1. **Problem statement:** We have a deck of cards with colors and numbers, each card worth $2^m$ points where $m$ is the card's number. - Original deck: 10 red (1 to 10), 10 blue (1 to 10), 10 yellow (1 to 10), 2 white (0). - Good group: sum of points = 2026. - Bad group: sum of points = 2025. We want to find the number of good and bad groups. 2. **Points per card:** Each card with number $m$ is worth $2^m$ points. 3. **Sum of points for all cards except white:** Each number 1 to 10 appears 3 times (red, blue, yellow), so total points for number $m$ is $3 \times 2^m$. Sum over $m=1$ to $10$: $$\sum_{m=1}^{10} 3 \times 2^m = 3 \times \sum_{m=1}^{10} 2^m = 3 \times (2^{11} - 2) = 3 \times (2048 - 2) = 3 \times 2046 = 6138.$$ 4. **Points for white cards:** Each white card is $2^0 = 1$ point, and there are 2 white cards, so total $2$ points. 5. **Total points in full deck:** $$6138 + 2 = 6140.$$ 6. **Goal:** Find number of subsets (groups) of cards with sum 2026 (good) and 2025 (bad). 7. **Key insight:** The white cards are special (2 cards worth 1 point each). The other cards come in triples for each number 1 to 10. 8. **Representing groups:** For each number $m$ from 1 to 10, we can choose 0 to 3 cards (red, blue, yellow) independently. For white cards, choose 0,1, or 2 cards. 9. **Sum of points for chosen cards:** $$S = \sum_{m=1}^{10} x_m 2^m + y \times 1,$$ where $x_m \in \{0,1,2,3\}$ is the number of cards chosen for number $m$, and $y \in \{0,1,2\}$ is the number of white cards chosen. 10. **We want to count the number of integer solutions $(x_1,...,x_{10}, y)$ to:** $$\sum_{m=1}^{10} x_m 2^m + y = T,$$ where $T=2026$ (good) or $T=2025$ (bad). 11. **Approach:** Use generating functions. - For each number $m$, the generating function is: $$f_m(z) = 1 + z^{2^m} + z^{2 \times 2^m} + z^{3 \times 2^m} = 1 + z^{2^m} + z^{2^{m+1}} + z^{3 \times 2^m}.$$ - For white cards: $$f_w(z) = 1 + z + z^2.$$ - Total generating function: $$F(z) = f_w(z) \times \prod_{m=1}^{10} f_m(z).$$ The coefficient of $z^T$ in $F(z)$ is the number of groups with sum $T$. 12. **Simplify problem using binary representation:** Note that $2^m$ are powers of two, and $x_m$ can be 0 to 3. Each $x_m$ can be represented as two bits (since max 3): $$x_m = 2a_m + b_m,$$ where $a_m, b_m \in \{0,1\}$. Then: $$\sum_{m=1}^{10} x_m 2^m = \sum_{m=1}^{10} (2a_m + b_m) 2^m = 2 \sum_{m=1}^{10} a_m 2^m + \sum_{m=1}^{10} b_m 2^m.$$ 13. **Rewrite sum:** $$S = 2A + B + y,$$ where $$A = \sum_{m=1}^{10} a_m 2^m, \quad B = \sum_{m=1}^{10} b_m 2^m,$$ with $a_m, b_m \in \{0,1\}$ and $y \in \{0,1,2\}$. 14. **Range of A and B:** Since $a_m, b_m$ are bits, $A$ and $B$ are sums of distinct powers of two from $2^1$ to $2^{10}$. The minimum is 0, maximum is: $$\sum_{m=1}^{10} 2^m = 2^{11} - 2 = 2046.$$ 15. **Rewrite target sums:** We want number of triples $(A,B,y)$ with $$2A + B + y = T,$$ where $T=2026$ or $2025$, and $y \in \{0,1,2\}$. 16. **For each $y$, solve:** $$2A + B = T - y.