MHB Find solutions in natural numbers

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The discussion centers on finding natural number solutions for the equation involving a series of fractions that sum to 5. Participants express enthusiasm for the problem, with one user congratulating another for their contributions. The equation features factorial terms in the numerators and products of linear expressions in the denominators. The focus remains on solving the equation accurately within the constraints of natural numbers. The thread highlights the collaborative effort in tackling complex mathematical problems.
anemone
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Find the solutions in natural numbers for the following equation:

$$\frac{10}{x+10}+\frac{10\cdot 9}{(x+10)(x+9)}+\cdots+\frac{10\cdot 9\cdot 8 \cdots\cdot 3 \cdot 2 \cdot 1}{(x+10)(x+9)(x+8)\cdots(x+3)(x+2)(x+1)}=5$$
 
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anemone said:
Find the solutions in natural numbers for the following equation:

$$\frac{10}{x+10}+\frac{10\cdot 9}{(x+10)(x+9)}+\cdots+\frac{10\cdot 9\cdot 8 \cdots\cdot 3 \cdot 2 \cdot 1}{(x+10)(x+9)(x+8)\cdots(x+3)(x+2)(x+1)}=5$$

Let $S$ be the sum Let $x=1$. Then we have
$$S=\dfrac{1}{11}\sum_{n=1}^{10}n=5$$ so $x=1$ is a solution. Increasing $x$ will result in a smaller sum (as the denominators of the fractions will be larger), so the only solution is $x=1$.
 
greg1313 said:
Let $S$ be the sum Let $x=1$. Then we have
$$S=\dfrac{1}{11}\sum_{n=1}^{10}n=5$$ so $x=1$ is a solution. Increasing $x$ will result in a smaller sum (as the denominators of the fractions will be larger), so the only solution is $x=1$.
Awesome! (Bow)

-Dan
 
Good job, greg1313!(Cool)
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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