Mathematics Olympiad questions, examining parity, elementary school Olympiad number theory questions

45 people. Regardless of the result, a game is worth 2 points and the total score is even, excluding 1985

There are x people participating in the gang, and the number of games is x(x-1) and the total score of 2 is x(x-1).

And x squared" x (x-1) > (x-1) squared.

Because 45 squares are greater than 1980, 1982, 1984, and 44 squares are smaller than 1980, 1982, 1984, so n=45

1980 20 people bar.

1) You have to know that every pair of games, whether it is a win or a draw, there will be 2 points, so you can first conclude that this total number of points must be a multiple of 2, which is (1980, 1982, 1984).

2) Now what we have to do is to exclude two of these numbers, and if you look closely, you can see that only 1980 is divisible by 5, so there are 5 times 2 = 10, which means that there are 10 pairs.

n-1+1)(n-1)/2 *2=1980+a

n(n-1)=1980+a

A: There were 45 participants. Correct total score of 1980.

In each chess match, regardless of whether you win or lose a tie, the total score is 2 points. The key is to calculate.

How many games were played. Assuming there are n players, (n-1)n 2 matches are played, and the total score is (n-1)n 2 2=(n-1)n points. Since the total score was an even number, 1985 was excluded, and only 1980 were eligible.

1980=44×45。So there were 45 people in total.

Parity Application Problem Olympiad Problem: There are 9 cups on the table, all mouth up, and 6 of them are "flipped" at the same time each timePlease explain: no matter how many times you go through this "flip", you can't make all 9 cups mouth down.

To make a cup mouth side down, it must go through an odd number of times"Flip".In order for 9 cups to be face down, the sum of 9 odd numbers must be passed"Flip".Namely"Flip"The total number of times is an odd number.

However, it is customary to turn 6 cups at a time, no matter how many times you pass"Flip", the total number of flips can only be an even number. So no matter how many times it passes"Flip"Purity, can't make all 9 cups mouth down. Dividend = 21 40 + 16 = 856.

A: The dividend is 856 and the divisor is 21. ;

1. In each of the following equations, there is at least one odd number and one even number, so how many even numbers are there among the 12 integers?

2. Take out 1234 consecutive natural numbers arbitrarily, is their sum odd or even?

3. A string of numbers is arranged in a row, and their law is: the first two numbers are 1, and from the third number, each number is the sum of the first two numbers. Such as: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ,...

Just ask: How many even numbers are in the first 100 numbers (including the 100th number) of this string?

4. Can 1010 be written as the sum of 10 consecutive natural numbers? If you can, write it out; If not, explain why.

5. Is it possible to divide the 25 natural numbers from 1 to 25 into groups, so that the numbers in each group are equal to the sum of the other numbers in the group?

6. In a chess match, the winner gets 1 point, the loser deducts 1 point, and if it is a draw, both sides get 0 points. There are a number of students competing today, and each of them is playing a game. It is now known that one of the students scored a total of 7 points, and the other Zhaokai Yu student scored a total of 20 points, which shows that there was at least one draw in the course of the game.

7. Write 1,2 ,... on the blackboard, 909, as long as there are two or more numbers on the blackboard, erase any two of the numbers a, b, and write a-b (where a b). Q: Is it odd or even left on the board at the end?

8. Set A1, A2 ,..., a64 is the natural number 1,2,..., 64, so that b1=a1-a2, b2=a3-a4,...，b32=a63-a64;c1=b1-b2，c2=b3-b4，…，c16=b31-b32;d1=c1-c2，d2=c3-c4，…，d8=c15-c16;……

If you keep doing this way, will you end up with an odd or even integer?

The parity of 99 numbers is.

Even, odd, odd, even, odd, odd, even, odd, ......

So there are 33*2=66 odd numbers.

Because 3 times an odd number is an odd number, and 3 times an even number is an even number, when the permutation is a, b, c, d, e.

If a is an even number and b is an odd number, then c must be an odd number (odd number - even number = odd number) Since the even number plus the odd number is an odd number, so d must be an even number, then e is an odd number, so the order of odd and even is.

