answer：Alright, so I have this problem about two arithmetic sequences, {a_n} and {b_n}. Their sums of the first n terms are S_n and T_n respectively. The ratio of these sums is given as (3n - 1)/(2n + 3). I need to find the ratio of the 10th terms, a_{10}/b_{10}.Hmm, okay. Let me recall what I know about arithmetic sequences. An arithmetic sequence has a common difference, so each term increases by a fixed amount. The sum of the first n terms of an arithmetic sequence can be calculated using the formula S_n = n/2 [2a_1 + (n - 1)d], where a_1 is the first term and d is the common difference.So, for sequence {a_n}, the sum S_n would be S_n = n/2 [2a_1 + (n - 1)d_a], and similarly for {b_n}, T_n = n/2 [2b_1 + (n - 1)d_b]. Given that S_n / T_n = (3n - 1)/(2n + 3), I can set up the equation:[n/2 (2a_1 + (n - 1)d_a)] / [n/2 (2b_1 + (n - 1)d_b)] = (3n - 1)/(2n + 3)Simplifying this, the n/2 cancels out from numerator and denominator, so we have:(2a_1 + (n - 1)d_a) / (2b_1 + (n - 1)d_b) = (3n - 1)/(2n + 3)Okay, so this gives a relationship between the terms of the two sequences. But I need to find a_{10}/b_{10}. Let me remember that in an arithmetic sequence, the nth term is given by a_n = a_1 + (n - 1)d. So, a_{10} = a_1 + 9d_a and b_{10} = b_1 + 9d_b.So, the ratio a_{10}/b_{10} is (a_1 + 9d_a)/(b_1 + 9d_b). Hmm, how can I find this ratio using the given information?Looking back at the equation I had:(2a_1 + (n - 1)d_a) / (2b_1 + (n - 1)d_b) = (3n - 1)/(2n + 3)I notice that if I set n = 19, then (n - 1) becomes 18, which is 2 times 9. That might be useful because in a_{10} and b_{10}, the coefficient of d is 9. Let me try that.So, plugging n = 19 into the equation:(2a_1 + 18d_a) / (2b_1 + 18d_b) = (3*19 - 1)/(2*19 + 3) = (57 - 1)/(38 + 3) = 56/41Simplify the numerator and denominator on the left:2(a_1 + 9d_a) / 2(b_1 + 9d_b) = 56/41The 2's cancel out:(a_1 + 9d_a)/(b_1 + 9d_b) = 56/41But wait, (a_1 + 9d_a) is exactly a_{10} and (b_1 + 9d_b) is b_{10}. So, this simplifies to:a_{10}/b_{10} = 56/41Wait, but hold on. Let me double-check that. When I set n = 19, I get S_{19}/T_{19} = 56/41, but earlier I thought S_{19} = 19a_{10} and T_{19} = 19b_{10}, which would make S_{19}/T_{19} = a_{10}/b_{10}. But according to my calculation, S_{19}/T_{19} = 56/41, so does that mean a_{10}/b_{10} = 56/41?Wait, let me verify. The sum S_n for an arithmetic sequence is also equal to n times the average of the first and nth term. So, S_n = n*(a_1 + a_n)/2. Therefore, S_{19} = 19*(a_1 + a_{19})/2. Similarly, T_{19} = 19*(b_1 + b_{19})/2.But a_{19} = a_1 + 18d_a, so a_1 + a_{19} = 2a_1 + 18d_a = 2(a_1 + 9d_a) = 2a_{10}. Similarly, b_1 + b_{19} = 2b_{10}. Therefore, S_{19} = 19*a_{10} and T_{19} = 19*b_{10}. Hence, S_{19}/T_{19} = a_{10}/b_{10}.But earlier, when I plugged n = 19 into the given ratio, I got 56/41. So, that would mean a_{10}/b_{10} = 56/41.Wait, but in my initial calculation, I had (2a_1 + 18d_a)/(2b_1 + 18d_b) = 56/41, which simplifies to (a_1 + 9d_a)/(b_1 + 9d_b) = 56/41, which is indeed a_{10}/b_{10} = 56/41.But hold on, in the problem statement, the ratio S_n / T_n is given as (3n - 1)/(2n + 3). When I plug n = 19, it's (57 - 1)/(38 + 3) = 56/41, which matches.Wait, but in the initial problem, the user wrote:"Given two arithmetic sequences {a_n} and {b_n} with the sum of the first n terms denoted by S_n and T_n respectively, if S_n / T_n = (3n - 1)/(2n + 3), find the ratio a_{10}/b_{10}."So, according to my calculations, a_{10}/b_{10} = 56/41.But wait, in the initial thought process, I thought of n = 19 because 10 is the middle term of 1 and 19. But let me think again.Alternatively, maybe I can express a_{10} and b_{10} in terms of S_n and T_n.Wait, another approach: since S_n = n/2 [2a_1 + (n - 1)d_a], and similarly for T_n.Given S_n / T_n = (3n - 1)/(2n + 3), so:[n/2 (2a_1 + (n - 1)d_a)] / [n/2 (2b_1 + (n - 1)d_b)] = (3n - 1)/(2n + 3)Simplify: (2a_1 + (n - 1)d_a) / (2b_1 + (n - 1)d_b) = (3n - 1)/(2n + 3)Let me denote x = 2a_1 and y = d_a, similarly u = 2b_1 and v = d_b.Then, the equation becomes:(x + (n - 1)y) / (u + (n - 1)v) = (3n - 1)/(2n + 3)Cross-multiplying:(x + (n - 1)y)(2n + 3) = (3n - 1)(u + (n - 1)v)Expanding both sides:x*(2n + 3) + y*(n - 1)*(2n + 3) = u*(3n - 1) + v*(n - 1)*(3n - 1)Let me compute each term:Left side:x*(2n + 3) = 2x n + 3xy*(n - 1)*(2n + 3) = y*(2n^2 + 3n - 2n - 3) = y*(2n^2 + n - 3) = 2y n^2 + y n - 3ySo, left side total: 2x n + 3x + 2y n^2 + y n - 3yRight side:u*(3n - 1) = 3u n - uv*(n - 1)*(3n - 1) = v*(3n^2 - n - 3n + 1) = v*(3n^2 - 4n + 1) = 3v n^2 - 4v n + vSo, right side total: 3u n - u + 3v n^2 - 4v n + vNow, equate coefficients of like terms on both sides.Left side:n^2 term: 2yn term: 2x + yconstant term: 3x - 3yRight side:n^2 term: 3vn term: 3u - 4vconstant term: -u + vSo, setting coefficients equal:For n^2: 2y = 3v --> equation (1)For n: 2x + y = 3u - 4v --> equation (2)For constants: 3x - 3y = -u + v --> equation (3)Now, we have three equations:1) 2y = 3v2) 2x + y = 3u - 4v3) 3x - 3y = -u + vLet me solve these equations step by step.From equation (1): v = (2/3)yPlugging v = (2/3)y into equation (2):2x + y = 3u - 4*(2/3)y = 3u - (8/3)yBring all terms to left:2x + y + (8/3)y - 3u = 0Combine like terms:2x + (1 + 8/3)y - 3u = 0Convert 1 to 3/3:2x + (3/3 + 8/3)y - 3u = 0 --> 2x + (11/3)y - 3u = 0Multiply both sides by 3 to eliminate denominators:6x + 11y - 9u = 0 --> equation (2a)Similarly, plug v = (2/3)y into equation (3):3x - 3y = -u + (2/3)yBring all terms to left:3x - 3y + u - (2/3)y = 0Combine like terms:3x + u - (3 + 2/3)y = 0Convert 3 to 9/3:3x + u - (11/3)y = 0Multiply both sides by 3:9x + 3u - 11y = 0 --> equation (3a)Now, we have:From (2a): 6x + 11y - 9u = 0From (3a): 9x + 3u - 11y = 0Let me write them as:6x + 11y = 9u --> equation (2a)9x + 3u = 11y --> equation (3a)Let me solve for u from equation (2a):From (2a): 9u = 6x + 11y --> u = (6x + 11y)/9Plug this into equation (3a):9x + 3*((6x + 11y)/9) = 11ySimplify:9x + (6x + 11y)/3 = 11yMultiply both sides by 3 to eliminate denominator:27x + 6x + 11y = 33yCombine like terms:33x + 11y = 33ySubtract 11y:33x = 22yDivide both sides by 11:3x = 2y --> x = (2/3)ySo, x = (2/3)yRecall that x = 2a_1 and y = d_a, so:2a_1 = (2/3)d_a --> a_1 = (1/3)d_aSimilarly, from equation (1): v = (2/3)y = (2/3)d_aBut v = d_b, so d_b = (2/3)d_aSo, now we have:a_1 = (1/3)d_ad_b = (2/3)d_aNow, let's find a_{10} and b_{10}.a_{10} = a_1 + 9d_a = (1/3)d_a + 9d_a = (1/3 + 9)d_a = (1/3 + 27/3)d_a = (28/3)d_aSimilarly, b_{10} = b_1 + 9d_bBut we need to find b_1. Let's see if we can find b_1 in terms of d_a.From earlier, we have u = 2b_1 and v = d_b = (2/3)d_aFrom equation (2a): u = (6x + 11y)/9We know x = (2/3)y, so x = (2/3)d_aSo, u = (6*(2/3)d_a + 11d_a)/9 = (4d_a + 11d_a)/9 = (15d_a)/9 = (5/3)d_aBut u = 2b_1, so 2b_1 = (5/3)d_a --> b_1 = (5/6)d_aTherefore, b_{10} = b_1 + 9d_b = (5/6)d_a + 9*(2/3)d_a = (5/6)d_a + 6d_a = (5/6 + 36/6)d_a = (41/6)d_aSo, a_{10} = (28/3)d_a and b_{10} = (41/6)d_aTherefore, the ratio a_{10}/b_{10} = (28/3)/(41/6) = (28/3)*(6/41) = (28*6)/(3*41) = (168)/(123) = Simplify by dividing numerator and denominator by 3: 56/41So, a_{10}/b_{10} = 56/41Wait, but earlier when I plugged n = 19, I got the same result. So, that seems consistent.Therefore, the ratio is 56/41.**Final Answer**The ratio (frac{a_{10}}{b_{10}}) is boxed{dfrac{56}{41}}.

answer：Okay, so I have this problem where I need to find the smallest positive integer ( n ) such that the sum ( f(1) + f(2) + dots + f(n) ) is at least 123,456. The function ( f(n) ) is defined as ( k^2 + k + 1 ), where ( n = 2^k(2l + 1) ) for some nonnegative integers ( k ) and ( l ). First, I need to understand what this function ( f(n) ) actually does. It seems like it's based on the highest power of 2 that divides ( n ). For example, if ( n ) is odd, then ( k = 0 ) because ( 2^0 = 1 ) divides every integer. If ( n ) is even, then ( k ) is at least 1, and so on. So, for each ( n ), we can express it as ( 2^k times ) an odd number, and then ( f(n) ) is just ( k^2 + k + 1 ).Let me try to compute ( f(n) ) for some small values of ( n ) to get a sense of how it behaves.- For ( n = 1 ): ( 1 = 2^0 times 1 ), so ( k = 0 ). Thus, ( f(1) = 0^2 + 0 + 1 = 1 ).- For ( n = 2 ): ( 2 = 2^1 times 1 ), so ( k = 1 ). Thus, ( f(2) = 1^2 + 1 + 1 = 3 ).- For ( n = 3 ): ( 3 = 2^0 times 3 ), so ( k = 0 ). Thus, ( f(3) = 0^2 + 0 + 1 = 1 ).- For ( n = 4 ): ( 4 = 2^2 times 1 ), so ( k = 2 ). Thus, ( f(4) = 2^2 + 2 + 1 = 7 ).- For ( n = 5 ): ( 5 = 2^0 times 5 ), so ( k = 0 ). Thus, ( f(5) = 1 ).- For ( n = 6 ): ( 6 = 2^1 times 3 ), so ( k = 1 ). Thus, ( f(6) = 3 ).- For ( n = 7 ): ( 7 = 2^0 times 7 ), so ( k = 0 ). Thus, ( f(7) = 1 ).- For ( n = 8 ): ( 8 = 2^3 times 1 ), so ( k = 3 ). Thus, ( f(8) = 3^2 + 3 + 1 = 13 ).So, the function ( f(n) ) alternates between 1, 3, 7, 13, etc., depending on the highest power of 2 dividing ( n ). It seems like numbers that are powers of 2 have higher values of ( f(n) ), while odd numbers always have ( f(n) = 1 ).Now, I need to sum these values from ( n = 1 ) to ( n ) and find the smallest ( n ) such that the sum is at least 123,456. This seems like it could be a bit tedious if done manually, so I need to find a pattern or formula to make this more manageable.Looking back at the problem statement, there's a formula given for the sum:[S(n) = sum_{k=0}^{infty} (k^2 + k + 1) leftlfloor frac{n + 2^k}{2^{k+1}} rightrfloor]This formula is a bit intimidating, but let me try to parse it. For each ( k ), we're multiplying ( (k^2 + k + 1) ) by the floor of ( frac{n + 2^k}{2^{k+1}} ). The floor function here essentially counts how many numbers up to ( n ) have ( k ) as the exponent in their highest power of 2 divisor. So, for each ( k ), ( leftlfloor frac{n + 2^k}{2^{k+1}} rightrfloor ) gives the number of integers up to ( n ) that are divisible by ( 2^k ) but not by ( 2^{k+1} ). This makes sense because each such number contributes ( k^2 + k + 1 ) to the sum.Therefore, the sum ( S(n) ) can be broken down into contributions from each ( k ), where each contribution is ( (k^2 + k + 1) ) multiplied by the number of times it occurs up to ( n ).To