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2 and 1 2 = 5 2

f(5/2)=2^3*(5/2)^5+2^2*(5/2)^3+2*(5/2)+1

First, deform the equation as follows:

3x^2+8xy-16y^2=4x^2-(x^2-8xy+16y^2)=(2x)^2-(x-4y)^2=8044

Because whether odd or even multiplied by 2 or multiples of 2 is even, the square of odd numbers is odd, the square of even numbers is even, and the subtraction of two even numbers is still even, so there is:

2x is an even number -->2x) 2 is an even number -->x-4y) 2 is an even number -->x-4y) is an even number - >x is an even number (because 4y is an even number).

Then decompose the factor:

3x^2+8xy-16y^2=(x+4y)(3x-4y)=8044

Since x, 3x, and 4y are all even, (x+4y) 2, (3x-4y) 2 are all positive integers, i.e.,

x+4y)/2]*[3x-4y)/2]=8044/4=2011

Since 2011 is a prime number that can only be decomposed into 1*2011, (x+4y) 2=2011, (3x-4y) 2=1

x=1006,y=754

x+y=1760

Wild goats are the heaviest, and they must be the least numerous, so let's make sure first.

Therefore, there are up to 5 wild goats, if there are 5, weight: 18 5 = 90 (catty).

Remaining: 99-90=9 (catty).

If the rest of the arguments are the lightest: 99-5) = catty) 9 catties No.

If there are 4, weight: 18 4 = 72 (catty).

Remaining: 99-72 = 27 (catty).

If the rest are the lightest: 99-4) = 38 (catty) 27 catties No.

If there are 3, weight: 18 3 = 54 (catty).

99-54=45 (jin).

If the rest are the lightest: 99-3) = catty) 45 catties can.

So, wild goats could be 1, 2, 3. It can't be 4 or more. Count them separately.

1. Wild goats are counted as 1, and the rabbit is A, the quail is B, and the turtledove is C

a+b+c=99-1=98 ①

3a+ 3 - Got:

This is an indefinite equation, which originally has an infinite number of solutions, but according to the meaning of the problem, it can only be a positive integer, which is limited.

To get an integer, b can only be in the range of 5 or multiples of 5.

b take 5Bring in c=80, bring b=5, c=80 into a=13

Try the calculation again, and find that b takes 30,Bring in c=54, bring b=30, c=54 into a=14

Rule: B plus 25, you can. 30 + 25 = 55, bring b = 55 into to obtain: c = 28, bring b = 55, c = 28 into a = 15

Add 25, b = 55 + 25 = 80, bring b = 80 in: c = 2, bring b = 80, c = 2 into a = 16

Wild goats are counted as 2, and rabbits are A, quails are B, and turtledoves are C

a+b+c=99-2=97 ①

3a+ 3 - Got:

Same as above, b is taken separately

obtained: b=5, c=86, a=6

b=30，c=60，a=7

b=55，c=34，a=8

b=80，c=8，a=9

Wild goats are counted as 3, and the rabbit is A, the quail is B, and the turtle dove is C

a+b+c=99-3=96 ①

3a+ 3 - Got:

Same as above, b is taken separately

Get: b=5, c=92, a= -1

b=30，c=66，a=0

b=55，c=40，a=1

b=80，c=14，a=2

The front a= -1, a=0 is unreasonable, rounded. There are a total of 10 solutions.

99 animals 99 catties, an average of one catty each; Shouhuai.

A muntjac weighs 18 catties and must be equipped with 34 turtledoves (35 for a total of 35 catties) ......

Or with 4 turtle doves and 25 quails at the same time (30 for a total of 30 catties) ......

Or with 10 turtle doves and 20 quails at the same time (31 for a total of 31 catties) ......

Or with 22 turtle doves and 10 quails at the same time (33 for a total of 33 catties) ......

Obviously, the number of muntjac cannot exceed 3 (the minimum number of animals involved in a muntjac is 30);

The two rabbits weigh 6 pounds and must be equipped with 8 turtledoves (enough for 10 for a total of 10 pounds) ......

Or with 2 turtle doves and 5 quails at the same time (enough for 9 for a total of 9 catties) ......

The above combinations remove the complete lack of a certain animal, there are many possibilities:

The first possibility: 3 combinations, a total of 3 muntjac, 2 rabbits, 20 turtle doves, 75 quails meet the requirements;

The second possibility: 2 4 combinations, a total of 2 muntjac, 8 rabbits, 40 turtle doves, 50 quails;

The third possibility: 7 combinations, a total of 1 muntjac, 14 rabbits, 60 turtle doves, 25 quails;

The fourth possibility: 2 combinations;

The fifth possibility: 2 ( combined;

Sixth possibility: 3 combinations;

The seventh possibility: 3 3 combinations;

Eighth possibility: 2 5 combinations;

The ninth possibility: 4 combinations;

There are also a number of returnees, and there are other circumstances ......;

Muntjac A rabbit B Quail C Turtle dove D

d must be an even number.

c must be a multiple of 5 with a chain.

b<33

a<6 infers c+d>60

Also know. 18a+3b+；a+b+c+d=99, and then only the stupid friend Sun Sumo can make up.

You can list equations, 18a+3b+; a+b+c+d=99.

Then consider all the only signs of the hand as the finger is a positive integer guess spine.

Draw a thick line from top left to bottom right, and the line should cover the top edge.

Counter-evidence can be attempted.

denote a(n) = a(n) s(n).

Assuming that the infinite summation of a(n) converges, then there is obviously an integer m, and all the natural numbers n and an infinitesimal e, such that .

a(m+1)+a(m+2)+.a(m+n)0, s(n) are monotonically increasing, so s(m+1)a(m+1) s(m+n)+a(m+n)/s(m+n)=[a(m+1)+a(m+2)+.

a(m+n)]/s(m+n)

1-s(m)/s(m+n)

Note that when a(n) diverges, when m is fixed, n increases, and s(m) s(m+n) tends to zero, i.e. the limit to the left of the above equation is 1

Thus contradictory.