💡Okay, so I have this problem where I need to find the smallest whole number that satisfies three different remainder conditions when divided by 4, 3, and 5. Let me try to break this down step by step.First, the problem says that when the number is divided by 4, it leaves a remainder of 2. So, if I call this number 'b', I can write this as:[ b equiv 2 pmod{4} ]Similarly, when 'b' is divided by 3, it also leaves a remainder of 2:[ b equiv 2 pmod{3} ]And finally, when 'b' is divided by 5, it leaves a remainder of 3:[ b equiv 3 pmod{5} ]Hmm, so I have three congruences here. I remember that when dealing with multiple congruences, especially when the moduli are pairwise coprime, the Chinese Remainder Theorem (CRT) can be useful. But wait, are 3, 4, and 5 pairwise coprime? Let me check:- GCD of 3 and 4 is 1.- GCD of 3 and 5 is 1.- GCD of 4 and 5 is 1.Yes, they are all pairwise coprime. So, CRT should apply here, and there should be a unique solution modulo 3×4×5 = 60. That means the smallest such number should be less than 60.But maybe I don't need to go all the way to 60. Let me see if I can solve these step by step.Looking at the first two congruences:[ b equiv 2 pmod{4} ][ b equiv 2 pmod{3} ]Since both have the same remainder, 2, I can think of 'b' as being 2 more than a multiple of both 4 and 3. The least common multiple (LCM) of 4 and 3 is 12. So, 'b' can be written as:[ b = 2 + 12m ]where 'm' is some integer.Now, I can substitute this expression for 'b' into the third congruence:[ b equiv 3 pmod{5} ]Substituting, we get:[ 2 + 12m equiv 3 pmod{5} ]Let me simplify this:[ 12m equiv 1 pmod{5} ]But 12 mod 5 is 2, so:[ 2m equiv 1 pmod{5} ]Now, I need to solve for 'm'. To do this, I can find the multiplicative inverse of 2 modulo 5. The inverse of 2 is a number 'x' such that:[ 2x equiv 1 pmod{5} ]Testing x=1: 2×1=2≡2≠1x=2: 2×2=4≡4≠1x=3: 2×3=6≡1 mod5. Yes, x=3.So, multiplying both sides of the equation by 3:[ 3×2m ≡ 3×1 pmod{5} ][ 6m ≡ 3 pmod{5} ]But 6 mod5 is 1, so:[ m ≡ 3 pmod{5} ]This means that 'm' can be written as:[ m = 3 + 5k ]where 'k' is some integer.Now, plugging this back into the expression for 'b':[ b = 2 + 12m = 2 + 12(3 + 5k) = 2 + 36 + 60k = 38 + 60k ]So, the smallest positive solution is when k=0, which gives:[ b = 38 ]Let me verify this:- 38 divided by 4: 38 ÷4=9 with remainder 2. Good.- 38 divided by 3: 38 ÷3=12 with remainder 2. Good.- 38 divided by 5: 38 ÷5=7 with remainder 3. Perfect.So, 38 is indeed the smallest number that satisfies all three conditions.

