How to use the simple method to calculate 1234 + 2345 + 3456 + 4567 + 5678 + 6789

How to use the simple method to calculate 1234 + 2345 + 3456 + 4567 + 5678 + 6789

1234 = 12342345 = 1234 + 11113456 = 2345 + 1111 = 1234 + 1111 × 24567 = 3456 + 1111 = 1234 + 1111 × 35678 = 1234 + 1111 × 46789 = 1234 + 1111 × 5, so the result = 1234 × 6 + 1111 × (1 + 2 + 3 + 4 + 5) = 24069 after learning arithmetic sequence: (1234 + 6789) * 6 / 2 = 24069

1.2345+2.3451+3.4512+4.5123+5.1234=?

The sum of each digit is 15,
So the total is 15 × 1.1111 = 11.111 + 5.5555 = 16.6665

1.2345, 2.3451, 3.4512, (), () to find the law

4.5123 5.1234

1234=0 2345=0 3456=1 5678=3 7890=4 6890=5
8887=6
Q: 6678 =?

Kindergarten topic, number circle = 4

1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 32 / 1 + 1 / 128 without 1 / 64

=1/2+1/4+1/8+1/16+1/32+1/32-1/32+1/128
=1-1/32+1/128
=1-4/128+1/128
=1-3/128
=125/128

2005 + 2006 times 20052005 minus 2005 times 20062006 plus 1
How much? We must go through the process!

2005+2006*20052005-2005*20062006+1
=2005+2006*2005*10001-2005*2006*10001+1
=2005+1
=2006

2006*20052005-2005*20062006 =

=2006*2005*10001-2005*2006*10001
=2006*2005*10001-2006*2005*10001
=0

2008 × (2006 / 2007) minus 2005 × (2007 / 2008) 2 + 4 + 6 +. + 2006 + 2008~
Come on,

2008 × (2006 / 2007) minus 2005 × (2007 / 2008)
2008×2006/2007-2005*2008/2007
=2008*(2006-2005)/2007
=2008/2007
2+4+6+.+2006+2008
=(2+2008)*2008/2/2
=1009020

There is a sequence of numbers 1,9,9,19,37,65121223. Each number is the sum of three adjacent numbers in front of him. What is the remainder of the number 2001 divided by three?

It is very simple to list the remainder of each number divided by 3 as follows:
1,0,0,1,1,2,1,1,1,0,2,0,2,1,0,0,1,1.
It can be seen that from item 14, there is a cycle of 14,
Since 2001 = 14 * 142 + 13,
So the remainder of item 2001 divided by 3 = the remainder of item 13 divided by 3 = 2

1. 3, 4, 7, after each number is the sum of the first two numbers, then the remainder of the number 2007 divided by 4 is

The remainder takes 6 as a cycle unit: 1,3,0,3,3,2
2007=6*334+3
So the remainder is 0