6th Grade Olympiad Math Problems Olympiad Math Olympiad 6th Grade

Fill an average of 5 boxes, and finally have an extra one, indicating that the single digit is 1 or 6 If you should use 6 boxes, and finally there are four more, it means that the single digit is an even number, so the single digit is 6 hundred digits, the single digit is 6, and the single digit is 6, and it is between 150-200.

Take 156,156 6=26 divisible, which is not in line with the topic.

166÷6=27...4 166÷7=23...5 Fit the topic.

So there are 166 parts in total.

If you switch to 6 boxes, you end up with 4 more; If you change to 7 boxes to be evenly packed, there will be 5 more in the end", that is, if you use 6 or 7 boxes to pack it evenly, it is 2 pieces worse, which is 2 less than the multiple of 42, and it can only be 166.

Pack an average of 5 boxes, with an extra one at the end"Don't, too. What kind of math problem do you think this is?

If the words "between 150,200" were replaced by "not more than 210", the answer would still be 166.

Divide by 5 and the remainder of 1 is definitely 1 or 6

Divide by 6 and the remainder is 4, and the last digit cannot be 1, it must be 6

Only 166 176 186 196150 to 200 are multiples of 7, there are 21 to 28, 1 to 8 7 single digits plus 5 to get 6, the number is 3, and the multiple is 23

23 times is 166

This question is actually "A certain number is divided by 5, and the remainder is 1; Divided by 6, 4 remains; Divided by 7, the remaining 5". The three remainders are all different from each other, what should I do? If we take a closer look at these conditions, it is not difficult to find that the last two of them can be restated as:

A certain number is divided by 6 and is less than 2; It is divided by 7 and is also less than 2. In this way, a breakthrough in solving the problem will be found.

The minimum number that meets the two conditions of "divided by 6 by 2 and divided by 7 by 2" is 6 7 2 40. But "40" is divided by 5 and 1 is not left. This requires adding "42" (the least common multiple of 6 and 7) one by one

40 42 82 (divided by 5 by 2 and rounded).

40 42 2 124 (divided by 5 by 4, rounded) 40 42 3 166 (5 divided by 1, eligible).

Pack an average of 5 boxes, with an extra one at the end, with a single digit of 1 or 6

The least common multiple of 5 and 6 is 30;

1, 3.

2. The total number is combined.

Layer 1 and Layer 2

Layer 3 4 Layer 4 Layer 4 8

Fifth layer 16

Sixth floor 32

There is a regular total number of layers of 2 (n-1).

So layer 10 is 2 9 = 512

3, there is 1 number on the first layer.

There are 3 numbers on the second layer.

There are 5 numbers on the third layer.

The nth layer has 2n-1 numbers.

The first 1 layer contains 1 number, and the first 2 layers contain 4 numbers.

The first 3 layers contain 9 numbers.

There are n*n numbers in the front n layer.

When n=44, 44*44=1936<2013<45*45=2025 so 2013 is in the 45th layer, 2013-1936=77 position, I hope it can help you--

1. If you make up the answer, the cuboid is in line with the topic, so the answer is 32, this is the coefficient of the formula of (1+x) to the nth power, so the tenth layer, n=9, and is the case of x=1, is 2 to the 9th power, 512

3. The last number in each row is a perfect square number, and it can be proved, so the last perfect square number before 2013 is 1936=44 2, and the next 2025=45 2

So 2013 is in line 45, number 77.

Question 1: Imagine a cube and you know that there are only three edges at each vertex.

Question 2: The sum of all numbers in the first is 1, the sum of the numbers in the second is 2, the sum of the numbers in the third is 4, and then 4 corresponds to 8

There's a rule that the number of layers minus one gives you the number of layers multiplied by two to the n-1 power of 2, so the 10th layer is the 9th power of 2, which is 512

Question 3 From the last number, you can see that it is the square number of 1 2 3 4 5, and how many squares is the last number, then it is the first row.

44x44=1936 45x45=2025 So that's the 77th number after the 44th row, so the answer is the 77th position in the 45th row.