💡Okay, so I have this problem: I need to compute the sum of powers of i from i^601 down to i^0, which is 1. That is, I need to find i^601 + i^600 + i^599 + ... + i + 1. Hmm, that's a lot of terms! But maybe there's a pattern or a shortcut I can use instead of calculating each term individually.First, I remember that i is the imaginary unit, where i squared equals -1. So, i^2 = -1. But what about higher powers of i? I think they cycle in a pattern. Let me recall: i^1 is i, i^2 is -1, i^3 is -i, and i^4 is 1. Then it repeats: i^5 is i again, i^6 is -1, and so on. So, the powers of i cycle every four terms: i, -1, -i, 1, and then back to i.If that's the case, maybe I can use this cyclical pattern to simplify the sum. Since the cycle repeats every four terms, perhaps I can break the sum into groups of four terms each and see if there's a pattern or a simplification.Let me write out the first few terms to see if I can spot a pattern:i^601 + i^600 + i^599 + ... + i + 1Since the exponents are decreasing from 601 down to 0, that's a total of 602 terms. That's a lot, but maybe I can find how many complete cycles of four are in 602 terms and then handle the remaining terms separately.First, let me figure out how many complete cycles of four there are in 602 terms. To do that, I can divide 602 by 4. Let's see: 4 times 150 is 600, so 602 divided by 4 is 150 with a remainder of 2. So, there are 150 complete cycles of four terms and 2 additional terms.Each cycle of four terms is i, -1, -i, 1. If I add these four terms together, what do I get?i + (-1) + (-i) + 1 = (i - i) + (-1 + 1) = 0 + 0 = 0Oh, interesting! Each cycle of four terms sums up to zero. That means that the 150 complete cycles will contribute 150 times zero, which is still zero. So, the entire sum reduces to just the remaining 2 terms after the 150 cycles.Now, I need to figure out what those remaining 2 terms are. Since we started at i^601 and went down to i^0, and we've accounted for 600 terms (150 cycles of 4), the remaining 2 terms are i^1 and i^0, which are i and 1, respectively.So, the sum simplifies to i + 1. But wait, let me make sure I'm counting correctly. The first term is i^601, which is the same as i^(4*150 + 1) = i^1 = i. Then, i^600 is i^(4*150) = i^0 = 1. So, actually, the first two terms of the sum are i and 1, and the rest of the terms from i^599 down to i^2 form the 150 cycles, each summing to zero.Therefore, the total sum is i + 1 + (sum of 150 cycles each summing to zero) = i + 1.But wait, let me double-check that. If I have 602 terms, starting from i^601 down to i^0, and each cycle is four terms, then the first term is i^601, which is i, then i^600 is 1, i^599 is -i, i^598 is -1, and so on. So, the first four terms would be i, 1, -i, -1, which sum to zero. Then the next four terms would be i, 1, -i, -1, and so on. So, actually, each group of four terms starting from i^601 would be i, 1, -i, -1, summing to zero.But wait, if I have 602 terms, and each group of four terms sums to zero, then the number of complete groups is 602 divided by 4, which is 150.5. Hmm, that doesn't make sense because you can't have half a group. So, perhaps I made a mistake earlier.Let me recast this. The sum is from n=0 to n=601 of i^n, right? Because i^601 is the highest power, and we go down to i^0, which is 1. So, the number of terms is 602. Now, the sum of a geometric series is given by S = (1 - r^(n+1))/(1 - r), where r is the common ratio, which in this case is i, and n is the number of terms minus one.Wait, maybe I should use the formula for the sum of a geometric series. The sum from k=0 to k=N of r^k is (1 - r^(N+1))/(1 - r), provided that r ≠ 1. In this case, r is i, and N is 601. So, the sum would be (1 - i^(602))/(1 - i).Let me compute that. First, compute i^(602). Since i cycles every 4, 602 divided by 4 is 150 with a remainder of 2. So, i^(602) = i^(4*150 + 2) = (i^4)^150 * i^2 = 1^150 * (-1) = -1. So, i^(602) = -1.Therefore, the sum is (1 - (-1))/(1 - i) = (1 + 1)/(1 - i) = 2/(1 - i). Now, to simplify this, I can multiply the numerator and denominator by the conjugate of the denominator, which is (1 + i):2/(1 - i) * (1 + i)/(1 + i) = 2*(1 + i)/(1 - i^2) = 2*(1 + i)/(1 - (-1)) = 2*(1 + i)/2 = (1 + i).So, the sum is 1 + i.Wait, but earlier I thought it was i + 1, which is the same as 1 + i. So, that matches. But earlier, when I tried to group the terms, I thought the first two terms were i and 1, but actually, the sum starts at i^601, which is i, and then i^600 is 1, and so on. So, the first two terms are i and 1, and the rest form 150 complete cycles of four terms each, which sum to zero. So, the total sum is i + 1.But using the geometric series formula, I also got 1 + i, which is the same. So, both methods agree.Therefore, the final answer is 1 + i, which can also be written as i + 1. But in the box, I think it's better to write it as 1 + i.Wait, but in the initial problem, the sum is written as i^601 + i^600 + ... + i + 1, which is the same as the sum from k=0 to k=601 of i^k, which is a geometric series with first term 1 and ratio i, summed over 602 terms. So, using the formula, we get (1 - i^602)/(1 - i) = (1 - (-1))/(1 - i) = 2/(1 - i) = 1 + i.Yes, that seems correct. So, the answer is 1 + i.But wait, let me check once more. If I compute the sum from k=0 to k=3 of i^k, that's 1 + i + (-1) + (-i) = 0. Similarly, from k=4 to k=7, it's 1 + i + (-1) + (-i) = 0, and so on. So, every four terms sum to zero. Since 602 divided by 4 is 150 with a remainder of 2, that means we have 150 complete cycles (each summing to zero) and two extra terms. The two extra terms are the first two terms of the next cycle, which are i^0 = 1 and i^1 = i. Therefore, the sum is 1 + i.Yes, that makes sense. So, the final answer is 1 + i.

