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

Solution posted below. Click on the button to view the solution.

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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