Ask for help with a few math Olympiad problems

999…9(n9) 999....9 (n 9s) + 1999....9 (n 9s).

999…9 (n 9s) [1000.]0 (n 0)-1]+1999....9 (n 9s).

999…9000…0 (n 9, n 0)-999....9 (n 9s) + 1999....9 (n 9s).

999…9000…0 (n 9, n 0) + [1999.]9(n9) 999....9 (n 9s).

999…9000…0 (n 9s, n 0s) + 1000....0 (n zeros).

1000…0 (2n zeros).

abc=2004

abc)^2=(ab)(bc)(ca)=(2004)^2

Due to symmetry, it is advisable to set ab>0 and c>0

Then because (ab)(bc)(ca) [ab+bc+ca) 3] 3=[(ab+bc+ca) 3] 27

ab=bc=ca, i.e., a=b=c=(2004) (1 3)).

So there is (ab+bc+ca) 3 27(ab)(bc)(ca) 27 (2004) 2

So ab+bc+ca 3 (2004) (2 3).

So the minimum value of ab+bc+ca is 3 (2004) (2 3) = 477

The largest number is (9 9) 9

(A, B and C depart at the same time).

When A and B meet, the distance between A and C is 15 (60 + 40) = 1500 meters, and this ---1500 meters is 1500 meters when A and B meet, and B is ahead of C's 1500 meters, and the time at this time is 1500 (50 - 40) = 150 minutes, and the AB distance is 150 (60 + 50) = 16500 meters.

These two questions are not Olympiad questions at all. Lajin.

1) Find the amount of the wheel dismantling unit 1 = the specific amount divided by the specific amount of the score that should be used to prevent fraud.

22+11) (1-3 7-2 5)=33 (6 35)=33*35 6=km.

2) 150 (5 16) = 480 people.

Because the second test 150 people to include the number of people in the first test.

480-150 = 230 people.

The question is like this, the difference between two barrels of oil is 56-34 = 22 kg, and the large barrel is 3 times that of the small barrel, indicating that the difference is 2 times. So 22 is 2 times, 1 times is 11, so the large barrel is 33 kilograms, and the small barrel is 11 kilograms.

Large barrels: (56-34) (3-1) 3-33 kg.

Keg: 33 3 11 kg.

Leftover oil in a keg.

Barrels of leftover oil.

List the equation: 56-x=3 (34-x).

Solution: x=23

Answer: 33,11

Let B be x years old A = 3x-4

3x-4-7=x+9

3x-x=9+4+7x=10

A and B are now 10 and 26 years old respectively.

If B's age is x years old, then A's age is 3x-4

A's age 7 years ago was 3x-4-7=3x-11, and B's age after 9 years was x+9

then 3x-11=x+9

2x=20x=10

3x-4=30-4=26

Set the current age A x year old and B y years old.

x=3y+4

x-7=y+9

Xie A is 26 years old and B is 10 years old.

Let's say that B's age is x

As far as is known: 3x-4-7 x+9

then x 10, 3x 4 26

A: Now A is 10 years old and B is 26 years old.

Let's talk about finite decimal numbers first, the finite decimal number with the denominator of prime numbers within 2005 is only 5, and these two numbers are multiplied by each other any number of times until the denominator is less than or equal to 2005, and in 2005, there are 10 multiples of 2, 4 multiples of 5, 19 multiples of 2 and 5, adding up to 33 finite decimals.

Then calculate the pure cyclic decimal, the denominator within 2005 is formed by the prime number in addition to 1 2 and 1 5, and the rest of the fractions are pure cyclic decimals, that is, the number of prime numbers less than 2005, which is actually beyond the scope of the Olympiad, this is dizzy, there are roughly more than 200.