answer:To prove that some knight was left without wine, we will proceed by contradiction. Assume that after the two claps, every knight still has exactly one glass of wine. We will show that this leads to a contradiction. 1. **Step 1: No two consecutive knights have red wine.** - Suppose there are two consecutive knights with red wine, say knights ( A ) and ( B ). - After the first clap, knight ( A ) will take the glass from the left neighbor of ( A ), and knight ( B ) will take the glass from the left neighbor of ( B ). - This means that the knight to the left of ( A ) will lose their glass, and the knight to the left of ( B ) will lose their glass. - Since ( A ) and ( B ) are consecutive, the knight to the left of ( B ) is the same as the knight to the left of ( A ), leading to a contradiction because one knight cannot lose two glasses. - Therefore, no two consecutive knights can have red wine. 2. **Step 2: No three consecutive knights have white wine.** - Suppose there are three consecutive knights with white wine, say knights ( C ), ( D ), and ( E ). - After the second clap, knight ( C ) will give their glass to the left neighbor of their left neighbor, which is knight ( E ). - Similarly, knight ( D ) will give their glass to the left neighbor of their left neighbor, which is knight ( F ) (the knight to the left of ( E )). - Knight ( E ) will give their glass to the left neighbor of their left neighbor, which is knight ( G ) (the knight to the left of ( F )). - This means that knight ( E ) will end up with two glasses, leading to a contradiction. - Therefore, no three consecutive knights can have white wine. 3. **Step 3: A knight with white wine does not sit between two knights with red wine.** - Suppose there is a knight with white wine, say knight ( H ), sitting between two knights with red wine, say knights ( I ) and ( J ). - After the first clap, knight ( I ) will take the glass from the left neighbor of ( I ), and knight ( J ) will take the glass from the left neighbor of ( J ). - This means that knight ( H ) will lose their glass to knight ( I ), and knight ( J ) will take the glass from the left neighbor of ( J ), leading to a contradiction because knight ( H ) will end up without a glass. - Therefore, a knight with white wine cannot sit between two knights with red wine. 4. **Step 4: The number of knights is divisible by 3.** - From Step 1, we know that the sequence of wine glasses must alternate between red and white, with no two consecutive red wines and no three consecutive white wines. - From Step 2, we know that the sequence must have at most two consecutive white wines. - From Step 3, we know that a knight with white wine cannot sit between two knights with red wine. - Therefore, the sequence of wine glasses must take the form ( RW^{k_1}RW^{k_2}RW^{k_3} cdots ), where ( k_i < 3 ) and ( k_i > 1 ). - Hence, ( k_i = 2 ) for all ( i ), so the sequence is ( RWW ) repeated over and over again. - Since there are 50 knights, and 50 is not divisible by 3, this leads to a contradiction. Since 50 is not divisible by 3, our assumption that every knight ends up with exactly one glass of wine is false. Therefore, some knight must be left without wine. (blacksquare)

answer:The 210 cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid that is one unit shorter in each dimension. If the original block has dimensions l times m times n, we must have (l - 1)times(m-1) times(n - 1) = 210. The prime factorization of 210 = 2cdot3cdot5cdot7 gives us several possibilities; for instance, l - 1 = 6, m - 1 = 5, and n - 1 = 7. We aim to minimize l cdot m cdot n by choosing dimensions close to each other, leading to a smaller product. If (l - 1, m - 1, n - 1) = (6, 5, 7), then l = 7, m = 6, n = 8. Thus, the total number of cubes in the block is l cdot m cdot n = 7 cdot 6 cdot 8 = boxed{336}.

answer:Given that we need to find a number which is both a multiple of 7 and a multiple of 5, and when 9 is added to it, the result is a prime number, we will go through the detailed steps to find the smallest such number. 1. Identify the common multiple of 7 and 5. [ text{The least common multiple (LCM) of 7 and 5 is } [7, 5] = 35. ] 2. Check if adding 9 to the smallest common multiple results in a prime number. [ 35 + 9 = 44 ] [ 44 text{ is not a prime number, as it is divisible by 2} ] 3. Proceed to the next multiples of 35 and check each one: [ 35 times 2 = 70 ] [ 70 + 9 = 79 ] [ 79 text{ is a prime number, as it has no positive divisors other than 1 and itself} ] 4. Verify if any smaller common multiples could give a correct result. Since 70 is the next multiple beyond 35, by confirmation: The smallest number which is a multiple of 35, and when increased by 9 gives a prime is indeed: [ 70 ] Conclusion: [ boxed{70} ]

answer:Rachel had to complete 8 pages of math homework. She had to complete 6 more pages of reading homework than math homework. So, the number of pages of reading homework she had to complete is: 8 pages (math homework) + 6 pages (additional reading homework) = 14 pages (reading homework) Therefore, Rachel had to complete boxed{14} pages of reading homework.