hey_thr67 wrote:abcd= 100
how many integral solution are there then.
Okay. I take back the formula.
It's not that simple.
Let me explain it for a general case.
But before that I have to explain the concept of calculating the number of integral solutions for linear equations like (x + y + z = 3) etc. Please try to follow carefully.
Let's say we have to calculate the number of non-negative integral solutions for x, y, and z such that (x + y + z = 2)
For such simple equation we can easily found the solution triplets {x, y, z} as follows
- {0, 0, 2}, {2, 0, 0}, {0, 2, 0}, {1, 1, 0}, {1, 0, 1}, and {0, 1, 1}
i.e. 6 such solutions
But how to calculate them in general?
In this case we can transform the problem as "In how many ways two 1s can be distributed in three buckets without any restrictions?". This means one bucket can receive more than one 1s.
Now visualize this : We have two 1s lying on the ground. And to distribute them in three buckets we are going to put two separators anywhere to separate the two 1s. Let's represent the separators by '|'.
Hence,
- |1|1 ---> Corresponds to {0, 1, 1}
||1 ---> Corresponds to {0, 0, 2}
1||1 ---> Corresponds to {1, 0, 1} etc
Now the problem is reduced to "In how many ways we can arrange two 1s and two separators?" And the answer to the question will be 4!/[(2!)*(2!)] = 6
Similarly if we are asked to find the number of non-negative integral solutions for x, y, z, and w such that (x + y + z + w = 2), the answer will be 5!/[(3!)*(2!)] = 10
Now coming to our problem, we need to find the number of positive integral solutions of abcd = 100 = (2²)*(5²)
It is obvious that a, b, c, and d will be made of 1, 2, and 5 only. Hence, we can assume that they will be of the form,
- a = (2^x1)*(5^y1)
b = (2^x2)*(5^y2)
c = (2^x3)*(5^y3)
d = (2^x4)*(5^y4)
As we have only two 2s and two 5s, we must satisfy the following
- x1 + x2 +x3 +x4 = 2
y1 + y2 + y3 + y4 = 2
We need to find the number of non-negative integral solutions of x1, x2, .., y1, y2... etc for the above equations.
As we have shown earlier number non-negative integral solutions to both of these two equations will be 10.
Hence, number of positive integral solution for abcd = 100 will be 10*10 = 100
Let's cross-check it.
We can group the factor of 100 as any of the following
- 1. {1, 1, 1, 100} --> 4 possible combinations
2. {1, 1, 2, 50} --> 12 possible combinations
3. {1, 1, 4, 25} --> 12 possible combinations
4. {1, 1, 5, 20} --> 12 possible combinations
5. {1, 1, 10, 10} --> 6 possible combinations
6. {1, 2, 2, 25} --> 12 possible combinations
7. {1, 2, 5, 10} --> 24 possible combinations
8. {1, 4, 5, 5} --> 12 possible combinations
9. {2, 2, 5, 5} --> 6 possible combinations
A total of (4 + 5*12 + 2*6 + 24) = 100 combinations
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