find the smallest ( n ) such that ( S(n) geq 123,456 ), I need to compute ( S(n) ) incrementally until it reaches or exceeds 123,456. However, doing this manually for each ( n ) would be impractical. Instead, I can try to find a pattern or an approximation for ( S(n) ) as a function of ( n ).Let me consider the contribution of each ( k ) to the sum ( S(n) ). For a given ( k ), the number of terms contributing ( (k^2 + k + 1) ) is ( leftlfloor frac{n}{2^k} rightrfloor - leftlfloor frac{n}{2^{k+1}} rightrfloor ). This is because ( leftlfloor frac{n}{2^k} rightrfloor ) counts all numbers up to ( n ) divisible by ( 2^k ), and subtracting ( leftlfloor frac{n}{2^{k+1}} rightrfloor ) removes those divisible by ( 2^{k+1} ), leaving exactly those divisible by ( 2^k ) but not ( 2^{k+1} ).Thus, the sum ( S(n) ) can also be expressed as:[S(n) = sum_{k=0}^{infty} (k^2 + k + 1) left( leftlfloor frac{n}{2^k} rightrfloor - leftlfloor frac{n}{2^{k+1}} rightrfloor right)]This seems like a more manageable expression. Now, to approximate ( S(n) ), I can note that for large ( n ), the floor functions can be approximated by their arguments, so:[S(n) approx sum_{k=0}^{infty} (k^2 + k + 1) left( frac{n}{2^k} - frac{n}{2^{k+1}} right) = n sum_{k=0}^{infty} (k^2 + k + 1) left( frac{1}{2^k} - frac{1}{2^{k+1}} right)]Simplifying the expression inside the sum:[frac{1}{2^k} - frac{1}{2^{k+1}} = frac{1}{2^{k+1}}]So,[S(n) approx n sum_{k=0}^{infty} (k^2 + k + 1) frac{1}{2^{k+1}} = frac{n}{2} sum_{k=0}^{infty} frac{k^2 + k + 1}{2^k}]Now, I need to compute the sum ( sum_{k=0}^{infty} frac{k^2 + k + 1}{2^k} ). Let's break this into three separate sums:[sum_{k=0}^{infty} frac{k^2}{2^k} + sum_{k=0}^{infty} frac{k}{2^k} + sum_{k=0}^{infty} frac{1}{2^k}]I know that:- ( sum_{k=0}^{infty} frac{1}{2^k} = 2 ) (geometric series)- ( sum_{k=0}^{infty} frac{k}{2^k} = 2 ) (arithmetic-geometric series)- ( sum_{k=0}^{infty} frac{k^2}{2^k} = 6 ) (can be derived using generating functions or known formulas)So, adding these up:[6 + 2 + 2 = 10]Therefore, the approximation becomes:[S(n) approx frac{n}{2} times 10 = 5n]This suggests that ( S(n) ) grows approximately linearly with ( n ), with a slope of 5. However, this is just an approximation, and the actual sum might differ, especially for smaller ( n ).Given that ( S(n) approx 5n ), to reach a sum of 123,456, we might estimate ( n ) to be around ( 123,456 / 5 approx 24,691.2 ). So, perhaps around 24,692. But since this is an approximation, I need to check the actual sum around this value to see if it meets or exceeds 123,456.However, computing ( S(n) ) for ( n ) around 24,692 manually is still impractical. Instead, I can try to find a more precise formula or use the given formula to compute ( S(n) ) more accurately.Looking back at the given formula:[S(n) = sum_{k=0}^{infty} (k^2 + k + 1) leftlfloor frac{n + 2^k}{2^{k+1}} rightrfloor]This formula essentially counts the number of times each ( k ) contributes to the sum. For each ( k ), ( leftlfloor frac{n + 2^k}{2^{k+1}} rightrfloor ) gives the number of integers up to ( n ) that have ( k ) as the exponent in their highest power of 2 divisor.To compute ( S(n) ), I can iterate over each ( k ) and compute the corresponding term, then sum them all up. However, this still requires knowing up to which ( k ) I need to compute, which depends on the largest power of 2 less than or