Preliminary round questions of the Hope Cup

1.Let n+20=x 2, n-21=y 2, and subtract the two formulas to get x 2-y 2=41, i.e. (x+y)(x-y)=41=41*1 so x=21 y=20 so n=441

2."A takes 4 minutes longer to get from A to C", the question is less conditional, right? Yes or no"A takes 4 minutes longer to get from A to C than C to B", otherwise the question will not work.

If the topic is as I described.

Then let A walk x meters per minute, then B's speed is (x + 30) meters, and let B take t minutes to walk from B to C.

Then there is a system of equations (xt+ and.

Solving the system of equations yields x=90, t=

So the time it takes for A to go from A to C is (minutes, and the distance from AB is (m.

1.Let n+20=p2 n-21=q2

Then n+60-n+21=41=(p+q)(p-q)41 is a prime number p-q=1 p+q=41 p=21 q=20 n=4212Suppose that the velocity of A is xm min then B is x+30 and then your problem is a little unclear, but it is roughly the solution of the system of equations.

(1500+317) (1500+1697)=1817 3197= Pre-Classical period accounted for the entire Mayan culture.

100%: x=92%:(3 and 5 6), x=(23 6) hours and 10 minutes.

4+1 6-3-5 6=1 3=20 minutes. 20 minutes passed between the time the computer was turned on and 92% battery remained.

19+a)/(31+a)=3/4，93+3a=76+4a，a=17.

a+b+c)/3+d=212

a+b+d)/3+c=184

a+c+d)/3+b=172

b+c+d)/3+a=200

Solution: a+b+c+d=384, (a+b+c+d) 4=96 (min)

1/2-1/3=1/6.Maximum value of 1/A minus 1/b: 1 6

Passing: 1 6 + 1 2 + 1 5 = 13 15;

Number of failers: 1-13 15=2 15

The ratio of the number of passers to the number of failings = (13 15) :(2 15) = 13:2=:1.

1. a+b+c+ab+bc+ac+abc=2006a(1+c)+ab(1+c)+b(1+c)+c=2006(1+c)(a+ab+b+1)=2007

1+c)(1+b)(1+a)=2007

2007 short division yields 3,3,223

So abc is 2, 2, 222

Area 8882. You check the ciphertext calculation method again, followed by x1+......That was a mess.