💡Okay, so I have this polynomial equation: ( x^3 + b_2 x^2 + b_1 x + 18 = 0 ). The problem is asking me to list all the possible integer roots of this polynomial. Hmm, I remember something about the Rational Root Theorem from algebra class. Let me try to recall how that works.First, the Rational Root Theorem says that any possible rational root, expressed in its lowest terms as ( frac{p}{q} ), has a numerator ( p ) that is a factor of the constant term and a denominator ( q ) that is a factor of the leading coefficient. In this polynomial, the constant term is 18, and the leading coefficient is 1 because the ( x^3 ) term has a coefficient of 1.Since the leading coefficient is 1, the denominator ( q ) must be 1, which means any rational root is actually an integer that divides the constant term 18. So, the possible integer roots are the divisors of 18. Let me list those out.The positive divisors of 18 are 1, 2, 3, 6, 9, and 18. But since we're considering integer roots, we also need to include the negative divisors. So, the negative divisors would be -1, -2, -3, -6, -9, and -18. Therefore, putting them all together, the possible integer roots are: -18, -9, -6, -3, -2, -1, 1, 2, 3, 6, 9, and 18.Wait, let me double-check that. If I plug in each of these values into the polynomial, they should satisfy the equation if they're roots. For example, if I take x = 1, plugging into the polynomial gives ( 1 + b_2 + b_1 + 18 = 0 ). That simplifies to ( b_2 + b_1 + 19 = 0 ). Hmm, but I don't know the values of ( b_2 ) and ( b_1 ), so I can't verify if 1 is actually a root. But the question is just asking for possible roots, not necessarily the actual ones. So, I think my list is correct.Another thing to consider is whether I missed any divisors. Let me list the factors of 18 again. 18 can be factored as 1×18, 2×9, 3×6. So, the divisors are indeed ±1, ±2, ±3, ±6, ±9, ±18. I don't think I missed any.Just to be thorough, let me think about why the Rational Root Theorem works. It's because if you have a polynomial with integer coefficients, any rational root must be such that the numerator divides the constant term and the denominator divides the leading coefficient. Since our leading coefficient is 1, the denominator is 1, so all possible rational roots are integers that divide 18.Therefore, I'm confident that the possible integer roots are all the divisors of 18, both positive and negative.

💡First, I need to determine the average pace of the first runner in a 10-mile race. The second runner has a known pace of 7 minutes per mile.After 56 minutes, the second runner has completed 8 miles (56 minutes ÷ 7 minutes per mile). This means the second runner is 2 miles away from the finish line.The second runner stops for 8 minutes. During this time, the first runner catches up. To find the first runner's pace, I calculate the distance the first runner covers in those 8 minutes to close the 2-mile gap.Let ( x ) be the first runner's pace in minutes per mile. The distance covered by the first runner in 8 minutes is ( frac{8}{x} ) miles. Setting this equal to the 2-mile gap:[frac{8}{x} = 2]Solving for ( x ):[x = frac{8}{2} = 4]Therefore, the first runner's average pace is 4 minutes per mile.

