Olympiad problems, help, Olympiad problems help

Solution: Factory A accounts for 2/5, and Factory B accounts for 2/7

So the parts are multiples of 35.

It could be 420, 455 or 490.

There are a total of x high-quality products in the whole venue.

If there are 420 of them.

Then 4x 5 = 420 2 5 4 21 + 420 2 7 3 10 = 32 + 36 = 68

x=85, so plant C produces a total of 420 parts (1-2, 5-2, 7)=132 pieces.

Among them, there are 85 high-quality products.

If there are 455, then 4x 5 = 455 2 5 4 21 + 455 2 7 3 10 is not an integer, so it is rounded.

If there are 490, then 4 x 5 = 490 2 5 4 21 + 490 2 7 3 10 is not an integer, so it is rounded.

Therefore, a total of 420 pieces were produced.

Among them, Plant C produced a total of 132 parts, of which 85 were high-quality products.

If you have any questions, you can hi、

Among them, factory A accounts for 2/5, factory B accounts for 2/7, in this batch of parts, factory A produces 21/4 of high-quality products, factory B has 3/10 of high-quality products, then the high-quality products produced by factory A account for 2 5 * 4 21 = 8 105 of the total, and the high-quality products produced by factory A account for 2 7 * 3 10 = 3 35 of the total

Therefore, the high-quality products produced by A and B account for 8 105 + 3 35 = 17 105 of the total, so the total number of parts should be a multiple of 105, and more than 400, and 105 * 4 = 420, which meets the requirements, so the total number of parts is 420.

The high-quality products produced in Field C account for one-fifth of the high-quality products of the whole factory. Then the high-quality products produced by A and B account for 1-1 5=4 5 of the total number of high-quality products

Thus C produced 420*(1-2 5-2 7)=132.

A and B produce a total of 420 * 17 105 = 68 high-quality products.

The total number of quality products is 68 4 5 = 85.

Therefore, the high-quality products produced by C are 85*1 5=17.

The estimation problem is a binary equation: let the total number be x, and the high-quality product is y, then: 2 5*4 21*x+3 10*2 7*x=4 5*y

420 is divisible by 35 and 105! Therefore, according to the title, the C field produced 420 * 11 35 = 132 parts, of which the high-quality products were 85 * 1 5 = 17.

The upstairs have already answered, so I won't talk about it......

A 3-digit number, after the exchange of single digits and hundreds of digits, it is still a 3-digit number, and the difference between it and the original 3 digits gets a single digit is 7, try to find what is the difference between them?

The original question is something like the one above!

The difference between the swapped three-digit number and the original three-digit number is a multiple of 9.

Let's deduce, let the original three-digit number be ABC, and the new three-digit number is CBA, assuming that the original three-digit number is larger, just a c, then there is abc-cba=(100a-a)-(100c-c)(10b-10b)=99(a-c).

Then we can know that the ten digits of the difference must be 9, and the sum of the digits of the multiple of 9 is a multiple of 9, so the hundred digit of the difference is 2, so the difference is 297

Analysis and solution: Let the original three-digit number be ABC, then the three-digit number obtained after the exchange of single digits and hundreds of digits is CBA, and their difference vertical form is:

a b c c b a

As can be seen from the above vertical type, A is larger than C, and the difference between 10 and C and A is 7, so it can be seen that C and A can only be the following cases: 1 and 4, 2 and 5, 3 and 6, 4 and 7, 5 and 8, 6 and 9. In these cases, A and C are 3 different, then, taking into account the borrowing, the difference between the above vertical can only be 297.

There is a problem with the question you provided. Read it for yourself.

There seems to be a problem with the title, please make it clear!

What does the title mean???

Your topic is not valid!!

10÷（1－1/2）÷（1－1/3）÷…1－1/10）＝10÷1/2÷2/3÷…÷9/10

10×2×3/2×…10 9 (a divided by b = a times b) 10 10

100 (only).

x+ sets the middle book to x

The equation will be solved by yourself = =

I won't ask again.

Let the minimum to the maximum be a,b,c,d,e

then a+b+c=3x18=54

e-d=5c+d+e=26x3=78

b-a=7a+e=22x2=44

The solution is a=13, b=20, c=21, d=26, e=31, so e-a=18

The difference between the largest heap and the smallest heap is 18

You don't have to count everything out. Let the second-fewest pile have x apples, and the second-most pile has y apples, then.

18*3+2y+5=26*3+2x-7 (the total number of apples is equal).

The solution is y=x+6, that is, the second most heap is 6 more than the second few, then the difference between the largest heap and the smallest heap is 5+6+7=18.