equal to ( n ).Given that ( n ) is around 24,692, the largest power of 2 less than or equal to ( n ) is ( 2^{14} = 16,384 ) and ( 2^{15} = 32,768 ). So, ( k ) would range from 0 to 14.Therefore, I can compute ( S(n) ) by summing over ( k = 0 ) to ( k = 14 ), each term being ( (k^2 + k + 1) times leftlfloor frac{n + 2^k}{2^{k+1}} rightrfloor ).Let me try to compute ( S(n) ) for ( n = 24,692 ) using this formula.First, I need to compute each term for ( k = 0 ) to ( k = 14 ).For each ( k ):1. Compute ( 2^k ).2. Compute ( frac{n + 2^k}{2^{k+1}} ).3. Take the floor of that value.4. Multiply by ( (k^2 + k + 1) ).Let's start with ( k = 0 ):- ( 2^0 = 1 )- ( frac{24,692 + 1}{2^{1}} = frac{24,693}{2} = 12,346.5 )- Floor: 12,346- Contribution: ( (0 + 0 + 1) times 12,346 = 12,346 )( k = 1 ):- ( 2^1 = 2 )- ( frac{24,692 + 2}{4} = frac{24,694}{4} = 6,173.5 )- Floor: 6,173- Contribution: ( (1 + 1 + 1) times 6,173 = 3 times 6,173 = 18,519 )( k = 2 ):- ( 2^2 = 4 )- ( frac{24,692 + 4}{8} = frac{24,696}{8} = 3,087 )- Floor: 3,087- Contribution: ( (4 + 2 + 1) times 3,087 = 7 times 3,087 = 21,609 )( k = 3 ):- ( 2^3 = 8 )- ( frac{24,692 + 8}{16} = frac{24,700}{16} = 1,543.75 )- Floor: 1,543- Contribution: ( (9 + 3 + 1) times 1,543 = 13 times 1,543 = 20,059 )( k = 4 ):- ( 2^4 = 16 )- ( frac{24,692 + 16}{32} = frac{24,708}{32} = 772.125 )- Floor: 772- Contribution: ( (16 + 4 + 1) times 772 = 21 times 772 = 16,212 )( k = 5 ):- ( 2^5 = 32 )- ( frac{24,692 + 32}{64} = frac{24,724}{64} = 386.3125 )- Floor: 386- Contribution: ( (25 + 5 + 1) times 386 = 31 times 386 = 11,966 )( k = 6 ):- ( 2^6 = 64 )- ( frac{24,692 + 64}{128} = frac{24,756}{128} = 193.40625 )- Floor: 193- Contribution: ( (36 + 6 + 1) times 193 = 43 times 193 = 8,300 )( k = 7 ):- ( 2^7 = 128 )- ( frac{24,692 + 128}{256} = frac{24,820}{256} ≈ 96.953125 )- Floor: 96- Contribution: ( (49 + 7 + 1) times 96 = 57 times 96 = 5,472 )( k = 8 ):- ( 2^8 = 256 )- ( frac{24,692 + 256}{512} = frac{24,948}{512} ≈ 48.7265625 )- Floor: 48- Contribution: ( (64 + 8 + 1) times 48 = 73 times 48 = 3,504 )( k = 9 ):- ( 2^9 = 512 )- ( frac{24,692 + 512}{1024} = frac{25,204}{1024} ≈ 24.61328125 )- Floor: 24- Contribution: ( (81 + 9 + 1) times 24 = 91 times 24 = 2,184 )( k = 10 ):- ( 2^{10} = 1,024 )- ( frac{24,692 + 1,024}{2,048} = frac{25,716}{2,048} ≈ 12.55859375 )- Floor: 12- Contribution: ( (100 + 10 + 1) times 12 = 111 times 12 = 1,332 )( k = 11 ):- ( 2^{11} = 2,048 )- ( frac{24,692 + 2,048}{4,096} = frac{26,740}{4,096} ≈ 6.52734375 )- Floor: 6- Contribution: ( (121 + 11 + 1) times 6 = 133 times 6 = 798 )( k = 12 ):- ( 2^{12} = 4,096 )- ( frac{24,692 + 4,096}{8,192} = frac{28,788}{8,192} ≈ 3.5107421875 )- Floor: 3- Contribution: ( (144 + 12 + 1) times 3 = 157 times 3 = 471 )( k = 13 ):- ( 2^{13} = 8,192 )- ( frac{24,692 + 8,192}{16,384} = frac{32,884}{16,384} ≈ 2.0078125 )- Floor: 2- Contribution: ( (169 + 13 + 1) times 2 = 183 times 2 = 366 )( k = 14 ):- ( 2^{14} = 16,384 )- ( frac{24,692 + 16,384}{32,768} = frac{41,076}{32,768} ≈ 1.25341796875 )- Floor: 1- Contribution: ( (196 + 14 + 1) times 1 = 211 times 1 = 211 )Now, let's sum up all these contributions:- ( k = 0 ): 12,346- ( k = 1 ): 18,519 → Total: 30,865- ( k = 2 ): 21,609 → Total: 52,474- ( k = 3 ): 20,059 → Total: 72,533- ( k = 4 ): 16,212 → Total: 88,745- ( k = 5 ): 11,966 → Total: 100,711- ( k = 6 ): 8,300 → Total: 109,011- ( k = 7 ): 5,472 → Total: 114,483- ( k = 8 ): 3,504 → Total: 117,987- ( k = 9 ): 2,184 → Total: 120,171- ( k = 10 ): 1,332 → Total: 121,503- ( k = 11 ): 798 → Total: 122,301- ( k = 12 ): 471 → Total: 122,772- ( k = 13 ): 366 → Total: 123,138- ( k = 14 ): 211 → Total: 123,349So, ( S(24,692) ≈ 123,349 ), which is just below 123,456. Therefore, I need to increase ( n ) slightly to see when the sum crosses 123,456.Let's try ( n = 24,693 ):I need to compute the contributions again, but only the terms where ( n ) affects the floor function. Since ( n ) is only increasing by 1, the floor functions for ( k ) where ( 2^k ) divides ( n ) might change.First, let's factorize 24,693 to see its highest power of 2 divisor. Since 24,693 is odd (ends with 3), it's not divisible by 2. Therefore, ( k = 0 ) for ( n = 24,693 ).Thus, the only term affected is ( k = 0 ):- ( k = 0 ): ( leftlfloor frac{24,693 + 1}{2} rightrfloor = leftlfloor 12,347 rightrfloor = 12,347 )- Contribution: ( 1 times 12,347 = 12,347 )Previously, for ( k = 0 ), the contribution was 12,346. So, the total sum increases by 1.Thus, ( S(24,693) = 123,349 + 1 = 123,350 ).Still below 123,456. Let's try ( n = 24,694 ):24,694 is even, so ( k = 1 ) for ( n = 24,694 ).Compute the contribution for ( k = 1 ):- ( leftlfloor frac{24,694 + 2}{4} rightrfloor = leftlfloor frac{24,696}{4} rightrfloor = 6,174 )- Contribution: ( 3 times 6,174 = 18,522 )Previously, ( k = 1 ) contributed 18,519. So, the total sum increases by 3.Thus, ( S(24,694) = 123,350 + 3 = 123,353 ).Still below. Next, ( n = 24,695 ):24,695 is odd, so ( k = 0 ).Contribution for ( k = 0 ):- ( leftlfloor frac{24,695 + 1}{2} rightrfloor = leftlfloor 12,348 rightrfloor = 12,348 )- Contribution: ( 1 times 12,348 = 12,348 )Previously, ( k = 0 ) contributed 12,347. So, total sum increases by 1.( S(24,695) = 123,353 + 1 = 123,354 ).Continuing this way is tedious, but I can see that each time ( n ) increases, the sum increases by 1 if ( n ) is odd, or by 3 if ( n ) is even (since ( k = 1 ) contributes 3). However, when ( n ) is a multiple of higher powers of 2, the contribution increases more.Wait, actually, when ( n ) is a multiple of ( 2^k ), the floor function for ( k ) increases by 1, which affects the contribution for that ( k ). So, for example, when ( n ) reaches a multiple of 4, the contribution for ( k = 2 ) increases by 7, and so on.Given that, perhaps I can find the exact point where the sum crosses 123,456 by considering the contributions from each ( k ) as ( n ) increases.Alternatively, since the approximate sum at ( n = 24,692 ) is 123,349, and we need 123,456, we need an additional 107. Since each increment of ( n ) adds at least 1 to the sum, and sometimes more, we can estimate that ( n ) needs to be around 24,692 + 107 = 24,799. But this is a rough estimate.However, to be precise, I need to compute the exact contributions as ( n ) increases beyond 24,692 until the sum reaches 123,456.Let me try to compute ( S(n) ) for ( n = 24,692 + x ), where ( x ) is the number of increments needed to reach the desired sum.Given that ( S(24,692) = 123,349 ), we need an additional ( 123,456 - 123,349 = 107 ).Each time ( n ) increases by 1, the sum increases by ( f(n) ). Since ( f(n) ) depends on the highest power of 2 dividing ( n ), we can track how much each increment adds.Let me list the values of ( f(n) ) for ( n ) from 24,693 onwards until the sum increases by 107.Starting from ( n = 24,693 ):- ( n = 24,693 ): odd, ( f(n) = 1 ). Sum becomes 123,350.