know must be important for the first student. So the first one must expect that the second one has 1 or 3 (if the first one has 2). And as the second student does not know, then he has certainly not 1. So the first pair is 2 and 3.
If the first one had 3, then he would expect the other one to have either 2 or 4. But if the second one had 2 (and the second one would have known that the first one does not have 1), then he would know the number of the first student. However, neither the second student knows the answer - so he has 4. The second pair of numbers is 3 and 4.
Solutions at the other end of interval are 9 and 8 or 8 and 7.
Complex Deduction
This is definitely one of the harder number puzzles on this site.
A teacher says: I'm thinking of two natural numbers greater than 1. Try to guess what they are.
The first student knows their product and the other one knows their sum.
First: I do not know the sum.
Second: I knew that. The sum is less than 14.
First: I knew that. However, now I know the numbers.
Second: And so do I.
What were the numbers?
The numbers were 2 and 9. And here comes the entire solution.
There shall be two natural numbers bigger than 1. First student knows their product and the other one knows their sum.
The sum is smaller than 14 (for natural numbers bigger than 1), so the following combinations are possible:
2 2 ... NO - the first student would have known the sum as well
2 3 ... NO - the first student would have known the sum as well
2 4 ... NO - the first student would have known the sum as well
2 5 ... NO - the first student would have known the sum as well
2 7 ... NO - the first student would have known the sum as well
2 11 ... NO - the first student would have known the sum as well
3 3 ... NO - the first student would have known the sum as well
3 5 ... NO - the first student would have known the sum as well
3 7 ... NO - the first student would have known the sum as well
3 8 ... NO - the product does not have all possible sums smaller than 14 (eg. 2 + 12)
3 9 ... NO - the first student would have known the sum as well
3 10 ... NO - the product does not have all possible sums smaller than 14
4 6 ... NO - the product does not have all possible sums smaller than 14
4 7 ... NO - the product does not have all possible sums smaller than 14
4 8 ... NO - the product does not have all possible sums smaller than 14
4 9 ... NO - the product does not have all possible sums smaller than 14
5 5 ... NO - the first student would have known the sum as well
5 6 ... NO - the product does not have all possible sums smaller than 14
5 7 ... NO - the first student would have known the sum as well
5 8 ... NO - the product does not have all possible sums smaller than 14
6 6 ... NO - the product does not have all possible sums smaller than 14
6 7 ... NO - the product does not have all possible sums smaller than 14
So there are the following combinations left:
2 6 ... NO - it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 (it is impossible to create a pair of numbers from sum 8, so that the product would have an alternative sum bigger than 14 ... eg. if 4 and 4, then there is no sum - created from their product 16 - bigger than 14 - eg. 2 + 8 = only 10)
3 4 ... NO - it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
3 6 ... NO - it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
4 4 ... NO - it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
4 5 ... NO - it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
The second student (knowing the sum) knew, that the first student (knowing the product) does not know the sum and he thought that the first student does not know that the sum is smaller than 14.
Only 3 combinations left:
2 8 ... product = 16, sum = 10
2 9 ... product = 18, sum = 11
2 10 ... product = 20, sum = 12
Let's eliminate the sums, which can be created using a unique combination of numbers - if the sum is clear when knowing the product (this could have been done earlier, but it wouldn't be so exciting) - because the second student knew, that his sum is not created with such a pair of numbers. And so the sum can not be 10 (because 7 and 3) - the second student knew, that the first student does not know the sum - but if the sum was 10, then the first student could have known the sum if the pair was 7 and 3.
The same reasoning is used for eliminating sum 12 (because 5 and 7).
So we have just one possibility - the only solution - 2 and 9.
And that's it.
An easier number puzzle is as follows. Two friends are chatting:
- Peter, how old are your children?
- Well Thomas, there are three of them and the product of their ages is 36.
- That is not enough ...
- The sum of their ages is exactly the number of beers we have drunk today.
- That is still not enough.
- OK, the last thing is that my oldest child wears a green cap.
How old were each of Peter's children?
Let's start with the known product - 36. Write on a sheet of paper the possible combinations giving the product of 36. Knowing that the sum is not enough to be sure, there are two possible combinations with the same sum (1-6-6 a 2-2-9). And as we learned further that the oldest son wears a cap, it is clear that the correct combination of ages is 2-2-9, where there is exactly one of them the oldest one.
What mathematical symbol can be placed between 5 and 9, to get a number greater than 5 and smaller than 9?
decimal point - 5.9
Can you arrange 9 numerals - 1, 2, 3, 4, 5, 6, 7, 8 and 9 - (using each numeral just once) above and below a division line, to create a fraction equaling to 1/3 (one third)?
5-digit Number
What 5-digit number has the following property? If we put numeral 1 in front of the number, we get a number three times smaller, than if we put the numeral 1 behind this number.
Using an easy equation: 3(x+100000) = 10x+1 we find out that the number is 42857.
9-digit Number
Find a 9-digit number, which you will gradually round off starting with units, then tenth, hundred etc., until you get to the last numeral, which you do not round off. The rounding alternates (up, down, up ...). After rounding off 8 times, the final number is . The original number is commensurable by 6 and 7, all the numbers from 1 to 9 are used, and after rounding four times the sum of the not rounded numerals equals 24.
- if rounding changes the next numeral character
10-digit Number
Find a 10-digit number, where the first figure defines the count of zeros in this number, the second figure the count of numeral 1 in this number etc. At the end the tenth numeral character expresses the count of the numeral 9 in this number.
