The question asks us to find a specific number. We are given that a series of fractional parts multiplied together, when applied to this number, result in 32. We need to reverse this process to find the original number.
Key terms: number, fractional parts, equation.
Let the unknown number be represented by $N$. The problem states:
$ \frac{1}{9} \times \frac{1}{8} \times \frac{1}{7} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times N = 32 $
Multiply all the individual fractions together:
$ \left( \frac{1}{9 \times 8 \times 7 \times 5 \times 4 \times 3 \times 2} \right) \times N = 32 $
Calculate the denominator:
$ 9 \times 8 = 72 $
$ 72 \times 7 = 504 $
$ 504 \times 5 = 2520 $
$ 2520 \times 4 = 10080 $
$ 10080 \times 3 = 30240 $
$ 30240 \times 2 = 60480 $
So the equation becomes:
$ \frac{1}{60480} \times N = 32 $
To find $N$, multiply both sides of the equation by 60480:
$ N = 32 \times 60480 $
Perform the multiplication:
$ N = 1935360 $
Therefore, the number is 1,935,360.
Which fraction among the following is the least ?
\(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)
Find the value of the following expression:
\(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\) is: