Each number is three times the sum of the first two numbers.
Look at the time and you will know that I am the fastest.
The correct answer to the fourth grade math problem is 20.
The first number 4 4*3= 12.
The second number 6 6*4=24
The third number 7 7*5=35
The seventh number is 15 3*5= 15.
Eighth Number 20 4*5=20
The ninth number 25 5*5=25
Fill in the numbers in the first book of the fourth grade mathematics: fill in the numbers in the first book of the fourth grade mathematics: 65283, 65294, 65308+06.50083.000000000005
It seems that the way to find mathematics in the fourth grade of primary school should be 8, 1 1.
The difference between the last number and the previous number is: 1, 1, 3, 1, 1, 3.
In the fourth grade of primary school mathematics, find a rule to fill in the numbers: 2.8, 7, 17.5, (), () This is more convenient for me.
2.8* 10/4=7
7* 10/4= 17.5
namely
a* 10/4=b
17.5* 10/4=43.75
43.75* 10/4= 109.375
Actually, I'm the same as the one upstairs.
n/2+n*2=5n/2= 10n/4
The solution of the fourth grade mathematics in primary school is 2.5. 10. 17.26.37. ().() 2+3 = 5.
5+5= 10
10+7= 17
17+9=26
26+ 1 1=37
37+ 13=50
50+ 15=65
Add odd numbers in turn ~ ~ ~
Mathematics in the fourth grade of primary school found the rule of filling numbers: 1, 2, 3, 3, 7, 5, 15, 9, 3 1, () Divide the meta-series into two series.
1, 3, 7, 15, 3 1, the difference law of two adjacent numbers: 2, 4, 8, 16, 2 times;
Or the last number is twice the previous number, 1.
2,3,5,9,( )。 The difference law of two adjacent numbers: 1, 2, 4, is also the law of twice.
The difference of the next one should be 8, () should be17;
Or the number at the back is twice that at the front, 1 less.
1,2,3,3, 7, 5, 15,9,3 1,( 17 )
The singular number is × 2+ 1.
Even number is × 2- 1.
What's the number in brackets 1 4 1 29 76 1 () for solving the fourth grade Olympic math problem? () Yes 1364.
Item 3 = Item 2×3- Item1:11= 3× 4-1.
Item 4 = Item 3×3- Item 2: 29=3× 1 1-4.
Item 5 = Item 4× 3-Item 3: 76=3×29- 1 1.
Item 6 = Item 5×3- Item 4: 199=3×76-29.
Item 7 = Item 6×3- Item 5: 52 1=3× 199-76.
Item 8 = Item 7× 3-Item 6: 3× 521-199 =1364.
Find the law of 0, 1, 2, 3, 5, and each item of the fourth grade math problem is equal to the sum of the first two items (starting from 3).
1+ 1=2
1+2=3
2+3=5
So I can see it.
0+ 1 on the second floor cannot be 2.