💡Alright, so I have this problem here: If ( x > 2m^2 - 3 ) is a necessary but not sufficient condition for ( -1 < x < 4 ), then I need to find the range of the real number ( m ). The options are A: ([-3, 3]), B: ((-infty, -3] cup [3, +infty)), C: ((-infty, -1] cup [1, +infty)), and D: ([-1, 1]).Hmm, okay. Let me break this down. First, what does it mean for ( x > 2m^2 - 3 ) to be a necessary condition for ( -1 < x < 4 )? I think necessary condition means that if ( -1 < x < 4 ) is true, then ( x > 2m^2 - 3 ) must also be true. So, in other words, every ( x ) that satisfies ( -1 < x < 4 ) must also satisfy ( x > 2m^2 - 3 ). So, if I think about it, the interval ( (-1, 4) ) must be entirely contained within the interval ( (2m^2 - 3, +infty) ). That makes sense because if ( x ) is between -1 and 4, it has to be greater than ( 2m^2 - 3 ). So, the lower bound of ( (-1, 4) ) is -1, which has to be greater than or equal to ( 2m^2 - 3 ). Let me write that down: ( -1 geq 2m^2 - 3 ). Solving for ( m ), I can add 3 to both sides: ( -1 + 3 geq 2m^2 ), which simplifies to ( 2 geq 2m^2 ). Dividing both sides by 2 gives ( 1 geq m^2 ). Taking square roots, this means ( |m| leq 1 ), so ( m ) is between -1 and 1. But wait, the problem also says that ( x > 2m^2 - 3 ) is not a sufficient condition for ( -1 < x < 4 ). That means there are some ( x ) values that satisfy ( x > 2m^2 - 3 ) but do not satisfy ( -1 < x < 4 ). So, the interval ( (2m^2 - 3, +infty) ) must include numbers outside of ( (-1, 4) ). Given that ( (-1, 4) ) is entirely within ( (2m^2 - 3, +infty) ), the only way for ( (2m^2 - 3, +infty) ) to include numbers outside ( (-1, 4) ) is if ( 2m^2 - 3 ) is less than or equal to -1. Because if ( 2m^2 - 3 ) were greater than -1, then ( (2m^2 - 3, +infty) ) would start above -1, and since ( (-1, 4) ) is entirely within it, but then ( x ) could be between ( 2m^2 - 3 ) and -1, which would not satisfy ( -1 < x < 4 ). Wait, no, actually, if ( 2m^2 - 3 ) is less than or equal to -1, then ( (2m^2 - 3, +infty) ) includes all numbers greater than ( 2m^2 - 3 ), which would include numbers less than -1 as well, right? But hold on, if ( 2m^2 - 3 leq -1 ), then ( x ) can be greater than ( 2m^2 - 3 ), which could be less than -1, so ( x ) could be less than -1 and still satisfy ( x > 2m^2 - 3 ). But ( -1 < x < 4 ) requires ( x ) to be greater than -1. So, if ( x > 2m^2 - 3 ) is not sufficient, that means there are ( x ) values that satisfy ( x > 2m^2 - 3 ) but do not satisfy ( -1 < x < 4 ). So, in order for ( x > 2m^2 - 3 ) to not be sufficient, ( 2m^2 - 3 ) must be less than or equal to -1. Because if ( 2m^2 - 3 ) were greater than -1, then ( x > 2m^2 - 3 ) would imply ( x > ) something greater than -1, which would make it a sufficient condition since ( x ) would have to be greater than -1, but since ( x ) is also less than 4, it would be sufficient. But since it's not sufficient, ( 2m^2 - 3 ) must be less than or equal to -1.So, putting it all together, ( 2m^2 - 3 leq -1 ) leads to ( 2m^2 leq 2 ), so ( m^2 leq 1 ), which gives ( |m| leq 1 ), meaning ( m ) is between -1 and 1. Therefore, the range of ( m ) is ([-1, 1]), which is option D.But wait, let me double-check. If ( m ) is in ([-1, 1]), then ( 2m^2 - 3 ) would be between ( 2(0) - 3 = -3 ) and ( 2(1) - 3 = -1 ). So, ( 2m^2 - 3 ) is between -3 and -1. Therefore, ( x > 2m^2 - 3 ) would mean ( x > ) something between -3 and -1. So, the interval ( (-1, 4) ) is entirely within ( (2m^2 - 3, +infty) ) because ( 2m^2 - 3 leq -1 ). But also, ( x > 2m^2 - 3 ) is not sufficient because ( x ) could be less than -1, which doesn't satisfy ( -1 < x < 4 ). So, yes, that makes sense. So, the range of ( m ) is indeed ([-1, 1]), which is option D.