- ( n = 24,694 ): even, ( f(n) = 3 ). Sum becomes 123,353.- ( n = 24,695 ): odd, ( f(n) = 1 ). Sum becomes 123,354.- ( n = 24,696 ): divisible by 4, ( f(n) = 7 ). Sum becomes 123,361.- ( n = 24,697 ): odd, ( f(n) = 1 ). Sum becomes 123,362.- ( n = 24,698 ): even, ( f(n) = 3 ). Sum becomes 123,365.- ( n = 24,699 ): odd, ( f(n) = 1 ). Sum becomes 123,366.- ( n = 24,700 ): divisible by 4, ( f(n) = 7 ). Sum becomes 123,373.- ( n = 24,701 ): odd, ( f(n) = 1 ). Sum becomes 123,374.- ( n = 24,702 ): even, ( f(n) = 3 ). Sum becomes 123,377.- ( n = 24,703 ): odd, ( f(n) = 1 ). Sum becomes 123,378.- ( n = 24,704 ): divisible by 16 (since 24,704 = 16 × 1,544), so ( k = 4 ), ( f(n) = 21 ). Sum becomes 123,378 + 21 = 123,399.- ( n = 24,705 ): odd, ( f(n) = 1 ). Sum becomes 123,400.- ( n = 24,706 ): even, ( f(n) = 3 ). Sum becomes 123,403.- ( n = 24,707 ): odd, ( f(n) = 1 ). Sum becomes 123,404.- ( n = 24,708 ): divisible by 4, ( f(n) = 7 ). Sum becomes 123,411.- ( n = 24,709 ): odd, ( f(n) = 1 ). Sum becomes 123,412.- ( n = 24,710 ): even, ( f(n) = 3 ). Sum becomes 123,415.- ( n = 24,711 ): odd, ( f(n) = 1 ). Sum becomes 123,416.- ( n = 24,712 ): divisible by 16 (24,712 = 16 × 1,544.5), but actually, 24,712 ÷ 16 = 1,544.5, which is not an integer. Wait, 24,712 ÷ 16 = 1,544.5, so it's not divisible by 16. Let me check: 24,712 ÷ 2 = 12,356; ÷2 = 6,178; ÷2 = 3,089, which is odd. So, ( k = 3 ), ( f(n) = 13 ). Sum becomes 123,416 + 13 = 123,429.- ( n = 24,713 ): odd, ( f(n) = 1 ). Sum becomes 123,430.- ( n = 24,714 ): even, ( f(n) = 3 ). Sum becomes 123,433.- ( n = 24,715 ): odd, ( f(n) = 1 ). Sum becomes 123,434.- ( n = 24,716 ): divisible by 4, ( f(n) = 7 ). Sum becomes 123,441.- ( n = 24,717 ): odd, ( f(n) = 1 ). Sum becomes 123,442.- ( n = 24,718 ): even, ( f(n) = 3 ). Sum becomes 123,445.- ( n = 24,719 ): odd, ( f(n) = 1 ). Sum becomes 123,446.- ( n = 24,720 ): divisible by 16 (24,720 ÷ 16 = 1,545), so ( k = 4 ), ( f(n) = 21 ). Sum becomes 123,446 + 21 = 123,467.Now, ( S(24,720) = 123,467 ), which is above 123,456. Therefore, the smallest ( n ) such that ( S(n) geq 123,456 ) is 24,720.But wait, let me check if ( n = 24,719 ) was 123,446, and ( n = 24,720 ) adds 21 to reach 123,467. So, the sum crosses 123,456 between ( n = 24,719 ) and ( n = 24,720 ). However, since we need the sum to be at least 123,456, and ( n = 24,720 ) is the first ( n ) where the sum exceeds 123,456, the answer should be 24,720.But let me verify the contributions again to ensure I didn't make a mistake in the calculations.Wait, when I computed ( S(24,692) ), I got 123,349. Then, as I incremented ( n ) up to 24,720, I added the contributions step by step. However, I might have missed some contributions from higher ( k ) when ( n ) crosses multiples of higher powers of 2.For example, when ( n = 24,704 ), which is divisible by 16, the contribution for ( k = 4 ) increases by 21. Similarly, when ( n = 24,720 ), which is divisible by 16, the contribution for ( k = 4 ) increases by another 21.But in my step-by-step addition, I only added the ( f(n) ) for each ( n ), which is correct because each ( f(n) ) is the contribution of that specific ( n ). Therefore, my calculation should be accurate.Thus, the smallest ( n ) such that ( S(n) geq 123,456 ) is 24,720.

