answer:1. **Initial Setup and Modulo 2 Analysis:** - We start with the numbers (1, 2, ldots, n) on the board. - By one move, we replace two numbers (a) and (b) with (a^2 - b). - We need to determine the values of (n) such that after (n-1) moves, it is possible to obtain (0). 2. **Sum Modulo 2:** - Note that (a^2 - b equiv a - b equiv a + b pmod{2}). - This implies that the sum of the numbers on the board is maintained modulo 2. - The sum of the first (n) natural numbers is (frac{n(n+1)}{2}). - For the sum to be even, (frac{n(n+1)}{2} equiv 0 pmod{2}). 3. **Modulo 4 Analysis:** - We need (n(n+1)) to be divisible by 4. - Since (n) and (n+1) are consecutive integers, one of them is even, and the product (n(n+1)) is always even. - For (n(n+1)) to be divisible by 4, either (n equiv 0 pmod{4}) or (n equiv 3 pmod{4}). 4. **Base Cases:** - **For (n = 3):** [ begin{aligned} &1, 2, 3 &1, 2^2 - 3 = 1 &1, 1 &1^2 - 1 = 0 end{aligned} ] Thus, it is possible to obtain 0 for (n = 3). - **For (n = 4):** [ begin{aligned} &1, 2, 3, 4 &1, 2, 3^2 - 4 = 5 &1, 2^2 - 5 = -1 &1, -1 &(-1)^2 - 1 = 0 end{aligned} ] Thus, it is possible to obtain 0 for (n = 4). 5. **Inductive Hypothesis:** - Assume that for (n = 4k) and (n = 4k - 1), it is possible to obtain 0 after (n-1) moves. 6. **Inductive Step:** - We need to show that it is possible to go from (4k) to (4k+4) and from (4k-1) to (4k+3). - Consider the sequence for (n = 4k): [ begin{aligned} &n, n-1, n-2, n-3, ldots &(n-1)^2 - n, n-2, n-3, ldots &n^2 - 3n + 1, n-2, n-3, ldots &(n-2)^2 - (n^2 - 3n + 1), n-3, ldots &-n + 3, n-3, ldots &(-n + 3)^2 - (n-3), ldots &(n-3)(n-4), n-4, ldots &(n-4)^2 - (n-3)(n-4), ldots &n-4, ldots end{aligned} ] This shows that we can reduce the sequence to (n-4) and continue the process. 7. **Conclusion:** - By induction, we have shown that for (n equiv 0 pmod{4}) and (n equiv 3 pmod{4}), it is possible to obtain 0 after (n-1) moves. (blacksquare) The final answer is ( boxed{ n equiv 0 pmod{4} } ) and (n equiv 3 pmod{4}).

answer:To determine which option is a perfect square, we need to prime factorize each term and ensure all exponents are even. 1. **Prime Factorization of Each Term**: - 3 = 3^1 - 4 = 2^2 - 7 = 7^1 2. **Expand Each Option and Combine**: - **(A) 3^4 4^5 7^7:** - 3^4 = 3^4 - 4^5 = (2^2)^5 = 2^{10} - 7^7 = 7^7 - Combined: 3^4 cdot 2^{10} cdot 7^7 - 7^7 has an odd exponent. - **(B) 3^6 4^4 7^6:** - 3^6 = 3^6 - 4^4 = (2^2)^4 = 2^8 - 7^6 = 7^6 - Combined: 3^6 cdot 2^8 cdot 7^6 - All exponents are even. - **(C) 3^5 4^6 7^5:** - 3^5 = 3^5 - 4^6 = (2^2)^6 = 2^{12} - 7^5 = 7^5 - Combined: 3^5 cdot 2^{12} cdot 7^5 - 3^5 and 7^5 have odd exponents. - **(D) 3^4 4^7 7^4:** - 3^4 = 3^4 - 4^7 = (2^2)^7 = 2^{14} - 7^4 = 7^4 - Combined: 3^4 cdot 2^{14} cdot 7^4 - All exponents are even. - **(E) 3^6 4^6 7^6:** - 3^6 = 3^6 - 4^6 = (2^2)^6 = 2^{12} - 7^6 = 7^6 - Combined: 3^6 cdot 2^{12} cdot 7^6 - All exponents are even. 3. **Conclusion**: - Options (B), (D), and (E) have all exponents even in their prime factorizations, making them perfect squares. Thus, the correct options are textbf{(B), textbf{(D)}, textbf{(E)}}. The final answer is boxed{textbf{(B)}, textbf{(D)}, textbf{(E)}}