💡Alright, so I've got this problem here with points A, B, and C. Let me try to understand what it's asking. First, the coordinates are given: A is at (3, 0), B is at (0, -3), and C is at (cosα, sinα). Okay, so point C is somewhere on the unit circle because its coordinates are cosine and sine of some angle α. That makes sense. Now, the problem says that vector OC is parallel to vector AB, and O is the origin. So, I need to find the value of α in the interval (π/2, 3π/2). Hmm, that interval is from 90 degrees to 270 degrees, so it's the left half of the unit circle, excluding the axes. Let me recall that if two vectors are parallel, one is a scalar multiple of the other. So, vector OC should be a scalar multiple of vector AB. First, let me find vector AB. Vector AB is from point A to point B, so it's B minus A. So, AB = (0 - 3, -3 - 0) = (-3, -3). Got that. Vector OC is from the origin to point C, so it's just the coordinates of C, which are (cosα, sinα). So, OC = (cosα, sinα). Since OC is parallel to AB, there must be some scalar k such that OC = k * AB. That means:cosα = k * (-3)sinα = k * (-3)So, both components of OC are equal to -3k. That suggests that cosα and sinα are equal because they both equal -3k. So, cosα = sinα. Wait, when does cosα equal sinα? I remember that happens at 45 degrees and 225 degrees, which are π/4 and 5π/4 in radians. But our interval is from π/2 to 3π/2, which is 90 degrees to 270 degrees. So, π/4 is 45 degrees, which is less than 90, so it's not in our interval. 5π/4 is 225 degrees, which is between 90 and 270, so that's in our interval. But wait, let me double-check. If cosα = sinα, then tanα = 1, right? Because tanα = sinα / cosα. So, tanα = 1. The solutions to tanα = 1 are α = π/4 + kπ, where k is any integer. So, in the interval (π/2, 3π/2), the solutions are π/4 + π = 5π/4. So, α = 5π/4. But wait, let me think again. The problem says that vector OC is parallel to vector AB. Vector AB is (-3, -3), which points to the southwest direction. Vector OC is (cosα, sinα), which is a unit vector. So, if OC is parallel to AB, it should point in the same direction as AB, but since it's a unit vector, it's just the direction. But AB is (-3, -3), which is the same direction as (-1, -1). So, OC should be in the direction of (-1, -1), which is 225 degrees or 5π/4 radians. Wait, but earlier I thought cosα = sinα, which gives 5π/4, but let me confirm. If OC is parallel to AB, then the direction ratios should be the same. So, the slope of AB is (-3)/(-3) = 1. The slope of OC is sinα / cosα = tanα. So, tanα should be equal to 1, which again gives α = π/4 + kπ. In the interval (π/2, 3π/2), the only solution is 5π/4. So, α = 5π/4. But wait, earlier I thought of 3π/4, which is 135 degrees, but that's in the second quadrant where cosα is negative and sinα is positive, so tanα would be negative, which doesn't equal 1. So, 3π/4 is not a solution here. Wait, maybe I made a mistake earlier. Let me go back. Vector AB is (-3, -3). Vector OC is (cosα, sinα). For them to be parallel, the components must be proportional. So, cosα / (-3) = sinα / (-3). Simplifying, cosα = sinα. So, tanα = 1. So, α = π/4 + kπ. In the interval (π/2, 3π/2), the solutions are π/4 + π = 5π/4. So, α = 5π/4. Wait, but in the initial problem, the user's solution said 3π/4. Maybe I'm missing something. Let me check again. If α = 5π/4, then cosα = -√2/2 and sinα = -√2/2. So, vector OC is (-√2/2, -√2/2). Vector AB is (-3, -3). So, OC is indeed a scalar multiple of AB, specifically, OC = (1/3√2) * AB. But wait, the user's solution said 3π/4. Let me see what happens if α = 3π/4. Then cosα = -√2/2 and sinα = √2/2. So, vector OC would be (-√2/2, √2/2). But vector AB is (-3, -3). So, OC is not a scalar multiple of AB because the y-component would have to be negative if the x-component is negative. So, that can't be. Therefore, the correct answer should be 5π/4, not 3π/4. Wait, maybe the user made a mistake in their solution. Let me see their steps again. They said vector AB is (-3, -3), vector OC is (cosα, sinα). Then they set sinα / -3 = cosα / -3, which simplifies to sinα = cosα. Then they concluded α = 3π/4. But as I saw, at 3π/4, sinα is positive and cosα is negative, so sinα ≠ cosα. So, that's incorrect. Therefore, the correct answer should be 5π/4. Wait, but let me think again. Maybe I'm misunderstanding the direction. If vector OC is parallel to vector AB, which is (-3, -3), then OC should point in the same direction, which is southwest. So, in the third quadrant, which is 5π/4. Yes, that makes sense. So, the correct α is 5π/4. I think the user's solution was incorrect because they might have confused the angle in the second quadrant where sinα is positive and cosα is negative, but in that case, sinα ≠ cosα. So, to summarize, the correct value of α is 5π/4. But wait, let me check once more. If α = 5π/4, then cosα = -√2/2 and sinα = -√2/2. So, vector OC is (-√2/2, -√2/2). Vector AB is (-3, -3). So, OC is indeed a scalar multiple of AB, specifically, OC = (1/(3√2)) * AB. Yes, that works. So, the correct answer is 5π/4. I think the initial solution had a mistake in the angle selection. So, the correct α is 5π/4.

