Junior 1 Math Fun Contest 20

The following poem is from the work of the Qing Dynasty mathematician Xu Ziyun, please count how many monks are in the poem?

The towering ancient temple is in the clouds, and I don't know how many monks are in the temple.

Three hundred and sixty-four bowls, see if you run them out.

Three people eat a bowl together, and four people eat a bowl of soup.

May I ask Mr. who is calculating, how many monks are in the temple?

1. Assuming that the question is x, then x = (1+2+3+4+5+6+7+8+9+.)2010）%9 = 6

As for why think for yourself.

2. Suppose 3xy = 100x+y

Two-digit integer solution of y = 100x (3x-1) x = 17, y = 34

So "Pencil Oak" = 173

3. The number chosen is this (1,2,3,4, 9,10,11,12, . 2009,2010）= 2008/2 +2 = 1006

Question 1: Because it is true that x takes any value, let x=10 first, and eliminate c, and obtain: (10+a)(10+b)=-11

Since a, b are both integers, so 10+a, 10+b are also integers, so 10+a, 10+b are divisors of -11. Here a and b have the same status, so only two cases need to be considered: 10+a=-1, 10+b=11; 10+a=1,10+b=-11.

So a=-11, b=1 or a=-9, b=-21

Let x=0 again, so that the polynomial calculation becomes a mononomial calculation, and the original formula is ab-10c=-11

Substituting a=-11, b=1 into the above equation yields: c=0

Substituting a=-9, b=-21 into the above equation yields: c=20

So c = 0 or c = 20

Question 2: You only need to prove DAC Fab and then get AC=AF. This results in acf=1 2 acb=45°

The process is as follows: Extend AB and DE intersect G.

d+∠g=90°,∠ebg+∠g=90° ∴d=∠ebg

EBG = ABF (equal to the vertex angle) D = ABF

dab=∠caf=90° ∴dac=∠baf

ad=ab ∴△dac≌△fab（asa）

ac=af ∵∠caf=90° 。acf=1/2∠afc=45°

CF bisects ACB

Solution: Known (x+a)(x+b)+c(x-10)=(x-11)(x+1).

x 2 + (a + b + c) x + ab-10c = x 2-10x-11

(a+b+c+10)x=10c-ab-11

Since the equation holds for any x, the corresponding coefficients of x are equal.

a+b+c+10=0

10c-ab-11=0

i.e. a+b=-10-c

ab=10c-11

So a, b are the two integer solutions of the equation x 2 + (10 + c) x + (10-11) = 0.

So the discriminant formula =(10+c) 2-4(10c-11)=(c-10) 2+44 is the perfect square.

Let (c-10) 2+44=d 2 and d>0 then d 2-(c-10) 2=44

i.e. (d+c-10)(d-c+10)=1 44=2 22=4 11

Since c and d are both integers, we get twelve sets of equations.

d+c-10=44, d-c+10=1 or d+c-10=22, d-c+10=2 or d+c-10=11, d-c+10=4

d+c-10=1, d-c+10=44 or d+c-10=2, d-c+10=22 or d+c-10=4, d-c+10=11

d+c-10=44, d-c+10=1 or d+c-10=22, d-c+10=2 or d+c-10=11, d-c+10=4

d+c-10=1, d-c+10=44 or d+c-10=2, d-c+10=22 or d+c-10=4, d-c+10=11

The integer solution is c=20, d=12 or c=0, d=12

Therefore, all the c values that satisfy the question are c=0 or c=20

1.Simplification x 2+(a+b+c)x+ab-10c=x 2-10x-11

So, a+b+c=-10, ab=-11, and a, b are integers, we get a=-11, b=1, c=0 or a=1, b=-11, c=0 or a=11, b=-1, c=-20 or, a=-1, b=11, c=-20

There is no picture for question 2.