💡Okay, so I have this problem where I'm given a function f(x) that's an odd function defined on all real numbers. For x greater than 0, it's given by f(x) = log base 2 of (x + 1) plus 3x. I need to find the range of real numbers x that satisfy f(x) > -4. The options are A: (-2,2), B: (-1,1), C: (-1, +∞), and D: (1, +∞). Hmm, okay.First, since f(x) is an odd function, I remember that means f(-x) = -f(x). So, if I know f(x) for positive x, I can find f(x) for negative x by just taking the negative of f(-x). That might be useful.Given that for x > 0, f(x) = log₂(x + 1) + 3x. So, for x < 0, f(x) should be equal to -f(-x). Let me write that down:For x < 0, f(x) = -f(-x) = -[log₂(-x + 1) + 3(-x)] = -log₂(1 - x) + 3x.Wait, let me check that. If x is negative, then -x is positive, so f(-x) = log₂(-x + 1) + 3*(-x). So, f(x) = -f(-x) = -log₂(1 - x) + 3x. Yeah, that seems right.Also, since f(x) is odd, f(0) should be 0. That's a property of odd functions. So, f(0) = 0.Now, I need to find where f(x) > -4. So, I need to consider two cases: x > 0 and x < 0.Starting with x > 0. For x > 0, f(x) = log₂(x + 1) + 3x. I need to find when this is greater than -4.So, log₂(x + 1) + 3x > -4.Hmm, let's see. Since x > 0, both log₂(x + 1) and 3x are increasing functions. So, as x increases, f(x) increases. Let me check the value at x = 0. Well, f(0) is 0, which is greater than -4. So, at x = 0, it's already greater than -4.But wait, for x > 0, f(x) is increasing because both log₂(x + 1) and 3x are increasing. So, if f(x) is increasing and f(0) = 0 > -4, then for all x > 0, f(x) > -4. So, for x > 0, all x satisfy f(x) > -4.Now, moving on to x < 0. For x < 0, f(x) = -log₂(1 - x) + 3x. I need to find when this is greater than -4.So, -log₂(1 - x) + 3x > -4.Let me rearrange this inequality:- log₂(1 - x) + 3x > -4Let me bring the log term to the other side:3x > -4 + log₂(1 - x)Hmm, this seems a bit tricky. Maybe I can substitute y = -x, so that y > 0 when x < 0. Let's try that.Let y = -x, so x = -y, and y > 0.Then, the inequality becomes:3*(-y) > -4 + log₂(1 - (-y)) -3y > -4 + log₂(1 + y)Multiply both sides by -1, which reverses the inequality:3y < 4 - log₂(1 + y)So, 3y + log₂(1 + y) < 4Hmm, okay. So, I have 3y + log₂(1 + y) < 4, where y > 0.I need to find the values of y such that 3y + log₂(1 + y) < 4.Let me define a function g(y) = 3y + log₂(1 + y). I need to find when g(y) < 4.Let me analyze g(y). Since y > 0, both 3y and log₂(1 + y) are increasing functions. Therefore, g(y) is also increasing.So, if I can find the value of y where g(y) = 4, then for all y less than that value, g(y) < 4.So, let's solve 3y + log₂(1 + y) = 4.This seems like a transcendental equation, which might not have an algebraic solution. Maybe I can solve it numerically.Let me try y = 1:g(1) = 3*1 + log₂(2) = 3 + 1 = 4. So, y = 1 is a solution.Therefore, since g(y) is increasing, g(y) < 4 when y < 1.Therefore, y < 1, which means x = -y > -1.So, for x < 0, f(x) > -4 when x > -1.Therefore, combining both cases:For x > 0, all x satisfy f(x) > -4.For x < 0, x must be greater than -1.At x = 0, f(x) = 0 > -4.So, overall, the solution is x > -1.Therefore, the range of x is (-1, +∞), which is option C.Wait, let me double-check. For x < 0, we found that x must be greater than -1. So, x is between -1 and 0. For x > 0, all x are allowed. So, combining these, x must be greater than -1. So, the range is (-1, +∞), which is option C.I think that makes sense. Let me just verify with some test points.For x = -0.5, which is greater than -1, let's compute f(-0.5):f(-0.5) = -log₂(1 - (-0.5)) + 3*(-0.5) = -log₂(1.5) - 1.5log₂(1.5) is approximately 0.58496, so f(-0.5) ≈ -0.58496 - 1.5 ≈ -2.08496, which is greater than -4.For x = -1, f(-1) = -log₂(1 - (-1)) + 3*(-1) = -log₂(2) - 3 = -1 - 3 = -4, which is equal to -4, so x = -1 is not included.For x = -2, which is less than -1, f(-2) = -log₂(1 - (-2)) + 3*(-2) = -log₂(3) - 6 ≈ -1.58496 - 6 ≈ -7.58496, which is less than -4, so x = -2 is not included.For x = 1, f(1) = log₂(2) + 3*1 = 1 + 3 = 4, which is greater than -4.So, everything checks out. The range is x > -1, so option C.**Final Answer**boxed{C}

💡Okay, so I have this problem here: it's about a graph of the function y equals a cosine of (bx plus c). They tell me that a, b, and c are positive constants, and the graph reaches its maximum at x equals 0. I need to find the smallest possible value of c. Hmm, okay, let's break this down step by step.First, I remember that the general form of a cosine function is y equals a cosine of (bx + c). In this form, 'a' affects the amplitude, 'b' affects the period, and 'c' is the phase shift. Since they mentioned that the graph reaches its maximum at x equals 0, that must have something to do with the phase shift, right? Because the maximum of a cosine function usually occurs at its starting point, which is when the argument inside the cosine is zero.Wait, let me recall. The cosine function, cos(theta), reaches its maximum value of 1 when theta is 0, 2π, 4π, and so on. So, for the function y equals a cosine of (bx + c), the maximum occurs when bx + c equals 0, 2π, 4π, etc. Since we're told that the maximum happens at x equals 0, let's plug that into the equation.So, if x is 0, then the argument inside the cosine becomes b*0 + c, which simplifies to just c. Therefore, for the function to reach its maximum at x equals 0, c must be equal to 0, 2π, 4π, and so on. But the problem asks for the smallest possible value of c, and since c is a positive constant, the smallest value would be 0.Wait, hold on. Is 0 considered a positive constant? Because sometimes in math, positive can mean strictly greater than 0, excluding 0. But the problem says positive constants a, b, and c. Hmm, so does that include 0 or not? I think in some contexts, positive can include 0, but in others, it's strictly greater than 0. Let me check.Looking back at the problem statement: it says "positive constants a, b, and c." So, in mathematics, positive constants usually mean greater than 0, not including 0. So, if c has to be positive, then 0 is not allowed. That means the next smallest value would be 2π. So, c equals 2π.But wait, let me think again. If c is 0, then the function is y equals a cosine of (bx), which does reach its maximum at x equals 0. So, is 0 allowed? The problem says positive constants, so if c has to be positive, then 0 is not allowed, and the next possible value is 2π. But if c can be zero, then 0 is the smallest possible value.I need to clarify this. Maybe I should consider both cases. If c is allowed to be zero, then the smallest possible c is 0. If c must be strictly positive, then the smallest c is 2π. But the problem says "positive constants," so I think they mean strictly greater than 0. Therefore, c cannot be zero, so the smallest possible value is 2π.But wait, let me double-check. In the context of phase shifts, c is often considered as a phase shift, and a phase shift of 0 would mean no shift. So, if c is zero, the graph starts at its maximum point, which is exactly what we want. So, is 0 considered a valid phase shift? Yes, it