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### Wordpandit Question of the day

Question:

In how many ways 300 can be written as the product of three integers

1. 108
2. 76
3. 18
4. 64

### 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.

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