answer:1. **Convert the given ratios to equations:** - The ratio of a to b is given as 5:2. This can be written as: [ frac{a}{b} = frac{5}{2} ] - The ratio of c to d is given as 7:3. This can be written as: [ frac{c}{d} = frac{7}{3} ] - The ratio of d to b is given as 5:4. This can be written as: [ frac{d}{b} = frac{5}{4} ] 2. **Find the ratio of c to b using the chain of ratios:** - Multiply the ratios frac{c}{d} and frac{d}{b} to find frac{c}{b}: [ frac{c}{b} = frac{c}{d} cdot frac{d}{b} = frac{7}{3} cdot frac{5}{4} = frac{7 times 5}{3 times 4} = frac{35}{12} ] 3. **Find the ratio of a to c using the ratios frac{a}{b} and frac{c}{b}:** - Divide the ratio frac{a}{b} by frac{c}{b}: [ frac{a}{c} = frac{frac{a}{b}}{frac{c}{b}} = frac{frac{5}{2}}{frac{35}{12}} = frac{5}{2} cdot frac{12}{35} = frac{60}{70} = frac{6}{7} ] - Thus, the ratio of a to c is 6:7. 4. **Conclusion:** - The ratio of a to c is frac{6{7}}. The final answer is boxed{text{(B)} frac{6}{7}}

answer:Let a, b, c, and d represent the number of nickels, dimes, quarters, and half-dollars, respectively. We are given: 1. The total number of coins is 120. 2. The total value of the coins is 12.40, equivalent to 1240 cents. Setting up equations, we have: [ a + b + c + d = 120 quad text{(Equation 1)} ] [ 5a + 10b + 25c + 50d = 1240 quad text{(Equation 2)} ] First, multiply Equation 1 by 5: [ 5a + 5b + 5c + 5d = 600 quad text{(Equation 3)} ] Subtract Equation 3 from Equation 2: [ (5a + 10b + 25c + 50d) - (5a + 5b + 5c + 5d) = 1240 - 600 ] [ 5b + 20c + 45d = 640 ] [ b + 4c + 9d = 128 quad text{(Equation 4)} ] For possible values of b, analyzing extremes: - If c = 0 and d = 0, then b = 128. - If b = 0, find integer values for c and d such that Equation 4 holds. For minimal b: [ 9 times 14 = 126 quad (d = 14, text{nearest integer satisfying } 9d text{ close to } 128) ] [ b = 128 - 126 = 2 quad (c = 0, text{no quarters used}) ] Thus, the smallest number of dimes b can be is 2 (when d = 14 and c = 0), and the largest number of dimes b can be is 128 (when c = 0 and d = 0). Difference: [ 128 - 2 = 126 ] The difference between the largest and smallest number of dimes in the bank is 126. The final answer is boxed{text{B: } 126}