Find a 10-digit number, where the first numeral character expresses the count of numeral 1 in this number, ..., the tenth numeral the count of zeros in this number.
Sum of all numerals must be ten because each numeral stands for the count of other numerals and because this number shall have ten numerals. Beginning to choose reasonable numerals for the first figure you can come across the correct number: .
Find the number if:
The cipher is made of 6 different numerals.
Even and odd digits alternate, including zero (in this case as an even number).
The difference between two adjacent numerals is always greater than one (in absolute value).
The first two numerals (as one number) as well as the two middle numerals (as one number) are a multiple of the last two numerals (as one number).
What is the cipher? There is more than 1 solution.
The possible 2 last numerals are as follows: 03, 05, 07, 09, 14, 16, 18, 25, 27, 29 and 30. At least two multiples less than 100 (this condition is already accomplished), which consist of even and odd numeral (respecting all other conditions) are for 03, 07, 09 and 18 as follows:
03 - 27, 63, 69, 81
07 - 49, 63
09 - 27, 63, 81
18 - 36, 72, 90
There are 5 numbers that can be made of these pairs of numerals to create the cipher: , 816309 and 903618. (If we assume, that also in the number 903618 is accomplished the requirement to alternate even and odd numbers, despite the opposite order.)
The Number Puzzle
The grid below is to be filled with six numbers (3 vertical and 3 horizontal) from the given list. You can use each number more than once. After completing the number puzzle sum up all digits in the grid. This is defined as the score. What is the maximum possible score?
the grid &&&&&&&&&&&&&&available numbers
Example: Using the numbers
78215 twice
The score is 73 (of course other solutions with higher scores are possible).
This one is not verified - I used 2 numbers 39543 and 89398. And this is what the grid looks like:
So the total score is 147.
Master Mind
Find the four-digit number designated by asterisks, given the following:
All four digits of the unknown number are different.
None of the digits is zero.
Each "0" on the right of each four-digit number below indicates that the number has a matching digit in a non-matching position with the unknown number.
Each "+" on the right of each four-digit number below indicates that the number has a matching digit in a matching position with the unknown number.
-------------
Using the numerals 1, 9, 9 and 6, mathematical symbols +, -, x, :, root and brackets create the following numbers:
29, 32, 35, 38, 70, 73, 76, 77, 100 and 1000.
All the numerals must be used in the given order (each just once) and without turning upside down.
I used [brackets] as the symbol for root.
Using four sevens (7) and a one (1) create the number 100. Except the five numerals you can use the usual mathematical operations (+, -, x, :), root and brackets ().
100 = 177-77 = (7+7)x(7+(1:7))
I do not know other solutions.
Rectify the following equality 101 - 102 = 1 by moving just one digit.
Move the numeral 2 half a line up to achieve 101 - 102 = 1.
Number Sequences
There are infinite formulas that will fit any finite series. Try to guess the following number in each sequence (using the most simple mathematical operations, because as I mentioned, there is more than one solution for each number sequence).
1, 4, 9, 18, 35, ?
23, 45, 89, 177, ?
7, 5, 8, 4, 9, 3, ?
11, 19, 14, 22, 17, 25, ?
3, 8, 15, 24, 35, ?
2, 4, 5, 10, 12, 24, 27, ?
1, 3, 4, 7, 11, 18, ?
99, 92, 86, 81, 77, ?
0, 4, 2, 6, 4, 8, ?
1, 2, 2, 4, 8, 11, 33, ?
1, 2, 6, 24, 120, ?
1, 2, 3, 6, 11, 20, 37, ?
5, 7, 12, 19, 31, 50, ?
27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ?
126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ?
4, 7, 15, 29, 59, 117, ?
2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ?
4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ?
, 2387, ? 7238 (moving of numerals)
1, 4, 9, 18, 35, ? 68 (x*2+2, +1, +0, -1, -2)
23, 45, 89, 177, ? 353 (x*2-1)
7, 5, 8, 4, 9, 3, ? 10, 2 (two series - every second number: 7, 8, 9, 10 and 5, 4, 3, 2)
11, 19, 14, 22, 17, 25, ? 20, 28 (two series - every second number: 11, 14, 17, 20 and 19, 22, 25, 28)
3, 8, 15, 24, 35, ? 48 (x+5, +7, +9, +11, +13)
2, 4, 5, 10, 12, 24, 27, ? 54, 58 (x*2, +1, *2, +2, *2, +3, *2, +4)
1, 3, 4, 7, 11, 18, ? 29 (a+b=c, b+c=d, c+d=e)
99, 92, 86, 81, 77, ? 74 (x-7, -6, -5, -4, -3)
0, 4, 2, 6, 4, 8, ? 6 (x+4, -2, +4, -2, +4, -2)
1, 2, 2, 4, 8, 11, 33, ?
1, 2, 6, 24, 120, ? 720 (x*2, *3, *4, *5, *6)
1, 2, 3, 6, 11, 20, 37, ?
5, 7, 12, 19, 31, 50, ? 81 (a+b=c, b+c=d, c+d=e)
27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ? 322, 161 (x*3+1, /2, *3+1, /2)
126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ? 485, 1456 (x/2, *3+1, /2, *3+1)
4, 7, 15, 29, 59, 117, ? 235 (x*2-1, *2+1, *2-1)
2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ? 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5
4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ? 4, 4
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