💡First, I need to calculate the factorials of 5 and 4. 5! is equal to 5 × 4 × 3 × 2 × 1, which equals 120. 4! is equal to 4 × 3 × 2 × 1, which equals 24.Next, I multiply these two factorials together: 120 × 24 = 2880.Then, I take the square root of 2880 and square it, which simplifies back to 2880.Finally, I multiply this result by 2 to get the final answer: 2880 × 2 = 5760.

💡Okay, so I'm trying to figure out how many starting values of at most six digits exist such that every number written by Anna is odd. Anna writes integers one below the other, and each subsequent number is either the sum or the product of the digits of the previous number. She notices that every number she writes is odd. First, I need to understand the rules clearly. If Anna starts with a number, say N, then the next number she writes is either the sum of the digits of N or the product of the digits of N. She keeps doing this, and all these numbers must be odd. So, my task is to find all such starting numbers N with at most six digits.Let me break this down. For every number Anna writes, it must be odd. That means both the sum and the product of the digits of any number in the sequence must be odd. Hmm, so both the sum and the product of the digits need to be odd. Wait, is that right? Or is it that each subsequent number is either the sum or the product, but it's not necessarily both? The problem says she writes either the sum or the product, so it's not necessarily both. But she observed that every number written is odd. So, regardless of whether she chooses the sum or the product, the result must be odd.So, for any number in the sequence, both the sum of its digits and the product of its digits must be odd. Because if either the sum or the product were even, then depending on which one she chooses, the next number could be even, which contradicts the observation that every number written is odd.Therefore, both the sum and the product of the digits of every number in the sequence must be odd. That gives us two conditions:1. The product of the digits must be odd.2. The sum of the digits must be odd.Let me analyze these conditions.First, the product of the digits is odd. For a product to be odd, all the factors must be odd. That means every digit in the number must be odd. So, if any digit is even, the product would be even, which is not allowed. Therefore, all digits in the starting number must be odd.Second, the sum of the digits must be odd. For the sum to be odd, the number of odd digits must be odd. Since all digits are already odd (from the first condition), the number of digits must be odd. Because the sum of an odd number of odd digits is odd, and the sum of an even number of odd digits is even.So, putting these together, the starting number must have all digits odd, and the number of digits must be odd. Therefore, the starting number must be an odd-digit number (1, 3, 5 digits) with all digits odd.Wait, but the problem says "at most six digits." So, we need to consider numbers with 1, 3, or 5 digits, each with all digits odd.But hold on, let me think again. If the starting number has an odd number of digits, all of which are odd, then the sum of its digits will be odd, and the product will also be odd. So, the next number, whether it's the sum or the product, will be odd. But then, we need to ensure that this property continues for all subsequent numbers.Wait, so it's not just the starting number that needs to have all digits odd and an odd number of digits, but every subsequent number generated by either summing or multiplying the digits must also satisfy these conditions.So, for example, if we start with a 1-digit number, say 3, then the next number could be the sum of its digits, which is 3, or the product of its digits, which is also 3. So, it stays