answer：First, I need to calculate the sums in the numerators and denominators of both fractions. For the first fraction, the numerator is (3 + 6 + 9), which equals 18. The denominator is (2 + 5 + 8), which equals 15. So, the first fraction simplifies to (frac{18}{15}).For the second fraction, the numerator involves multiplying 4 by the sum of (1 + 2 + 3). The sum is 6, and multiplying by 4 gives 24. The denominator is (5 + 10 + 15), which equals 30. So, the second fraction simplifies to (frac{24}{30}).Next, I simplify each fraction to its lowest terms. The fraction (frac{18}{15}) can be simplified by dividing both the numerator and the denominator by 3, resulting in (frac{6}{5}). Similarly, the fraction (frac{24}{30}) can be simplified by dividing both the numerator and the denominator by 6, resulting in (frac{4}{5}).Now, I subtract the second fraction from the first fraction: (frac{6}{5} - frac{4}{5}). Since the denominators are the same, I can subtract the numerators directly, which gives (frac{2}{5}).Therefore, the result of the expression is (frac{2}{5}).

answer：First, I need to determine the total number of animals Erica saw on Saturday and Monday.On Saturday, she saw 3 lions and 2 elephants, which adds up to 5 animals.On Monday, she saw 5 rhinos and 3 warthogs, totaling 8 animals.Adding the animals from Saturday and Monday gives me 13 animals.Since Erica saw a total of 20 animals during her safari, I can find out how many animals she saw on Sunday by subtracting the 13 animals from the total:20 total animals - 13 animals (Saturday and Monday) = 7 animals on Sunday.On Sunday, she saw 2 buffaloes and some leopards. To find the number of leopards, I subtract the number of buffaloes from the total animals seen on Sunday:7 animals (Sunday) - 2 buffaloes = 5 leopards.Therefore, Erica saw 5 leopards on Sunday.

answer：To determine the fraction of the time that Amy and Lily tie in their modified game of tic-tac-toe, I start by analyzing the given probabilities of each player winning.First, I note that Amy wins (frac{2}{5}) of the time and Lily wins (frac{1}{4}) of the time. To combine these probabilities, I convert them to have a common denominator. The least common multiple of 5 and 4 is 20, so I rewrite the fractions:[frac{2}{5} = frac{8}{20} quad text{and} quad frac{1}{4} = frac{5}{20}]Next, I add these probabilities to find the total probability of either Amy or Lily winning:[frac{8}{20} + frac{5}{20} = frac{13}{20}]To find the probability of a tie, I subtract the total win probability from 1:[1 - frac{13}{20} = frac{20}{20} - frac{13}{20} = frac{7}{20}]Therefore, the probability that Amy and Lily tie is (frac{7}{20}).

answer：First, I need to identify the two types of coffee packages Carina has. She has 4 packages of the second type, each containing 10 ounces. This means the total ounces of the second type of packages is 4 multiplied by 10, which equals 40 ounces.Next, I know that the total amount of coffee Carina has is 70 ounces. By subtracting the ounces from the second type of packages, I can find out how many ounces are in the first type of packages: 70 ounces minus 40 ounces equals 30 ounces.Carina has 2 more of the first type of package than the second type. Since she has 4 of the second type, she must have 4 plus 2, which equals 6 of the first type.Finally, to find the size of each first type of package, I divide the total ounces of the first type by the number of first type packages: 30 ounces divided by 6 packages equals 5 ounces per package.