The most difficult problem to find the law has a solution

25 times 8 = ?25 times 12 = ?16 times 25 = ?25 times 24 = ?28 times 25 = ?32 times 25 = ?

36 times 25 = ?36 times 25 = ?25 times 44 = ?18 times 25 = ?25 times 52 = ?56 times 25 = ?

Each of these answers is 100 more than the previous one

Reason: (x+4)*25=25*x+100

16 times 125 = ?125 times 24 = ?

Each of these answers is 1000 more than the previous one

Reason: 125 (8 + x) = 125 x + 1000

25 times 8 = 25 * 4 * 2 = 200

25 times 12 = 25 * 4 * 3 = 300

16 times 25 = 25 * 4 * 4 = 400

20 times 25 = 25 * 4 * 5 = 500

25 times 24 = 25 * 4 * 6 = 600

28 times 25 = 25 * 4 * 7 = 700

32 times 25 = 25 * 4 * 8 = 800

36 times 25 = 25 * 4 * 9 = 900

40 times 25 = 25 * 4 * 10 = 1000

25 times 44 = 25 * 4 * 11 = 1100

28 times 25 = 25 * 4 * 12 = 1200

25 times 52 = 25 * 4 * 13 = 1300

56 times 25 = 25 * 4 * 14 = 1400

16 times 125 = 25 * 4 * 15 = 1500

125 times 24 = 25 * 4 * 16 = 1600

1. Bu Roll, (Eggplant Bureau.)

Answer Analytic surplus: Four numbers in a group, based on the first group and twice as many as the second group, but in reverse order. The third group is tripled, and the order becomes positive, so the first one is empty:

3 9 = 27, the second empty: 16 3 = 48

Watchtower Lord!

can be found. Numbers are made.

1, 3, 5, 7, 9

and is growing regularly.

That is, the order of 2n+1, then You Nian judged that the increase of n to the first is close to 2008.

So the number of sons is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19... 2n+1 then add these numbers together.

2n+1+1)+(2n+2) Because n=0 there is a number, 2n+1 is 2n+2 numbers.

2（n+1）（n+1）<=2008

n is an integer. Not rounded).

n = 30, then 2(n+1)(n+1) = 1992, because the next set of maximum numbers is 31

Then check 2008-1992=16, so the last report is 16

x+y=xy=m, which means that x,y are the two roots of the equation x 2-mx+m=0.

There are two roots of the equation vertical plum, using the root finding formula x=m 2+sqr(m 2-4m) 2, y=m 2-sqr(m 2-4m) 2

All you need is m>=4 or m<=0 in the group.

For example, when m=0, x=y=0

m=6, x=3+sqr(3), y=3-sqr(3), and so on.

Note: sqr(3) indicates the remainder or 3 of the root number

Five-of-a-kind: 16 parts;

n: [n(n+1) 2]+1;

Regularity: 1234 5....n add 1 each time

16 Add 1 more each time than the last, that is, 2, 3, 4, 5....

Consider 1+2+3+4+...

The first n items and. n(n+1)/2=2008

n is an integer. The solution is n=63

Assuming there are 62 in front of , 62*(62+1) 2=1953 should be 63 next

So there are 62

I did this yesterday and no one could do it at the time. Too off-the-slew, don't do it.

If my guess is correct, it should be like this: when the numbers of each figure are added and summed, it is 14;13；9；9+？；13+?;

So let's guess that the sequence is 14 13 9 13 14, so the two question marks are 4 1

I don't know if it's correct or not, it's been a long time since I've done this kind of question.

The first empty 20

The second empty 6n

The third empty 6n+2

When n is odd: (n+1) 1+(n+1) 2=3n+3 When n is even: (n+2) 1+n 2 3n+2

When n is odd: (n+1) 1+(n+1) 2=3n+3

When n is even: (n+2) 1+n 2 3n+2

So fill in 6, 8, 12, 14, 18, 20 ,.. in turn6n,6n+2

I used to feel the same way, I used to be good at math, but I had a headache when I saw it. However, after years of hard work in the examination room, I still summed up a little experience. But first of all, I'm only a sophomore junior high school student, and I may not have a lot of questions to find patterns.

One. Look at it individually. Generally, the problem of finding the pattern is related to the number of terms (that is, the number of terms he is).

You just look at them one by one, and generally find multiples or square numbers. I have encountered the question 1 3 8 15 before....Or 2 5 10 17....

What else 1 3 7 15 31....None of this seems to matter, but it's actually the square of the 123456, or the power of 2 or something. If you do more, you will feel like it again--

It's the truth. Two. Look at it together in a column. That is, there is a correlation between each number, for example, the second number is how many times or how many times the first number is adding or subtracting.

I have been exposed to this kind of junior high school very little, I often encountered it in elementary school, and I used to do Olympiad math problems as if I had encountered 1357 items of number is a kind of law, 2468 is a kind of law - this kind of question is a bit old-fashioned, challenging students' brain cells.

It's pretty much the same.,If you don't understand, you can ask me -- it's purely hand-beaten.,Please adopt.。

Simplify, add, subtract, multiply and divide, find the law, and look at the problem.

Dizzy! Then what kind of IQ measurement is to find a pattern! Not so high IQ.