High School Mathematics Olympiad Questions Number Theory High School Mathematics Olympiad Competit

Upstairs did it wrong again.

How can there be no solution, (4,4) is a set of solutions.

Proof of correctness:

Suppose (a,b)=d, a=a1*d, b=b1*d, then (a1,b1)=1, [a,b]=a1*b1*d because (a,b)*[a,b]=ab

a,b)+9[a,b]+9(a+b)=7ab

Divide the two sides of the equation by d, 1+9a1*b1+9a1+9b1=7a1*b1*d

1+9a1+9b1=a1*b1*(7d-9)

7d-9=1/a1*b1+9(1/a1+1/b1) <= 1+9(1+1)=19

d<=4

d=4, the inequality takes the equal sign, so there is a unique solution a1=b1=1 a=b=4

d=1, 1 a1*b1+9(1 a1+1 b1)<0 No solution.

d=2, 1+9a1+9b1=5a1*b1

5a1-9)b1=9a1+1 Note that 5a1-9<9a1+1 so b1 cannot be 1

b1>=2, so 2(5a1-9)<=(5a1-9)b1=9a1+1

solution, a1<=19The enumeration is examined one by one, and it is found that there are only two solutions in total.

a1=2, b1=19, in the same way, a1=19, b1=2, a,b are 4 and 38, 38 and 4

d=3, 1+9a1+9b1=12a1*b1 because 3|12a1*b1 so 3|1+9a1+9b1, so 3|1. Contradiction, so there is no solution.

In summary, there are three sets of solutions, a=b=4;a=4, b=38；a=38, b=4

Let a=xn,b=yn,(x,y)=1

a,b)=n,[a,b]=xyn

a,b)+9[a,b]+9(a+b)

n+9xyn+9n(x+y)=7xyn 2 Since 0 is not considered in the least common multiple

So 1+9xy+9(x+y)=7xyn

1+9(x+y)=xy(7n-1)

7n-1=1/xy+9(1/y+1/x)

If x,y>0,7n-1<9

n=1,1+9(x+y)=6xy

x=(9y+1)/(6y-9)=3/2+1/2*[29/(6y-9)]

29 (6y-9) cannot be a positive integer.

So there is no solution.

Let (a,b)=r.[a,b]=abr

Original test = R plus 9 * Abr plus 9 * (A plus B) = ABR 2

Do it along this line.

1.Each AI contains 30 elements; 2.For each pair of i,j:1 i j n,ai aj is a unit set; Null set.

From the three properties, it can be concluded that there are only 30 sets with the same element at most (which can be proved by the counter-proof method), in addition to the same elements in these 30 sets, there are 29 elements in each set, which are different from each other, and it can be seen from property 2 that the 30 elements in the other sets are composed of one of each of the 29 elements in these 30 sets except for the same elements, a total of 29 30.

The maximum positive integer that exists is n=30+29 30

Answer: 871 has a problem, wrong right).

Let a1 a2=a(1,2).

a1∩a3=a（1，,3）

an-1∩an=a（n-1，n）

When a(1,2), a(1,3), a(1,4) ......a(n-1,n) has different sets of elements.

Since AI contains 30 elements, it can be composed of a total of 31 different cell sets, which is the largest at this time.

n max = 30 + 1 = 31

First, let f(x)=x 2+(m-2)x+2m-1 have a solid root between 0 and 1.

Therefore, as long as f(0)*f(1)<0 is satisfied, it means that these two points should have different values of the function for the trace.

Solution: 1 22 3), the root is 7-2, which is between 0 and 1, so it satisfies the question and should be taken, and the answer should be more than 6-2 7

【Proof】Among the 21 numbers, there are four numbers a, b, c, and d, which satisfy a+b=c+d, that is, a-c=d-b, and the problem is equivalent to that there must be four numbers, the difference between two of which is equal to the difference between the other two numbers!

The inversion does not work, that is, within 1 100, 21 numbers can be extracted, so that the difference between any two numbers is not the same! (These differences can be 1, 2, 3, 4, 5,...)

The largest set of two adjacent numbers that are not the same from 1 100 is (the difference between the two neighbors increases in turn).

There are 14 numbers in total, and if there are 21 numbers, you can find four numbers m, n, s, and t, where m-n = s-t

Contradict the counter-setup!

Thus the proposition is proven!

ps: Take 21 numbers arbitrarily, if you want the sum of two pairs to be different, then there must be 21 * 20 2 = 210 kinds (this is the permutation and combination of high school, that is, two numbers are drawn from 21 numbers, there are 210 kinds of drawing), and the sum of the two in 1 100 is nothing more than 3 199 (the minimum sum is 1 + 2, the maximum sum is 100 + 99), a total of less than 200 kinds, less than 210, by the drawer principle, there must be two and is the same!

<> "This Shichang is an additional test question of the National High School Mathematics League a few years ago, I don't know what kind of Qi Hui rot you want.