### 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 = 2^{2} × 3 × 5^{2}

Let xyz = 2^{2} × 3 × 5^{2} ……..(i)

Now x will be of the form 2^{a1} × 3^{b1} × 5^{c1}, y will be of the form 2^{a2} × 3^{b2} × 5^{c2} and z will be of the form

2^{a3} × 3^{b3} × 5^{c3}.

Hence (i) becomes

2^{a1+a2+a3} × 3^{b1+b2+b3} × 5^{c1+c2+c3} = 2^{2} × 3 × 5^{2}

Hence,

a_{1} + a_{2} + a_{3} = 2 which gives us ^{4}C_{2} = 6 solutions

b_{1} + b_{2} + b_{3} = 1 which gives ^{3}C_{2} = 3 solutions

c_{1} + c_{2} + c_{3} = 2 which gives ^{4}C_{2} = 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 ^{3}C_{1} 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.

300 = 2^2 x 3 x 5^2

now no. of ways power of 2 can be distributed on 3 integers is (2+3-1)C2 ways = 4C2 = 6.

no of way power of 3 i.e 1 can be distributed on 3 integers = (1+3-1)C1 ways = 3 ways .

similarly as power of five can be distributed on 3 integers in 6 ways .

hence total no. of ways 6x6x3 = 108 .

1) A ..

1.

180…

Let the 3 numbers be x,y,z.

xyz = 1500 = 2^2*3*5^3

Let x = 2^a3^b5^c

y = 2^d 3^e 5^f

z = 2^g 3^h 5^i

Accordingly a+d+g = 2

b+e+h = 1

c+f+i = 3

Non negative integral solutions for these equations are 4C2*3C2*5C2 = 6*3*10 = 180.

3)18

2)