$$ 17. **Counting solutions:** $A$ and $B$ are sums of distinct powers of two from the set $\{2^1, 2^2, ..., 2^{10}\}$. Each can be any subset sum of these 10 powers of two. Number of subsets of 10 elements is $2^{10} = 1024$. So $A$ and $B$ each can be any integer in $[0,2046]$ that is a sum of distinct powers of two. 18. **Key:** $A$ and $B$ are independent subsets of the same set of powers of two. 19. **Rewrite equation:** $$B = T - y - 2A.$$ For fixed $y$, count number of $A$ such that $B$ is a valid subset sum (i.e., $0 \leq B \leq 2046$ and $B$ is a sum of distinct powers of two). 20. **Check constraints:** - $A$ in $[0,2046]$. - $B = T - y - 2A$ in $[0,2046]$ and must be a sum of distinct powers of two. 21. **Since $A$ and $B$ are sums of distinct powers of two, their binary representations have no repeated bits.** 22. **We want to count pairs $(A,B)$ with $A,B$ subsets of $\{2^1,...,2^{10}\}$ satisfying $B = T - y - 2A$. 23. **Rewrite $2A$ in binary:** Multiplying $A$ by 2 shifts its binary representation left by 1 bit. So if $A$ has bits $a_{10}...a_1$, then $2A$ has bits $a_{10}...a_1 0$ (shifted left). 24. **Since $B$ is a subset sum of $2^1$ to $2^{10}$, its bits correspond to $b_1,...,b_{10}$. 25. **Equation in bits:** $$B = T - y - 2A.$$ We can write $T - y$ in binary and try to find $A$ and $B$ satisfying this. 26. **Calculate $T - y$ for each $y$:** - For $T=2026$: - $y=0$: $2026$ - $y=1$: $2025$ - $y=2$: $2024$ - For $T=2025$: - $y=0$: $2025$ - $y=1$: $2024$ - $y=2$: $2023$ 27. **Binary of these numbers:** Calculate binary of $2026, 2025, 2024, 2023$: - $2024 = 111111001000_2$ - $2025 = 111111001001_2$ - $2026 = 111111001010_2$ - $2023 = 111111000111_2$ (We can verify these by decimal to binary conversion.) 28. **Since $2A$ shifts $A$ left by 1 bit, the least significant bit of $2A$ is 0.** Therefore, the least significant bit of $B$ equals the least significant bit of $T - y$. 29. **Because $B$ is a subset of $2^1$ to $2^{10}$, its least significant bit corresponds to $2^1=2$, so bit 0 (the $2^0$ bit) of $B$ is always 0.** Hence, the $2^0$ bit of $B$ is 0. 30. **Check if $T - y - 2A$ can have $2^0$ bit set:** If $T - y$ has $2^0$ bit set, then $B$ would have $2^0$ bit set, which is impossible. 31. **Check $2^0$ bit of $T - y$ for each $y$:** - For $T=2026$: - $y=0$: $2026$ binary ends with 0 (even number), so $2^0$ bit = 0. - $y=1$: $2025$ ends with 1, $2^0$ bit = 1 (impossible for $B$). - $y=2$: $2024$ ends with 0, $2^0$ bit = 0. - For $T=2025$: - $y=0$: $2025$ ends with 1, $2^0$ bit = 1 (impossible). - $y=1$: $2024$ ends with 0. - $y=2$: $2023$ ends with 1 (impossible). 32. **Therefore, only $y$ values with $2^0$ bit of $T - y$ equal to 0 are possible:** - For $T=2026$: $y=0,2$. - For $T=2025$: $y=1$. 33. **Rewrite problem for these cases:** Count number of $A$ such that $B = T - y - 2A$ is a subset sum (bits in $\{2^1,...,2^{10}\}$). 34. **Since $B$ and $A$ are subsets of the same set, and $2A$ shifts $A$ left by 1 bit, the bits of $B$ and $2A$ do not overlap in the same bit positions.** 35. **This implies $A$ and $B$ have disjoint bit sets shifted by 1 bit.** 36. **Because $B = T - y - 2A$, and $T - y$ is fixed, the bits of $T - y$ are partitioned into bits of $B$ and bits of $2A$.** 37. **Since $2A$ is $A$ shifted left by 1, the bits of $A$ correspond to bits 1 to 10 of $2A$.