answer:1. We start with the sequences ({a_n}) and ({b_n}) defined recursively by: [ a_0 = 2, quad b_0 = 2 ] [ a_{n+1} = a_n sqrt{1 + a_n^2 + b_n^2} - b_n ] [ b_{n+1} = b_n sqrt{1 + a_n^2 + b_n^2} + a_n ] 2. The goal is to find the ternary representations of (a_4) and (b_4). 3. First, let's determine (sqrt{1 + a_n^2 + b_n^2}): - Induction base case: For (n = 0): [ a_0 = 2, quad b_0 = 2 ] [ sqrt{1 + a_0^2 + b_0^2} = sqrt{1 + 2^2 + 2^2} = sqrt{1 + 4 + 4} = sqrt{9} = 3 = 3^{2^0} ] - Inductive step: Assume (sqrt{1 + a_n^2 + b_n^2} = 3^{2^n}) holds for (n). Prove for (n + 1): [ sqrt{1 + a_{n+1}^2 + b_{n+1}^2} = sqrt{1 + left(a_n sqrt{1 + a_n^2 + b_n^2} - b_nright)^2 + left(b_n sqrt{1 + a_n^2 + b_n^2} + a_nright)^2} ] Simplifying, we get: [ 1 + a_{n+1}^2 + b_{n+1}^2 = 1 + a_n^2(1 + a_n^2 + b_n^2) + b_n^2 - 2a_n b_n sqrt{1 + a_n^2 + b_n^2} + b_n^2(1 + a_n^2 + b_n^2) + a_n^2 + 2a_n b_n sqrt{1 + a_n^2 + b_n^2} ] [ = 1 + (a_n^2 + b_n^2)(1 + a_n^2 + b_n^2) + a_n^2 + b_n^2 ] [ = (1 + a_n^2 + b_n^2)^2 ] Hence: [ sqrt{1 + a_{n+1}^2 + b_{n+1}^2} = 3^{2^{n+1}} ] 4. Now consider the sequence of complex numbers ({z_n}) defined by (z_n = a_n + b_n i): - Initial conditions: (z_0 = 2 + 2i) - Recursive relation: (z_{n+1} = z_n (3^{2^n} + i)) - Specifically: [ z_1 = (2 + 2i)(3 + i) ] [ z_2 = z_1 (3^2 + i) = z_1 (9 + i) ] [ z_3 = z_2 (3^4 + i) = z_2 (81 + i) ] [ z_4 = z_3 (3^8 + i) = z_3 (6561 + i) ] 5. To find (a_4) and (b_4), we convert the product ((2 + 2i)(3^{2^0} + i)(3^{2^1} + i)(3^{2^2} + i)(3^{2^3} + i)) into ternary: - Using balanced ternary, where digits are (-1), (0), and (1): Let: [ x + yi = (3^1 + i)(3^2 + i)(3^4 + i)(3^8 + i) ] - For (x): [ x_j = begin{cases} -1 & text{if } b(j) equiv 2 mod 4 0 & text{if } b(j) equiv 1 mod 2 1 & text{if } b(j) equiv 0 mod 4 end{cases} ] - For (y): [ y_j = begin{cases} -1 & text{if } b(j) equiv 1 mod 4 0 & text{if } b(j) equiv 0 mod 2 1 & text{if } b(j) equiv 3 mod 4 end{cases} ] 6. Converting to standard ternary (digits 0, 1, 2): - (x = 221211221122001_3) - (y = 110022202212120_3) 7. Therefore: - (a_4 = 2x - 2y) - (b_4 = 2x + 2y) 8. Performing the arithmetic in ternary, we find: - (a_4 = 1000001100111222_3) - (b_4 = 2211100110000012_3) # Conclusion: The ternary representations of (a_4) and (b_4) are: [ a_4 = boxed{1000001100111222} ] [ b_4 = boxed{2211100110000012} ]

answer:1. We start with the given expression ( f(x, y) = x^2 + y^2 + ax + by + c ). 2. To simplify the problem, we perform a change of variables to eliminate the linear terms ( ax ) and ( by ). Specifically, we complete the square for the terms involving ( x ) and ( y ). Consider the transformation: [ x' = x + frac{a}{2} quad text{and} quad y' = y + frac{b}{2} ] Substituting these into the original expression, we get: [ f(x, y) = left( x' - frac{a}{2} right)^2 + left( y' - frac{b}{2} right)^2 + ax + by + c ] 3. Expanding the squares and simplifying: [ f(x, y) = left( x'^2 - ax' + frac{a^2}{4} right) + left( y'^2 - by' + frac{b^2}{4} right) + ax + by + c ] [ = x'^2 + y'^2 - ax' - by' + frac{a^2}{4} + frac{b^2}{4} + ax + by + c ] [ = x'^2 + y'^2 + left( frac{a^2}{4} + frac{b^2}{4} + c right) ] 4. Notice that the terms ( ax ) and ( by ) cancel out, leaving us with: [ f(x, y) = x'^2 + y'^2 + left( c + frac{a^2}{4} + frac{b^2}{4} right) ] 5. Let ( k = c + frac{a^2}{4} + frac{b^2}{4} ). The problem now reduces to finding integers ( x' ) and ( y' ) such that: [ p mid x'^2 + y'^2 + k ] 6. Since ( p ) is an odd prime, we can use the fact that every quadratic residue modulo ( p ) can be expressed as a sum of two squares. This is a consequence of the theorem that states: "An odd prime ( p ) can be expressed as a sum of two squares if and only if ( p equiv 1 pmod{4} ) or ( p = 2 )." 7. Given that ( p ) is an odd prime, we can always find integers ( x' ) and ( y' ) such that: [ x'^2 + y'^2 equiv -k pmod{p} ] 8. Therefore, there exist integers ( x ) and ( y ) such that: [ p mid x^2 + y^2 + ax + by + c ] Thus, we have shown that for any given integers ( a, b, c ) and an odd prime ( p ), there exist integers ( x ) and ( y ) such that ( p ) divides ( x^2 + y^2 + ax + by + c ). (blacksquare)