the same. So, 3 is a valid starting number.Similarly, if we start with a 3-digit number, say 111, the sum of its digits is 3, and the product is 1. Both are odd, so the next number is either 3 or 1, which are both 1-digit numbers, which are odd. Then, from there, it would just stay at 3 or 1. So, 111 is a valid starting number.But what about a 3-digit number like 113? The sum of its digits is 1 + 1 + 3 = 5, which is odd, and the product is 1 * 1 * 3 = 3, which is also odd. So, the next number is either 5 or 3, both of which are 1-digit numbers, which are odd. So, 113 is also valid.Wait, but what if we have a 3-digit number where the sum or the product leads to a number with an even number of digits? For example, if we have a 3-digit number where the sum is 11, which is a 2-digit number. But 11 has two digits, which is even, but all digits are odd (1 and 1). However, the sum of the digits of 11 is 2, which is even, so that would be a problem because the sum would be even, which would make the next number even. Therefore, 11 is not a valid number because the sum of its digits is even.Wait, but in our case, the starting number must be such that all subsequent numbers are odd. So, if the starting number is 113, which leads to 5 or 3, which are both 1-digit numbers, which are odd, and then they stay at 5 or 3. So, 113 is okay.But if we have a starting number that leads to a number with an even number of digits, which could potentially have an even sum, then that starting number would not be valid.Therefore, we need to ensure that not only the starting number has all digits odd and an odd number of digits, but also that any number generated by summing or multiplying the digits also has all digits odd and an odd number of digits.Wait, but the sum or product of digits is a single number, which could be multi-digit. So, for example, if we have a 3-digit number where the sum is 11, which is 2-digit, but as I saw earlier, 11 has an even sum of digits, which would lead to an even number next. So, that starting number would not be valid.Therefore, to ensure that all subsequent numbers are odd, we need to ensure that the sum and product of digits of any number in the sequence are such that they themselves have all digits odd and an odd number of digits.This seems recursive. So, perhaps the only numbers that satisfy this are the 1-digit odd numbers, because if you start with a 1-digit odd number, the sum and product of its digits are itself, so it just stays there, which is odd.But wait, earlier I thought that 111 is valid because it leads to 3 or 1, which are 1-digit numbers. But 111 is a 3-digit number, all digits odd, and the sum is 3, which is odd, and the product is 1, which is odd. So, it seems like 111 is valid.But then, if we have a 3-digit number like 113, which leads to 5 or 3, which are 1-digit numbers, which are odd, so that's okay.But what about a 3-digit number like 135? The sum is 1 + 3 + 5 = 9, which is odd, and the product is 1 * 3 * 5 = 15, which is a 2-digit number. 15 has digits 1 and 5, both odd, but the number of digits is 2, which is even. Then, the sum of digits of 15 is 1 + 5 = 6, which is even, so that would lead to an even number, which is not allowed. Therefore, 135 is not a valid starting number because it can lead to 15, which can lead to 6, which is even.Therefore, 135 is invalid. So, not all 3-digit numbers with all digits odd are valid. Only those 3-digit numbers where the sum and product of digits lead to numbers that are either 1-digit or multi-digit numbers that themselves satisfy the condition of all digits odd and an odd number of digits.But this seems complicated. Maybe it's easier to consider that the only valid starting numbers are the 1-digit odd numbers because any multi-digit number could potentially lead