answer:Starting with the least significant (rightmost) column: - Since E + E results in H in base-6, 2E equiv H pmod{6}. E and H must be less than 6. Possible pairs (E, H) considering no carry from third to fourth column: - E = 1, H = 2 - E = 2, H = 4 - E = 3, H = 0 (carry of 1) - E = 4, H = 2 (carry of 1) - E = 5, H = 4 (carry of 1) Checking for the second column: - H + H equiv E pmod{6} and could potentially include a carryover from the first column. Testing (E, H) = (2, 4): - From the first column: 2 + 2 = 4 - Second column: 4 + 4 = 8 equiv 2 pmod{6} (no carry needed for this case, but it doesn't match since it suggests E = 2, contradicting the given E = S) Testing (E, H) = (3, 0): - From the first column: 3 + 3 = 6 equiv 0 pmod{6}, and a carry of 1. - Second column: 0 + 0 + 1 = 1 pmod{6}, This suggests E = 1 (conflicts, as it must be S for third column) From the given setup, it appears (E, H) = (2, 4) with S = 2 is the only non-conflicting choice: begin{array}{c@{}c@{}c@{}c} &2&4&2_6 &+&4&2_6 cline{2-4} &4&2&4_6 end{array} Sum of S + H + E = 2 + 4 + 2 = 8. In base-6, the sum is boxed{12_6}.

answer:Let d, e, and f represent the ID numbers of Dan, Emily, and Fiona, respectively. Given each ID is a two-digit prime and they are consecutive primes: Let's choose d = 11, e = 13, and f = 17 (considering the smallest set of two-digit consecutive primes). From the conversation: - Today's date corresponds to Emily and Fiona's sum: e + f. - The day before Emily's birthday corresponds to Dan and Fiona's sum: d + f. - The day after today corresponds to Dan and Emily's sum: d + e. 1. Solve e + f = 13 + 17 = 30 (Today's date). 2. Solve d + f = 11 + 17 = 28 (Day before Emily's birthday). 3. Solve d + e = 11 + 13 = 24 (Day after today). The equations hold true for the chosen primes and the sums match realistic day dates (not exceeding 31). Conclusion: Fiona's ID number is 17. The final answer is boxed{17}.

answer:Let's calculate this step by step: 1. Increase 150 by 40%: 40% of 150 = 0.40 * 150 = 60 So, 150 + 60 = 210 2. Decrease the result by the square root of 25%: First, find the square root of 25%, which is the square root of 0.25. The square root of 0.25 = 0.5 (or 50%) Now, decrease 210 by 50%: 50% of 210 = 0.50 * 210 = 105 So, 210 - 105 = 105 3. Increase the result by 15% raised to the power of 2: First, calculate 15% raised to the power of 2: (15%)^2 = (0.15)^2 = 0.0225 (or 2.25%) Now, increase 105 by 2.25%: 2.25% of 105 = 0.0225 * 105 = 2.3625 So, 105 + 2.3625 = 107.3625 The final number after all the increases and decreases is boxed{107.3625} .

answer:1. Let ( P_1, P_2, P_3, P_4 ) be the graphs of four quadratic polynomials. Suppose ( P_1 ) is tangent to ( P_2 ) at the point ( q_2 ), ( P_2 ) is tangent to ( P_3 ) at the point ( q_3 ), ( P_3 ) is tangent to ( P_4 ) at the point ( q_4 ), and ( P_4 ) is tangent to ( P_1 ) at the point ( q_1 ). Assume that all the points ( q_1, q_2, q_3, q_4 ) have distinct ( x )-coordinates. 2. Let ( q_i = (x_i, y_i) ) for ( 1 leq i leq 4 ). We can express the tangency conditions as: [ P_1(x) - P_2(x) = c_1 (x - x_1)^2, ] [ P_2(x) - P_3(x) = c_2 (x - x_2)^2, ] [ P_3(x) - P_4(x) = c_3 (x - x_3)^2, ] [ P_4(x) - P_1(x) = c_4 (x - x_4)^2, ] where ( c_1, c_2, c_3, c_4 ) are non-zero constants. 3. We need to show that ( sum_{i=1}^4 c_i = sum_{i=1}^4 c_i x_i = sum_{i=1}^4 c_i x_i^2 = 0 ). This can be written in matrix form as: [ begin{pmatrix} c_1 & c_2 & c_3 & c_4 end{pmatrix} begin{pmatrix} 1 & x_1 & x_1^2 1 & x_2 & x_2^2 1 & x_3 & x_3^2 1 & x_4 & x_4^2 end{pmatrix} = begin{pmatrix} 0 & 0 & 0 end{pmatrix}. ] 4. The matrix ( A := begin{pmatrix} 1 & x_1 & x_1^2 1 & x_2 & x_2^2 1 & x_3 & x_3^2 1 & x_4 & x_4^2 end{pmatrix} ) is a ( 4 times 3 ) matrix. It is a submatrix of the full rank ( 4 times 4 ) Vandermonde matrix ( begin{pmatrix} 1 & x_1 & x_1^2 & x_1^3 1 & x_2 & x_2^2 & x_2^3 1 & x_3 & x_3^2 & x_3^3 1 & x_4 & x_4^2 & x_4^3 end{pmatrix} ), which has rank 4. Therefore, the rank of ( A ) is 3. 5. The set of vectors ( mathbf{z} ) that can be written as ( mathbf{z} = A mathbf{v} ) forms a 3-dimensional subspace of ( mathbb{R}^4 ). Each of these vectors lies in the 3-dimensional subspace of vectors ( mathbf{z} in mathbb{R}^4 ) with ( begin{pmatrix} c_1 & c_2 & c_3 & c_4 end{pmatrix} mathbf{z} = 0 ). 6. Therefore, for ( mathbf{z} in mathbb{R}^4 ), there exists a ( mathbf{v} in mathbb{R}^3 ) such that ( mathbf{z} = A mathbf{v} ) if and only if ( begin{pmatrix} c_1 & c_2 & c_3 & c_4 end{pmatrix} mathbf{z} = 0 ). 7. It suffices to prove that ( c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4 = 0 ). Let ( P_1(x) = a_2 x^2 + a_1 x + a_0 ). Then we have: [ c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4 = c_1 P_1(x_1) + c_2 P_2(x_2) + c_3 P_3(x_3) + c_4 P_4(x_4). ] 8. Using the tangency conditions, we can write: [ c_1 P_1(x_1) + c_2 P_2(x_2) + c_3 P_3(x_3) + c_4 P_4(x_4) = left( sum_{i=1}^4 c_i P_1(x_i) right) - sum_{1 leq i < j leq 4} c_i c_j (x_j - x_i)^2. ] 9. Since ( sum_{i=1}^4 c_i = sum_{i=1}^4 c_i x_i = sum_{i=1}^4 c_i x_i^2 = 0 ), we have: [ a_2 sum_{i=1}^4 c_i x_i^2 + a_1 sum_{i=1}^4 c_i x_i + a_0 sum_{i=1}^4 c_i + left( sum_{i=1}^4 c_i x_i right)^2 - left( sum_{i=1}^4 c_i right) left( sum_{i=1}^4 c_i x_i^2 right) = 0. ] 10. Therefore, ( c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4 = 0 ), which implies that the points ( q_1, q_2, q_3, q_4 ) lie on the graph of an at most quadratic polynomial. (blacksquare)