is. So, if c is allowed to be zero, then that's the smallest possible value.But the problem says "positive constants a, b, and c." So, if c is zero, is that considered positive? Hmm, this is a bit confusing. In some contexts, positive numbers are those greater than zero, and non-negative includes zero. So, if the problem specifies positive constants, then c must be greater than zero. Therefore, the smallest possible c is 2π.Wait, but let me think about the function. If c is zero, the function is y equals a cosine of (bx), which does reach its maximum at x equals 0. So, if c is allowed to be zero, that's the smallest possible c. But if c has to be positive, then the next possible value is 2π. So, I need to figure out whether c can be zero or not.Looking back at the problem statement: "positive constants a, b, and c." So, if they are positive, then c must be greater than zero. Therefore, the smallest possible c is 2π. So, I think that's the answer.But just to be thorough, let's consider the general solution. The maximum occurs when bx + c equals 2πk, where k is an integer. At x equals 0, this becomes c equals 2πk. Since c is positive, the smallest k is 1, so c equals 2π. Therefore, the smallest possible value of c is 2π.Wait, but hold on. If k is zero, then c is zero, which is not positive. So, the next integer is k equals 1, giving c equals 2π. So, yes, that makes sense.Therefore, the smallest possible value of c is 2π.But wait, let me think again. If c is allowed to be zero, then it's smaller than 2π. But since the problem says positive constants, c must be greater than zero. So, 2π is the smallest possible value.Alternatively, if the problem had said non-negative constants, then c could be zero. But since it's positive, c must be greater than zero, so 2π is the answer.Okay, I think I've thought this through enough. The smallest possible value of c is 2π.

💡First, I need to determine the total number of apples Mrs. Hilt ate. I know that she ate apples over a period of 3.0 hours and that her average consumption rate was 1.66666666699999 apples per hour.To find the total number of apples, I can multiply the average number of apples she ate per hour by the total number of hours she spent eating apples.So, I will calculate:1.66666666699999 apples/hour × 3.0 hoursPerforming the multiplication gives me:5.00000000099997 applesSince the number of apples should be a whole number, I will round the total to the nearest whole number.Therefore, the total number of apples Mrs. Hilt ate is approximately 5 apples.

💡Okay, so I have this problem where I need to shift the graph of the function y = √3 cos x + sin x to the left by m units, and after shifting, the graph should be symmetric about the origin. I need to find the minimum value of m. The options are given as A: π/12, B: π/6, C: π/3, D: 2π/3.First, I remember that when you shift a graph to the left by m units, you replace x with (x + m) in the function. So, the shifted function would be y = √3 cos(x + m) + sin(x + m). But wait, maybe it's easier if I rewrite the original function in a different form. I recall that expressions like a cos x + b sin x can be written as a single sine or cosine function using the amplitude-phase form. The formula is something like R sin(x + φ) or R cos(x + φ), where R is the amplitude and φ is the phase shift.Let me try that. The original function is y = √3 cos x + sin x. To write this as a single sine function, I can use the identity:a cos x + b sin x = R sin(x + φ)Where R = √(a² + b²). Here, a is √3 and b is 1. So, R = √((√3)² + 