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# PLS HELP :(

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1. For non-zero real numbers $a,$ $b,$ and $c,$ find all possible values of the expression $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}.$Enter your values separated by commas. For example, if you think the possible values are 4, 5, and 6, then enter you answer as "4,5,6".

2. For an integer $n > 1,$ let $f(n) = \frac{1}{\log_n (10!)}.$Find $f(3!) + f(5!) + f(7!).$

3. Find the positive value of $x$ that satisfies $x \lfloor x \rfloor = 28.$

4. Compute $\sum_{n = 1}^{100} \lfloor \log_2 n \rfloor.$

5. The positive real numbers $a$ and $b$ satisfy $\frac{\log a}{\log b} = \frac{a}{b} = \frac{3}{4}.$Find $a + b.$

6. Find all solutions to $\Bigg| \bigg| \Big| |x - 1| - 1 \Big| - 1 \bigg| - 1 \Bigg| = 0.$Enter your values separated by commas. For example, if you think the possible values are 4, 5, and 6, then enter you answer as "4,5,6".

7. If $\log_2 x,$ $1 + \log_4 x,$ and $\log_8 4x$ are nonzero consecutive terms in a geometric sequence, find all possible values of $x.$ Enter your values separated by commas. For example, if you think the possible values are 4, 5, and 6, then enter your answer as "4,5,6".

8. Find all values of $x$ for which $2^{\log_{10} (x^2)} = 3 (2^{1 + \log_{10} x}) + 16.$Enter your values separated by commas. For example, if you think the possible values are 4, 5, and 6, then enter you answer as "4,5,6".

9. Find the smallest positive integer $n$ that satisfies $\lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n + 34} \rfloor.$Find the smallest positive integer $n$ that satisfies $\lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n + 34} \rfloor.$

10. Find the smallest positive real number $x$ such that $\lfloor x^2 \rfloor - x \lfloor x \rfloor = 6.$

Nov 4, 2023

#1
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Problem 1

Case 1: All three numbers are positive. Then ∣a∣=a, ∣b∣=b, and ∣c∣=c, so the expression is equal to [a/a + b/b + c/c + abc/abc = 1 + 1 + 1 + 1 = \boxed{4}.]

Case 2: Two of the numbers are negative and one is positive. Without loss of generality, assume that a and b are negative and c is positive. Then ∣a∣=−a, ∣b∣=−b, and ∣c∣=c, so the expression is equal to [\frac{-a}{-a} + \frac{-b}{-b} + \frac{c}{c} + \frac{-abc}{-abc} = 1 + 1 + 1 - 1 = \boxed{2}.]

Case 3: All three numbers are negative. Then ∣a∣=−a, ∣b∣=−b, and ∣c∣=−c, so the expression is equal to [\frac{-a}{-a} + \frac{-b}{-b} + \frac{-c}{-c} + \frac{-abc}{-abc} = 1 + 1 + 1 + 1 = \boxed{4}.]

Therefore, the possible values of the expression are 2 and 4.

Nov 5, 2023
#2
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Problem 2

We have \begin{align*} f(3!) &= \frac{1}{\log_3 (10!)} \ &= \frac{1}{\log_3 (10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)} \ &= \frac{1}{\log_3 (3^6\cdot 2^6\cdot 5)} \ &= \frac{1}{6\log_3(3) + 6\log_3(2) + \log_3(5)} \ &= \frac{1}{6+6\cdot 0.699 + 0.431} \ &= \frac{1}{7.130} \ &= \boxed{\frac{100}{713}}. \end{align*}Similarly, \begin{align*} f(5!) &= \frac{1}{\log_5 (10!)} \ &= \frac{1}{\log_5 (5^6\cdot 2^6\cdot 3\cdot 2\cdot 1)} \ &= \frac{1}{6\log_5(5) + 6\log_5(2) + \log_5(3)} \ &= \frac{1}{6+6\cdot 0.431 + 0.176} \ &= \frac{1}{6.607} \ &= \boxed{\frac{150}{991}}. \end{align*}Finally, \begin{align*} f(7!) &= \frac{1}{\log_7 (10!)} \ &= \frac{1}{\log_7 (7^6\cdot 2^6\cdot 5\cdot 3\cdot 2\cdot 1)} \ &= \frac{1}{6\log_7(7) + 6\log_7(2) + \log_7(5)} \ &= \frac{1}{6+6\cdot 0.301 + 0.176} \ &= \frac{1}{6.477} \ &= \boxed{\frac{150}{972}}. \end{align*}

Therefore, f(3!)+f(5!)+f(7!) = 400/2777​.

Nov 5, 2023