to a number with an even number of digits, which could then lead to an even sum.Wait, but earlier I saw that 111 is valid because it leads to 3 or 1, which are 1-digit numbers. So, maybe some 3-digit numbers are valid.Let me think again. If a 3-digit number has all digits odd, then the sum of its digits is odd (since 3 odd digits sum to odd), and the product is also odd (since all digits are odd). So, the next number is either the sum or the product, which are both odd. But the sum or product could be a 1-digit or 2-digit number.If the sum is a 1-digit number, then it's fine because it's odd. If the sum is a 2-digit number, then we have to check if that 2-digit number has all digits odd and an odd number of digits. Wait, but 2 is even, so a 2-digit number has an even number of digits, which would mean that the sum of its digits must be odd, but the number of digits is even, which would make the sum of digits even or odd?Wait, no. The number of digits is separate from the sum of digits. For example, 15 is a 2-digit number with digits 1 and 5, both odd, so the sum is 6, which is even. Therefore, 15 is a 2-digit number with all digits odd, but the sum of its digits is even, which would lead to an even number next. Therefore, 15 is invalid.Therefore, if a 3-digit number leads to a 2-digit number, that 2-digit number must have all digits odd and an odd number of digits, but 2 is even, so it's impossible. Therefore, any 3-digit number that leads to a 2-digit number is invalid because the 2-digit number would have an even number of digits, which would require the sum of its digits to be odd, but the number of digits is even, which could lead to an even sum.Wait, no. The number of digits being even doesn't directly affect the sum of the digits. For example, 11 is a 2-digit number with all digits odd, and the sum is 2, which is even. So, 11 is invalid because the sum is even.Therefore, any 3-digit number that leads to a 2-digit number is invalid because the 2-digit number would have an even sum of digits, leading to an even number next. Therefore, the only valid 3-digit numbers are those where the sum of digits is a 1-digit number, i.e., the sum is less than 10.Similarly, the product of digits must also be a 1-digit number, because if the product is a 2-digit number, then it would have an even number of digits, leading to an even sum of digits, which is invalid.Wait, but the product could be a 2-digit number. For example, 111 has a product of 1, which is 1-digit. 113 has a product of 3, which is 1-digit. 135 has a product of 15, which is 2-digit. So, 135 is invalid because the product is 15, which is 2-digit, leading to an even sum.Therefore, to ensure that the product is a 1-digit number, the product of the digits must be less than 10. Since all digits are odd, the product must be 1, 3, 5, 7, or 9.Similarly, the sum of the digits must be less than 10 to be a 1-digit number.Therefore, for a 3-digit number to be valid, it must satisfy two conditions:1. The sum of its digits is less than 10 (so that the next number is a 1-digit number).2. The product of its digits is less than 10 (so that the next number is a 1-digit number).Similarly, for a 5-digit number, the sum and product of its digits must be less than 10, so that the next number is a 1-digit number.Wait, but 5-digit numbers have more digits, so the sum and product would be larger. For example, the smallest 5-digit number with all digits odd is 11111. The sum is 5, which is less than 10, and the product is 1, which is less than 10. So, 11111 is valid.But what about 11113? The sum is 1+1+1+1+3=7, which is less than 10, and the product is 1*1*1*1*3=3, which is less than 10. So, 11113 is valid.Similarly, 11131, 11311, 13111, 31111 are all valid.What about 11133? The sum is 1+1+1+3+3=9, which