Question
Question: find number of integerd solutions of \(x \cdot y \cdot z \cdot w \cdot = 2^{10} \cdot 3^5 \cdot 5^8...
find number of integerd solutions of
x⋅y⋅z⋅w⋅=210⋅35⋅58⋅79

4651046400
Solution
We are to count the number of 4‐tuples of integers (x,y,z,w) (which may be positive or negative) satisfying
x⋅y⋅z⋅w=210⋅35⋅58⋅79.Step 1. Count the ways to distribute the prime factors (ignoring signs):
For any prime p with exponent r in the number, we assign its exponent among the four variables by solving
a1+a2+a3+a4=r,ai≥0.The number of solutions is (4−1r+4−1)=(3r+3).
- For 210: (310+3)=(313)=286.
- For 35: (35+3)=(38)=56.
- For 58: (38+3)=(311)=165.
- For 79: (39+3)=(312)=220.
Thus, the total number of ways (for the absolute values) is:
286×56×165×220.Step 2. Include the sign distributions:
Since the product x⋅y⋅z⋅w is positive, an even number of the variables must be negative.
For 4 variables:
- 0 negatives: (04)=1 way,
- 2 negatives: (24)=6 ways,
- 4 negatives: (44)=1 way.
So, there are 1+6+1=8 ways to assign signs.
Step 3. Multiply everything together:
The total number of solutions is:
8⋅(286×56×165×220).Let’s compute step‐by‐step:
- 286×56:
286×50=14300,
286×6=1716,
Sum = 14300+1716=16016.
- 165×220:
165×200=33000,
165×20=3300,
Sum = 33000+3300=36300.
- Now, 16016×36300:
First, note that 36300=363×100.
Calculate 16016×363:
- 16016×300=4804800,
- 16016×60=960960,
- 16016×3=48048,
Sum = 4804800+960960+48048=5813808.
Thus, 16016×36300=5813808×100=581380800.
- Finally, multiply by 8:
581380800×8=4651046400.
Final Answer:
4651046400