Finding number of integral solutions

Question

I am really getting confused in this question.

Number of integral solutions of the equation.

$x_1x_2x_3x_4=770$ options-

1. $2^{11}$

2. $2^{10}$

3. $4^4$

4. $5^5$

I attemtemted it by saying that $7*11*5*2=770$

so solutions should be $4!$ then we can say $77*5*2*1=770$ so $4!$

and $77*10*1*1=770$ so $4!/2$

and then $770*1*1*1=770$ so $4!/3!$

now multiplying them i get $27648$. i am sure i have made a big mistage here but i dont know what. please help

Answer 1

You have four buckets, $x_1$, $x_2$, $x_3$, and $x_4$, and four numbers, $2, 5, 7, 11$, to put in those buckets. As the buckets are ordered, the numbers unique, and you can put several numbers into the same bucket, there's $4^4$ ways of performing this task.

However, you're interested in inegral solution, so you ought to consider positive and negative. Let us choose signs of $x_1, \dots, x_4$. If we have chosen the signs of $x_1$, $x_2$, $x_3$, then the sign of $x_4$ is uniquely determined. There is $2\cdot 2 \cdot 2$ ways of choosing the signs of $x_1, \dots x_3$.

That is, the answer should be $4^4\cdot 2^3 = 2^{11}$.

Answer 2

Case 1: $2\cdot 5\cdot 7\cdot 11 = 770 \to 4! = 24$

Case 2: $10\cdot 7\cdot 11\cdot 1 = 770 \to 4!$

Case 3: $14\cdot 5 \cdot 11\cdot 1 = 770 \to 4!$

Case 4: $22 \cdot 5\cdot 7\cdot 1 = 770 \to 4!$

Case 5: $35 \cdot 2\cdot 11\cdot 1 = 770 \to 4!$

Case 6: $55 \cdot 2\cdot 7\cdot 1 = 770 \to 4!$

Case 7: $77\cdot 2\cdot 5\cdot 1 = 770 \to 4!$

Case 8: $70\cdot 11\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$

Case 9: $110\cdot 7\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$

Case 10: $154\cdot 5\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$

Case 11: $385\cdot 2\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$

Case 12: $770\cdot 1\cdot 1\cdot 1 = 770 \to 4$.

Thus the total number of solutions is: $7\cdot 24 + 4\cdot 12 + 4 = 220$

Answer 3

770=$7\cdot11\cdot2\cdot5$

for, $x_1x_2x_3x_4$we can place 5,7,2 and 11 all in $4 \choose 1$ ways.

So total no of solutions are $4 \choose 1$$4 \choose 1$$4 \choose 1$$4 \choose 1$=$4^4$