answer:1. **Understanding the Setup**: - We have 8 members each needing to have the same number of friends among themselves. - Possible numbers of friends per person are 1 to 6 due to the restriction that not all are friends (thus excluding 0 and 7). 2. **Analyzing Symmetry**: - The cases for n=1 friend and n=6 friends are symmetric, similarly for n=2 and n=5, and n=3 and n=4. 3. **Counting Configurations for n=1**: - Each person pairs with exactly one other, forming 4 disjoint pairs. - Choose a friend for the first person: 7 choices. - The second pair has 5 choices (from the remaining 6 people). - The third pair has 3 choices (from the remaining 4 people). - The last pair has no choice but to pair up with each other. - Total configurations for n=1: 7 times 5 times 3 = 105. 4. **Counting Configurations for n=2**: - **Case 1: Square Groups** - Divide into two groups of 4, each forming a square. - Choose 4 people from 8: binom{8}{4} = 70. - Each square arrangement is unique, so we divide by 2 to avoid overcounting the symmetric groups: frac{70}{2} = 35 configurations. - **Case 2: Octagonal Configuration** - Each person forms a vertex of an octagon. - Fix one person, choose 2 friends from the remaining 7: binom{7}{2} = 21 ways. - The remaining 5 arrange in 5! = 120 ways. - Total octagonal configurations: 21 times 120 = 2520. - Total configurations for n=2: 35 + 2520 = 2555. 5. **Using Symmetry for n=3 and n=4 and n=5 and n=6**: - Configurations for n=3 and n=4: Assume symmetric to n=2. - Configurations for n=5 and n=6: Symmetric to n=1 and n=2 respectively. 6. **Total Configurations**: - 105 times 2 + 2555 times 2 = 210 + 5110 = 5320. # Conclusion: The total number of different ways the friendships can be configured among the eight people, such that each person has the same number of friends, is 5320. The final answer is boxed{B}

answer:1. **Identify the sets of even and odd numbers:** - Odd numbers from 1 to 99: 1, 3, 5, 7, ldots, 99 - Even numbers from 2 to 100: 2, 4, 6, 8, ldots, 100 2. **Sum of single-digit numbers:** - For odd numbers: [ 1 + 3 + 5 + 7 + 9 = 25 ] - For even numbers: [ 2 + 4 + 6 + 8 = 20 ] Thus, the difference in the sum of single-digit numbers is: [ 25 - 20 = 5 ] 3. **Consider the numbers with tens digit: 1 to 9.** The sum of digits will be larger for the group whose sum of units' digits is greater. As calculated above: [ text{Sum of units' digits for odd numbers} = 25 ] [ text{Sum of units' digits for even numbers} = 20 ] Therefore, the sum of digits in the group of odd numbers will exceed that of the even numbers by: [ 5 times 10 = 50 ] 4. **Account for the number 100 in the even group:** - The sum of the digits of 100 is: [ 1 + 0 + 0 = 1 ] Thus, adding 100 affects the difference by 1: [ 50 - 1 = 49 ] # Conclusion: The sum of all digits used in writing the odd numbers from 1 to 100 is larger than the sum of all digits used in writing the even numbers from 1 to 100 by (49). [ boxed{49} ]