Wordpandit Question of the day
Question:Â
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.
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)