Wordpandit Question of the day
Question:
In how many ways 300 can be written as the product of three integers
Answer Options:
- 108
- 76
- 18
- 64
Solution posted below. Click on the button to view the solution.
Answer and Explanation
We have 300 = 22 × 3 × 52
Let xyz = 22 × 3 × 52 ……..(i)
Now x will be of the form 2a1 × 3b1 × 5c1, y will be of the form 2a2 × 3b2 × 5c2 and z will be of the form
2a3 × 3b3 × 5c3.
Hence (i) becomes
2a1+a2+a3 × 3b1+b2+b3 × 5c1+c2+c3 = 22 × 3 × 52
Hence,
a1 + a2 + a3 = 2 which gives us 4C2 = 6 solutions
b1 + b2 + b3 = 1 which gives 3C2 = 3 solutions
c1 + c2 + c3 = 2 which gives 4C2 = 6 solutions
Hence, total solutions are 6 × 3 × 6 = 108.
Now out of these 108 solutions, there are some solutions in which two integers are equal. These are
1 ×1 × 300, 2 × 2 × 75, 5 ×5 × 12, 10 × 10 × 3. Now all these solutions must have occurred 3 times in the above 108 solutions. Hence, there are 108 – 12 = 96 solutions in which x, y and z are distinct.
Now x, y and z can be arranged in 3! = 6 ways. Hence the number of distinct solutions are 96/6 = 16. Hence there are 4 solutions in which two integers are equal and 16 solutions in which all the integers are unequal.
Now in case of 4 solutions, where 2 integers are equal, we have the cases where two integers are negative. For example, corresponding to the solution, 1 ×1 × 300, there are two other cases like
– 1 ×- 1 × 300 and 1 × – 1 × – 300. Hence, there are 4 × 3 = 12 solutions where two of the integers are equal.
Again for every solution where all the three integers are unequal, we can have 2 negative integers. This can be done in 3C1 or 3 ways. For example, corresponding to the solution say 4 × 3 × 25 we have – 4 × – 3 × 25, – 4 × 3 × – 25, 4 × – 3 × – 25.
Hence, corresponding to each solution where x, y and z are distinct, we have 3 more solutions.
Hence, in total we have 16 × 4 = 64 solutions where all the three integers are distinct.
Hence, the total number of solutions is 64 + 12 = 76.