The fifth grade Olympiad math problem, the master helps

It is 2008*(1+10001+100010001) 2007*(1+10001+100010001).

It's 2008 2007.

The format of the second question looks too messy.

Luck * Luck At the end is luck, that is, 0, 1, 5, 6 these four kinds.

There are 3 lines in the calculation process, the format is a bit messy, I look like the line of Beijing is gone, so Beijing is 0 and the north * luck = Austria, and because the final result is 7 places, then the north can only be 1, 2, 3, 1 is impossible, because the Olympics will be equal to the luck, 2 * 5 = 0, 2 * 1 = 2, 2 * 6 = 2, 2 * 0 = 0, all do not meet.

So north = 3, so Ao = 8, the final result is 3086?

The third question is to shout 1 first, and the rest of the opponents shout n, the number of your 11-n.

2008 + 20082008 + 200820082008 = 2007 (1 + 10001 + 100010001) 2008 (1 + 10001 + 100010001) = 2007 2008

2.The maximum plus minimum of the number reported by the two is 10 + 1 = 11, to ensure that the number added up each time does not exceed 11,100-11 = 89, divided by 11 and the remainder of 1, the first person to report the number of 1, the second person no matter how many to report, the first person to report the difference between this number and 11, you can ensure that the first person to report the number wins.

The molecule is simplified into 2007+2007 10001+2007 100010001

The denominator is simplified into 2008+2008 10001+2008 100010001

So the final result was 2007 2008

2.Is the Beijing Olympics empty underneath.

40% was dug on the first day

The next day dug 40% *7 12

On the third day, I pressed 1 - 40% -40%*7 12

On the third day, I dug 1- 40% more than the second day, -40%*7 12 - 40%*7 12

The part of the excavation is 92 meters, so the total length is 92 (1 - 40% -40% * 7 12 - 40% * 7 12 ).

I pressed it for a total of 3 days, so I dug an average of 92 ( 1 - 40% -40% * 7 12 - 40% *7 12 ) 3 per day

In summary, the equation is: 92 ( 1 - 40% -40%*7 12 - 40%*7 12 ) 3 = 230 meters.

On the first day, 40% of the total length was dug, on the second day it was 7/12 of the first day = 40% of the total length * 7 12 = percent (70 3), on the third day the whole cloth was dug and on the third day the whole length was dug 1 - 40% -70 3 % = 110 3 % of the total length

It is known that 92 m was dug on the third day compared to the second day ==> total length = 92 (110 3 % 70 3 %) = 690 m.

An average of 230 meters are dug per day.

Let the overall length be a meter.

Then the first day the rice was dug up.

The next day 7 meters were dug.

On the third day, all the digging is done, that is, (m.

On the third day, 92 meters more were dug than the second, ie.

The solution is a = 690 meters.

So digging a 3=230 meters per day on average.

The overall length is x, the first day is, the second day is, and the third day is 7 30x+90, so it adds up.

The solution is x=690 meters, and the average is 690 3=230 meters.

The total length of the canal is 690 meters, and an average of 230 meters are dug per day.

The length of the canal is x meters.

40%x+40%x*7/12+40%x*7/12+92=x2/15x=92

x = 690 digging an average of 690 3 = 230 m per day.

Solution: In 2002 numbers, it can be divided into 3 types: a, by 3 integers; b, divided by 3 by 1; c. 2 divided by 3

Obviously, the sum of any 3 A's, any 3 B's, or any 3 C's numbers is divisible by 3.

The question is converted to find out which of the three types of numbers A, B, and C is the most out of 2002 numbers.

Since 2002 is divided by 3 by 1, it is the most class B, and the number = 2001 3 + 1 = 668

A maximum of 668 such numbers can be selected (the selected numbers are 1, 4, 7, 10, 13,......).1999,2002）

Let's start low.

First of all, the last digit of the original number is 4

Because the new number is 4 times the original number, the last digit of the new number is 6, that is, the last two digits of the original number are 6464 4=256, so the last two digits of the new number are 56, and the last four digits of the original number are 564564 4=2256, so the last three digits of the new number are 256, and the last four digits of the original number are 2564

2564 4 = 10256, so the last four digits of the new number are 0256, and the last five digits of the original number are 02564

02564 4=10256, so the last five digits of the new number are 10256, and the last six digits of the original number are 102564

102564 4=410256, which meets the requirements of the question, so the original number is 102564

Idea: Let this number be 10x+4

I know it by the meaning of the title.

If the number is a two-digit number.

Then 40+x=4(10x+4) x has no solution (no integer solution) if this is a three-digit number.

Then 400+x=4(10x+4)39x=384 x no solution, if it is 4 digits.

Then 4000+x=4(10x+4) 39x=4000-16=3984 has no solution.

If it's 5 digits.

40000+x=4(10x+4)39x=40000-16=39984 No solution.

If it's 6 digits.

400000+x=4（10x+4） 39x=399984 x=10256

So this number is 102564

Let this number be 10x+4

I know it by the meaning of the title.

If the number is a two-digit number.

Then 40+x=4(10x+4) x has no solution (no integer solution) if this is a three-digit number.

Then 400+x=4(10x+4)39x=384 x no solution, if it is 4 digits.

Then 4000+x=4(10x+4) 39x=4000-16=3984 has no solution.

If it's 5 digits.

40000+x=4(10x+4)39x=40000-16=39984 No solution.

If it's 6 digits.

400000+x=4（10x+4） 39x=399984 x=10256

So this number is 102564

102564, either 102564102564 or 102564102564102564, theoretically has an infinite number of primitives.

Also, maybe the title should be changed to: What is the smallest original number?