1²) = √(3 + 1) = √4 = 2.Now, to find φ, I know that:sin φ = a / R = √3 / 2cos φ = b / R = 1 / 2Looking at these values, sin φ = √3 / 2 and cos φ = 1 / 2. That corresponds to φ = π/3 because sin(π/3) = √3 / 2 and cos(π/3) = 1/2.So, the original function can be rewritten as y = 2 sin(x + π/3).Now, if I shift this graph to the left by m units, the new function becomes y = 2 sin((x + m) + π/3) = 2 sin(x + m + π/3).The problem states that after shifting, the graph is symmetric about the origin. A graph is symmetric about the origin if it's an odd function, meaning f(-x) = -f(x).So, let's check if the shifted function is odd. Let me denote the shifted function as f(x) = 2 sin(x + m + π/3). Then, f(-x) should equal -f(x) for all x.Let's compute f(-x):f(-x) = 2 sin(-x + m + π/3) = 2 sin(-(x - m - π/3)) = -2 sin(x - m - π/3)On the other hand, -f(x) = -2 sin(x + m + π/3)For f(-x) to equal -f(x), we need:-2 sin(x - m - π/3) = -2 sin(x + m + π/3)Divide both sides by -2:sin(x - m - π/3) = sin(x + m + π/3)Hmm, when does sin(A) = sin(B)? That happens when A = B + 2πk or A = π - B + 2πk for some integer k.So, let's set up the equations:1. x - m - π/3 = x + m + π/3 + 2πkSimplify this:x - m - π/3 = x + m + π/3 + 2πkSubtract x from both sides:-m - π/3 = m + π/3 + 2πkBring m to the left and constants to the right:-2m = 2π/3 + 2πkDivide both sides by -2:m = -π/3 - πkBut m has to be positive, so let's see for which k this is possible.If k = -1:m = -π/3 - (-π) = -π/3 + π = 2π/3That's positive. If k = 0, m = -π/3, which is negative. So, the smallest positive m from this case is 2π/3.Now, let's check the other case:2. x - m - π/3 = π - (x + m + π/3) + 2πkSimplify the right side:π - x - m - π/3 + 2πkSo, the equation becomes:x - m - π/3 = π - x - m - π/3 + 2πkBring all terms to the left:x - m - π/3 - π + x + m + π/3 - 2πk = 0Simplify:2x - π - 2πk = 0Which gives:2x = π + 2πkx = π/2 + πkBut this has to hold for all x, which is only possible if the coefficient of x is zero, but here it's 2, so this case doesn't provide a solution that works for all x.Therefore, the only valid solution comes from the first case, which gives m = 2π/3 when k = -1.So, the minimum positive value of m is 2π/3, which is option D.Wait, but let me double-check. If I shift the function y = 2 sin(x + π/3) to the left by 2π/3, the new function becomes y = 2 sin(x + 2π/3 + π/3) = 2 sin(x + π). But sin(x + π) = -sin x, so the function becomes y = -2 sin x, which is indeed an odd function because -2 sin(-x) = 2 sin x = -(-2 sin x). So, yes, it is symmetric about the origin.Alternatively, if I had chosen a smaller m, say π/3, then the shifted function would be y = 2 sin(x + π/3 + π/3) = 2 sin(x + 2π/3). Is this function odd?Let's check: f(-x) = 2 sin(-x + 2π/3) = -2 sin(x - 2π/3). On the other hand, -f(x) = -2 sin(x + 2π/3). These are not the same unless sin(x - 2π/3) = sin(x + 2π/3), which isn't generally true. So, m = π/3 doesn't work.Similarly, for m = π/6, shifted function is y = 2 sin(x + π/6 + π/3) = 2 sin(x + π/2). Then f(-x) = 2 sin(-x + π/2) = 2 cos x, and -f(x) = -2 sin(x + π/2) = -2 cos x. So, f(-x) = 2 cos x and -f(x) = -2 cos x, which are negatives of each other, not equal. So, it's not symmetric about the origin.Same with m = π/12: shifted function is y = 2 sin(x + π/12 + π/3) = 2 sin(x + 5π/12). Checking f(-x) = 2 sin(-x + 5π/12) = -2 sin(x - 5π/12), and -f(x) = -2 sin(x + 5π/12). These aren't equal unless sin(x - 5π/12) = sin(x + 5π/12), which isn't generally true.Therefore, the smallest m that works is indeed 2π/3.**Final Answer**boxed{D}