1 If p and q are positive integers, how many logarithms can satisfy 2p 3q 25? Process.

1. p=2, q=7 is the smallest positive integer solution, and the other solutions are 2+3k, 7-2k, so other positive integer solutions also have p=5, q=5;p=8,q=3；p=11,q=1

2. The three-digit number that meets the requirements may be 1,1,2;1,1,3；1,1,5；1,1,7；Each of these possibilities can form 3 different three-digit numbers, so there are 12 three-digit numbers, and the product of each digit number is a prime number.

3. From the meaning of the title, there must be a number of 2 (because 1 3 + 1 3 + 1 3 = 1, and a b c), if b and c are greater than 3, then b = c = 4, which is not in line with the topic, so there is only one possibility: a = 2, b = 3, c = 6

4. n 5-n = n(n 4-1) = n (n 2 + 1) (n 2-1) = (n 2 + 1) (n - 1) n (n + 1), because (n - 1) n (n + 1) is the product of three consecutive integers, so the maximum value that can be divided by the number series is 6

According to the intention of burying the bent hole of the dry and envy ant question: 5*3=25+4=29, then 30*(5*3)=30*29=900+2=902

So the answer is: 902

Because p q=4 q-(p+q) 2, so:

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So: Answer: 5 (6 4) is 36 lots difference.

Let p and q be two numbers, stipulating: p q=3 p-(p+q) 2,2 4=3*2-(2+4) 2=6-3=3

7 (2 4) = 7 3

A new definition of an arithmetic method for counting rocks.

Because p q=4 q-(p+q) 2, so:

So: Answer: 5 (6 4) is 36

Integer, the left side of the equation is an even number, the right side is an odd number, there is no solution.

Two unknowns, one equation, are indefinite equations, and indefinite equations are solved integer methods:

The coefficients of p,q are copolymeric, and the two sides of the non-coprime are divided by the greatest common factor to make them co-prime.

Write p as an expression of q.

Solve according to the divisible properties of number theory.

The first step of this problem is not passable, the common factor 2 is extracted on the left, and the odd number on the right is not divisible, so there is no solution to this problem.

Hope, thank you.

Left even numbers and right odd numbers are not a no-go. So p,q has and only one is equal to 0...

If (2q-1) p =2,(2p-1) q =2, then 2q-1 =2p,2p-1 =2q, the sum of the two search stools gives 2p+2q-2 =2p+2q is obviously contradictory, so (2q-1) p,(2p-1) q has at least a hunger loss less than 2 let (2q-1) p 2 because (2q-1) p is an integer, and p 1 q is rotten god 1, then (2q-1) p=1, i.e., 2q-1=p and (2p-1) q=(4q-3) q is an integer, i.e., 4-(3). q) is an integer, so q=1 or q=3 and q 1, then q=3 p=5 then q+p=8