Even, odd, odd, even, odd, odd...

So the odd number accounts for 2 3 columns

Then the odd number of 99 is 99*2 3=66.

This arrangement has the following rules:

0 (even) 1; (odd) 3*1-0=3; (odd) 3*3-1=8; (even) 8*3-3=21; (odd) 21*3-8=55(odd number).

We can find that this set of numbers is exactly an even number with two odd number cycles, and 69 is divisible by 3 and is completely round, so there are two-thirds of the odd numbers, which is 66.

Answer: (1) Odd Odd = Even.

2) Even number Even phone = even number.

3) Odd Number Even = Odd Number.

4) Odd Odd = Odd.

5) Odd Even = Even.

Odd Odd = Even.

Odd Even = Even.

Even Even - Even.

Odd Odd = Odd.

Odd Even = Even.

Even Even - Even.

Hope to help the landlord.

The description of the conditions is omitted.

Odd function: f(x)=f(-x).

Even function: f(x)=-f(-x).

An even number is divisible by 2.

What can't be is an odd number.

Counter-evidence.

Let abcdef not be even, it is all odd, that is, abcdef is all odd.

Let a=2k +1, b=2k +1, c=2k +1, d=2k +1, e=2k +1, f=2k +1;

where k1 and k6 are all integers.

The equation is: (2k1+1) +2k2+1) +2k3+1) +2k4+1) +2k5+1) =(2k6+1) ;

Square out, subtract 1 from left and right, divide by 4 left and right, and continue: k1*(k1+1)+k2*(k2+1)+k3*(k3+1)+k4*(k4+1)+k5*(k5+1)+1

k6*(k6+1);

Analysis: In the case of k as an integer, k(k+1) is an even number (regardless of whether k is odd or even).

So the left side of the above equation is: even number + even number + even number + even number + even number + even number + 1, which is an odd number;

To the right of the equation is an even number.

The left and right cannot be equal, and the equation cannot be established, so the previous "abcdef is all odd" is wrong and proven.

Solution: The big box contains a total of 1001 + 1000 = 2001 (pieces) of black and white chess pieces.

Because 2 pieces are drawn and 1 piece is put back each time, there is 1 less chess piece for each touch, and after 1999 touches, there are 2001-1999 = 2 (pieces) of chess pieces left.

There are two situations when you touch 2 pieces from inside the box at a time:

1) The two pieces touched are of the same color. At this time, take a black chess piece from the small box and put it into the large box. When the two pieces touched are both black, there is one black piece missing from the box; When the two pieces touched are both white, there is an extra black piece in the box.

2) The two pieces touched are of different colors, i.e. one black and one white. At this time, the white chess pieces taken out should be put back into the big box, and there is one black chess piece missing in the big box.

Combined (1) and (2), each time you touch, the total number of black pieces in the big box is either one less or one more, that is, the parity of the number of black pieces is changed. It turned out that there were 1,000 even black chess pieces in the big box, and after touching 1999 times, that is, after changing the parity of 1999 times, there were still odd black chess pieces. Because there are only 2 pieces left in the box, the last two pieces left are one black and one white.

2.The number of people who hit ** is always even.

is an even number. Simplicity is the truth.

2.Hit ** once, the number of calls between two people = 2, several times, the number of calls = 2 * several = even number, and the person who has played the odd number ** is an even number.

Answer the second question first.

Suppose there are a total of n people who have played a **, that is, a total of a ** has been received, and the total number of hits ** is an even number of 2a

If the number of people who have played even ** is b, the total number must be even.

Then the total number of people who have played the base number should be even.

Then there should be an even number of people who played the number of times**.

So should be even.

Question 1: Pick out 2 pieces each time, put 1 piece back, and finally have 2 pieces left.

Each time you touch it, you either add one sunspot or one less sunspot, that is, the number of sunspots changes the parity by 1999 times.

It turned out to be an even number, and it was changed to a base number, so it was one black and one white.