1.Three of a kind (cube or cuboid).

2.It is not difficult to conclude from the known numbers: the first row adds = 1, the second row adds = 2, and so on:

..It is then concluded that the sum of all the numbers in row nth is: 2 (n-1), and in row 10 substituting n=10 into the above equation=512

3.The last number in the first row is 1 2, the last number in the second row is 2 2, the last number in the third row is 3 2, and so on, 45 2 = 2025, 2025-2013 = 12, so 2013 is the 13th last number in the 45th row of the graph.

This problem is a Yang Hui triangle, and the Yang Hui triangle can be used to solve the problem.

The title of the book's after-class exercise (4), right?

6 games, a total of 12 points. An average of 3 points per person.

A first, must be greater than 4 points. A maximum of 6 points is awarded.

At the end, it must be less than 3 points. A maximum of 2 points and a minimum of 0 points are awarded.

A + D has a maximum of 8 points and a minimum of 4 points.

So B + C has a maximum of 8 points and a minimum of 4 points.

B = C, so B has a maximum of 4 points and a minimum of 2 points.

When B is 2 minutes, D must take the maximum value of 2, but B should be greater than D, so it is not true.

When B is 4 minutes, A must take the minimum value of 4, but B should be less than A, so it is not valid.

B can only take 3 points.

Test: A 4 points 2 wins and 1 loss, 1 win 2 draws B 3 points 1 win, 1 draw and 1 loss, 3 draws.

C 3 points 1 win, 1 draw and 1 loss, 3 draws.

D 2 points 1 win 2 losses, 1 loss 2 draws A 5 points 2 wins 1 draw, 2 wins 1 draw B 3 points 1 win, 1 draw and 1 loss, 3 draws.

C 3 points 1 win, 1 draw and 1 loss, 3 draws.

Ding 1 point, 1 draw, 2 losses, 1 draw, 2 losses, 6 points, three wins.

B 3 points, 1 win, 1 draw, 1 loss.

C 3 points 1 win, 1 draw and 1 loss.

Ding 0 points and three losses.

A 6 3 wins.

B C 3 1 wins, 1 draw, 1 loss.

Ding 0 3 losses.

I'm in a hurry, so I wrote it simply, and I can list charts.

3 points, because it is a total of six rounds, so there are 12 points in total. Then, A cannot be greater than 6 points, and B cannot be greater than 3 points. So it's 3 points.

80 grams of salt water with a mass of 20%. How many grams of salt do I need to add to turn this brine into 75% wt?

20% salt water 80 g of which the mass of water is 80* (1-20%), i.e. 64g

Turning into table brine with a mass fraction of 75% indicates that the mass of brine is 64 (1-75%)=256 (g).

The amount of salt added = 256-80 = 176 (g).

The road repair team planned to build a road in 4 days, and the first day was completed 4 20 of the total. The second day was completed on the first day of 4 5, the first.

The ratio of meters to three or four days is 3:2 ...On the fourth day, 146 meters were repaired, which was just the completion of the task. Find out how many meters the road is long.

The remaining amount in three or four days is (1-4, 20-4, 20*4, 5), that is, 16 25, so on the fourth day, 16 25*2 (3+2) of the total amount is completed

i.e. 32 125 so the road length = 146 (32 125) =

Children go up from the first floor to the fourth floor is 3 floors, 6 3 = 2 minutes, children go up to the first floor for 2 minutes, adults go upstairs twice as fast as children, 2 2 = 1, adults take 1 minute to go to the first floor, adults from the first floor to the sixth floor, is 5 floors, 1 * 5 = 5 minutes.

So 5 minutes.

The speed is 2 times that of a child, the child actually climbed 3 floors from the first floor to the fourth floor, and the adult climbed 5 floors from the first floor to the 6th floor, so it takes 6*5 3=10 minutes.

It takes 6 4 * 6 = 9 minutes for a child to go up to the 6th floor, and the speed of an adult is 2 times that of a child, and the time is naturally 4 and a half minutes.