is less than 10, and the product is 1*1*1*3*3=9, which is less than 10. So, 11133 is valid.But what about 11333? The sum is 1+1+3+3+3=11, which is greater than or equal to 10, so the sum would be 11, which is a 2-digit number. Then, the sum of digits of 11 is 2, which is even, leading to an even number next. Therefore, 11333 is invalid.Similarly, the product of 11333 is 1*1*3*3*3=27, which is a 2-digit number. The sum of digits of 27 is 9, which is odd, but the number of digits is 2, which is even, so the sum of digits is 9, which is odd, but the number of digits is even, so the next number would be 9, which is odd. Wait, but 27 is a 2-digit number, so the sum of its digits is 9, which is odd, but the number of digits is 2, which is even. So, does that matter?Wait, the number of digits being even doesn't directly affect the sum of digits. The sum of digits of 27 is 9, which is odd, so the next number would be 9, which is odd. So, 27 is a valid number because the sum of its digits is odd. But wait, 27 is a 2-digit number, which has an even number of digits, but the sum of its digits is odd. So, is 27 a valid number?Wait, according to our earlier reasoning, any number must have all digits odd and an odd number of digits. But 27 has two digits, which is even, so it doesn't satisfy the condition of having an odd number of digits. Therefore, 27 is invalid because it has an even number of digits, even though the sum of its digits is odd.Therefore, any number with an even number of digits is invalid because the number of digits must be odd. So, even if the sum of digits is odd, if the number has an even number of digits, it's invalid.Therefore, going back to 11333, the product is 27, which is a 2-digit number, which is invalid because it has an even number of digits. Therefore, 11333 is invalid.So, for a 5-digit number to be valid, both the sum and the product of its digits must be less than 10, so that the next number is a 1-digit number, which is valid.Similarly, for a 3-digit number, the sum and product must be less than 10.Therefore, the valid starting numbers are:1. 1-digit odd numbers: 1, 3, 5, 7, 9. So, 5 numbers.2. 3-digit numbers with all digits odd, sum of digits < 10, and product of digits < 10.3. 5-digit numbers with all digits odd, sum of digits < 10, and product of digits < 10.Now, let's count how many 3-digit numbers satisfy these conditions.For 3-digit numbers:All digits are odd: 1, 3, 5, 7, 9.Sum of digits < 10.Product of digits < 10.Let's list all possible 3-digit numbers with all digits odd and sum < 10.The smallest 3-digit number with all digits odd is 111.Sum = 1+1+1=3 < 10.Product = 1*1*1=1 < 10.So, 111 is valid.Next, 113:Sum = 1+1+3=5 < 10.Product = 1*1*3=3 < 10.Valid.131:Sum = 1+3+1=5 < 10.Product = 1*3*1=3 < 10.Valid.311:Sum = 3+1+1=5 < 10.Product = 3*1*1=3 < 10.Valid.115:Sum = 1+1+5=7 < 10.Product = 1*1*5=5 < 10.Valid.151:Sum = 1+5+1=7 < 10.Product = 1*5*1=5 < 10.Valid.511:Sum = 5+1+1=7 < 10.Product = 5*1*1=5 < 10.Valid.117:Sum = 1+1+7=9 < 10.Product = 1*1*7=7 < 10.Valid.171:Sum = 1+7+1=9 < 10.Product = 1*7*1=7 < 10.Valid.711:Sum = 7+1+1=9 < 10.Product = 7*1*1=7 < 10.Valid.133:Sum = 1+3+3=7 < 10.Product = 1*3*3=9 < 10.Valid.313:Sum = 3+1+3=7 < 10.Product = 3*1*3=9 < 10.Valid.331:Sum = 3+3+1=7 < 10.Product = 3*3*1=9 < 10.Valid.So, how many 3-digit numbers do we have here?Let's count:111, 113, 131, 311, 115, 151, 511, 117, 171, 711, 133, 313, 331.That's 13 numbers.Now, let's check if there are any more 3-digit numbers with all digits odd, sum < 10, and product < 10.What about 135?Sum = 1+3+5=9 < 10.Product = 1*3*5=15 >= 10.So, product is 15, which is invalid because it's a 2-digit number. Therefore, 135 is invalid.Similarly, 153, 315, 351, 513, 531 are all invalid because their products are 15, which is 2-digit.What