Let a , b , c and d be non-negative integers satisfying a + b + c + d = 3 6 0 , and that b , c , d are multiplies of 2, 3, 5, respectively.
Let
P
denote the total number of solutions satisfying the condition above, and
Q
denote the total number of solutions satisfying the condition above with an additional constraint of
a
=
0
.
Find the sum of the first 7 digits of P Q after the decimal point.
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Yes, there's a technique to determine the exact probability of 2 7 1 2 4 3 2 2 2 1 , but the working gets very tedious.
First, your question simplifies to 2 B + 3 C + 5 D = 3 6 0 . So we want to find the value of coefficient of x 3 6 0 in the Maclaurin series of ( 1 − x 2 ) ( 1 − x 3 ) ( 1 − x 5 ) 1 . And it turns out that it's 2221.
Next, we find the total number of non-negative integer triplets ( B , C , D ) satisfying the given equation 2 B + 3 C + 5 D = 3 6 0 . By using linear Diophantine equations techniques , in particular, by applying the Bezout's Identity . And this comes out to be 271243.
Take their ratio and you get your answer.
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Amazing. I need to spend some time digesting this. Btw, did you compute the coefficient of X^360 using a software.
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Yes. Otherwise, it gets super duper tedious. Your question is best left for a computer to solve, or for someone who has nothing else better to do.
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