** 38. **Therefore, the bits of $T - y$ at positions 1 to 10 are split between $A$ and $B$, and bit 0 of $T - y$ must be in $B$ (which must be 0).** 39. **Count number of ways to split bits of $T - y$ at positions 1 to 10 into two disjoint subsets $A$ and $B$.** 40. **Number of such splits is $2^{k}$ where $k$ is the number of bits set in $T - y$ at positions 1 to 10.** 41. **Calculate number of set bits in $T - y$ at bits 1 to 10:** - For $T=2026, y=0$: - Binary of 2026 (bits 10 to 1): count set bits. - For $T=2026, y=2$: - Binary of 2024. - For $T=2025, y=1$: - Binary of 2024. 42. **Binary of 2024 (decimal):** $2024 = 111111001000_2$ Bits 10 to 1 (from right to left, bit 1 is $2^1$): Positions with 1: bits 10,9,8,7,6,5,3 Count: 7 bits set. 43. **Binary of 2026:** $2026 = 111111001010_2$ Bits 10 to 1: bits 10,9,8,7,6,5,3,1 Count: 8 bits set. 44. **Number of good groups:** - For $y=0$, $T-y=2026$, bits set = 8, ways = $2^8=256$. - For $y=2$, $T-y=2024$, bits set = 7, ways = $2^7=128$. Total good groups = $256 + 128 = 384$. 45. **Number of bad groups:** - For $T=2025$, only $y=1$ possible. - $T-y=2024$, bits set = 7, ways = $2^7=128$. 46. **Answer for part (a):** - Good groups: 384 - Bad groups: 128 --- 47. **Part (b):** Add 10 purple cards numbered 1 to 10. Now each number 1 to 10 has 4 cards (red, blue, yellow, purple). 48. **Repeat steps with $x_m \in \{0,1,2,3,4\}$.** Represent $x_m$ in base 2 with 3 bits: $$x_m = 4c_m + 2a_m + b_m,$$ where $c_m, a_m, b_m \in \{0,1\}$. Sum: $$S = \sum_{m=1}^{10} x_m 2^m + y = \sum_{m=1}^{10} (4c_m + 2a_m + b_m) 2^m + y = 4C + 2A + B + y,$$ where $$C = \sum c_m 2^m, A = \sum a_m 2^m, B = \sum b_m 2^m,$$ with $c_m,a_m,b_m \in \{0,1\}$ and $y \in \{0,1,2\}$. 49. **Range of $C,A,B$:** each from 0 to 2046. 50. **Equation:** $$4C + 2A + B + y = T,$$ with $T=2026$ or $2025$. 51. **Rewrite:** $$B = T - y - 4C - 2A.$$ 52. **Constraints:** $B$ is subset sum of $2^1$ to $2^{10}$, so bits 1 to 10. 53. **$4C$ shifts $C$ left by 2 bits, $2A$ shifts $A$ left by 1 bit.** 54. **Bits of $B$, $2A$, and $4C$ occupy disjoint bit positions:** - $B$: bits 1 to 10 - $2A$: bits 2 to 11 - $4C$: bits 3 to 12 55. **$T - y$ must have zero bits in positions 0, 11, 12 to avoid conflicts.** 56. **Check $2^0$ bit of $T - y$ as before:** Only $y$ values with $2^0$ bit zero are possible. For $T=2026$, $y=0,2$; for $T=2025$, $y=1$. 57. **Check bits 11 and 12 of $T - y$:** - $T=2026$ max bit is 11 (since $2^{11}=2048$), bit 12 is 0. - Bit 11 of $T - y$ must be 0 for $B$ and $2A$ to not overlap. 58. **Count number of set bits in $T - y$ at bits 1 to 10, 11, 12:** - Bits 11 and 12 must be 0. 59. **Number of ways to split bits of $T - y$ at bits 1 to 10, 11, 12 into three disjoint subsets $B, A, C$ is:** $$3^{k},$$ where $k$ is the number of set bits in $T - y$ at bits 1 to 10 (since bits 11 and 12 are zero). 60. **Calculate $k$ for each $T - y$ as before:** - For $T=2026, y=0$: $k=8$ bits set. - For $T=2026, y=2$: $k=7$ bits set. - For $T=2025, y=1$: $k=7$ bits set. 61. **Number of good groups:** $$3^8 + 3^7 = 6561 + 2187 = 8748.$$ 62. **Number of bad groups:** $$3^7 = 2187.$$ 63. **Final answers:** - (a) Good groups: 384, Bad groups: 128 - (b) Good groups: 8748, Bad groups: 2187