about 177?Sum = 1+7+7=15 >= 10.So, sum is invalid.Similarly, 717, 771 have sum >=10.What about 333?Sum = 3+3+3=9 < 10.Product = 3*3*3=27 >=10.So, product is invalid.Similarly, 355:Sum = 3+5+5=13 >=10.Invalid.555:Sum = 5+5+5=15 >=10.Invalid.So, no more 3-digit numbers beyond the 13 we have.Now, let's move on to 5-digit numbers.For 5-digit numbers:All digits are odd: 1, 3, 5, 7, 9.Sum of digits < 10.Product of digits < 10.This is more restrictive because 5 digits, all odd, summing to less than 10.The smallest 5-digit number with all digits odd is 11111.Sum = 1+1+1+1+1=5 < 10.Product = 1*1*1*1*1=1 < 10.Valid.Next, let's consider numbers with four 1s and one 3.11113:Sum = 1+1+1+1+3=7 < 10.Product = 1*1*1*1*3=3 < 10.Valid.Similarly, 11131, 11311, 13111, 31111.Each of these has four 1s and one 3.Sum = 7, product = 3.Valid.So, that's 5 numbers.Next, four 1s and one 5.11115:Sum = 1+1+1+1+5=9 < 10.Product = 1*1*1*1*5=5 < 10.Valid.Similarly, 11151, 11511, 15111, 51111.Each has four 1s and one 5.Sum = 9, product = 5.Valid.Another 5 numbers.Next, four 1s and one 7.11117:Sum = 1+1+1+1+7=11 >=10.Invalid.Similarly, any number with four 1s and one 7 will have sum >=10, so invalid.Similarly, four 1s and one 9:Sum = 1+1+1+1+9=13 >=10.Invalid.Next, three 1s and two 3s.11133:Sum = 1+1+1+3+3=9 <10.Product = 1*1*1*3*3=9 <10.Valid.Similarly, 11313, 11331, 13113, 13131, 13311, 31113, 31131, 31311, 33111.Wait, how many are there?The number of permutations of three 1s and two 3s is 5! / (3!2!) = 10.So, 10 numbers.Each has sum =9, product=9.Valid.Next, three 1s and two 5s.11155:Sum =1+1+1+5+5=13 >=10.Invalid.Similarly, any number with three 1s and two 5s will have sum >=10.Similarly, three 1s and two 7s:Sum =1+1+1+7+7=17 >=10.Invalid.Three 1s and two 9s:Sum =1+1+1+9+9=21 >=10.Invalid.Next, two 1s and three 3s.11333:Sum =1+1+3+3+3=11 >=10.Invalid.Similarly, any number with two 1s and three 3s will have sum >=10.Similarly, two 1s and three 5s:Sum =1+1+5+5+5=17 >=10.Invalid.And so on for higher digits.What about numbers with more than one digit greater than 1?For example, 11111 is valid.11113, 11131, etc., are valid.11133, etc., are valid.But numbers like 11333 are invalid because their sum is >=10.Similarly, numbers with more than two 3s will have sum >=10.Therefore, the only valid 5-digit numbers are:- 11111- 11113, 11131, 11311, 13111, 31111 (5 numbers)- 11115, 11151, 11511, 15111, 51111 (5 numbers)- 11133, 11313, 11331, 13113, 13131, 13311, 31113, 31131, 31311, 33111 (10 numbers)So, total 1 + 5 + 5 + 10 = 21 numbers.Now, let's check if there are any other 5-digit numbers with all digits odd, sum <10, and product <10.What about numbers with one 1 and four 3s?13333:Sum =1+3+3+3+3=13 >=10.Invalid.Similarly, any number with four 3s will have sum >=10.What about numbers with one 1, one 3, and three 5s?Sum =1+3+5+5+5=19 >=10.Invalid.So, no.What about numbers with two 3s and three 1s, which we already counted as 10 numbers.Yes, we already have those.Therefore, total 21 5-digit numbers.Now, let's check if there are any 5-digit numbers with digits other than 1, 3, 5, 7, 9, but since all digits must be odd, we only have 1,3,5,7,9.Therefore, we have 21 valid 5-digit numbers.Now, let's summarize:- 1-digit numbers: 5- 3-digit numbers: 13- 5-digit numbers: 21Total valid starting numbers: 5 + 13 + 21 = 39.Wait, but let me double-check the counts.For 3-digit numbers, I listed 13 numbers:111, 113, 131, 311, 115, 151, 511, 117, 171, 711, 133, 313, 331.Yes, that's 13.For 5-digit numbers, I have:1. 111112. 11113, 11131, 11311, 13111, 31111 (5)3. 11115, 11151, 11511, 15111, 51111 (5)4. 11133, 11313, 11331, 13113, 13131, 13311, 31113, 31131, 31311, 33111 (10)Total: 1 + 5 + 5 + 10 = 21.Yes, that's correct.So, total valid starting numbers: 5 + 13 + 21 = 39.Therefore, the answer is 39.