answer:Let's go through the solutions for both (a) and (b) and provide more detailed steps and explanations. Part (a): 1. **Construct the Sequence:** - Consider the sequence of numbers formed by repeating "1989" several times. [ 1989, ; 19891989, ; 198919891989, ; ldots, ; 1989 cdots 1989 quad (text{where } 1989 text{ is repeated } k text{ times}) ] 2. **Apply Dirichlet's Principle:** - Dirichlet's principle (or the pigeonhole principle) states that if there are more items than containers, at least one container must contain more than one item. 3. **Analyze Remainders:** - Let's consider the remainders when each of these numbers is divided by 1988. Since there are fewer possible remainders (1988) than numbers in our sequence, there must be at least two numbers that give the same remainder. - Suppose we have two such numbers (1989a) and (1989b) (where (a) and (b) are formed by repeating "1989" (m) and (n) times, and (m > n)) which leave the same remainder when divided by 1988. 4. **Compute the Difference:** - The difference between these two numbers will be: [ 1989underbrace{1989 cdots 1989}_{m text{times}} - 1989underbrace{1989 cdots 1989}_{n text{times}} = 1989 underbrace{1989 cdots 1989}_{(m-n) text{times}} ] - This difference is a large number formed by repeating "1989" (m-n) times followed by a series of zeros. 5. **Divisibility by 1988:** - Since the remainders are the same, their difference is divisible by 1988. - This implies that: [ 1989 cdot 10^{4(m-n)} equiv 0 ; (text{mod} ; 1988) ] - Here, (10^{4(m-n)}) is just a large power of ten. Hence, the constructed number which has "1989" repeated followed by several zeros will be divisible by 1988. **Conclusion**: A number of the form (19891989cdots19890cdots0) is divisible by (1988). [boxed{text{Proven}}] Part (b): 1. **Construct the Sequence:** - Consider the sequence of numbers formed by repeating "1988" several times. [ 1988, ; 19881988, ; 198819881988, ; ldots, ; 1988 cdots 1988 quad (text{where } 1988 text{ is repeated } k text{ times}) ] 2. **Apply Dirichlet's Principle:** - Similar to part (a), consider the remainders when each of these numbers is divided by 1989. By Dirichlet's principle, since there are more sequences than possible remainders, we must have at least two numbers with the same remainder. 3. **Analyze Remainders:** - Suppose (x) and (y) are two such numbers formed by repeating "1988" (m) and (n) times respectively (where (m > n)) that leave the same remainder when divided by 1989. 4. **Compute the Difference:** - The difference between these two numbers will be: [ 1988underbrace{1988 cdots 1988}_{m text{times}} - 1988underbrace{1988 cdots 1988}_{n text{times}} = 1988 underbrace{1988 cdots 1988}_{(m-n) text{times}} ] - This difference is a large number formed by repeating "1988" (m-n) times followed by a series of zeros. 5. **Divisibility by 1989:** - Because the remainders are the same, their difference is divisible by 1989. - This implies that: [ 1988 cdot 10^{4(m-n)} equiv 0 ; (text{mod} ; 1989) ] - Thus, a number of the form "1988" repeated several times followed by zeros is divisible by 1989. **Conclusion**: A number of the form (19881988cdots19880cdots0) is divisible by (1989). [boxed{text{Proven}}]

answer:The diagonals of a rhombus intersect each other at right angles and bisect each other. So, each diagonal is split into two halves when they meet. For one triangle, the half-lengths of the diagonals are 10 feet (half of 20) and 8 feet (half of 16). These half-lengths form the legs of right triangles inside the rhombus. Using the Pythagorean theorem to find the length of a side of the rhombus (hypotenuse of the triangle): [ text{Side length} = sqrt{(10^2 + 8^2)} = sqrt{(100 + 64)} = sqrt{164} ] Since 164 = 4 times 41, we can simplify this as: [ sqrt{164} = sqrt{4} times sqrt{41} = 2sqrt{41} ] The perimeter of the rhombus, having four equal sides, is: [ 4 times 2sqrt{41} = boxed{8sqrt{41} text{ feet}} ]

answer:1. **Weekday Distribution**: A month with 30 days includes 4 complete weeks and 2 extra days. Every day of the week will occur exactly 4 times in the first 28 days. 2. **Days Calculation**: Since (30 equiv 2 pmod{7}), the two extra days of the month depend on the weekday it starts. 3. **Examining Each Starting Day**: - **Monday**: Extra days are Monday and Tuesday. Fridays and Sundays both occur 4 times. - **Tuesday**: Extra days are Tuesday and Wednesday. Fridays and Sundays both occur 4 times. - **Wednesday**: Extra days are Wednesday and Thursday. Fridays and Sundays both occur 4 times. - **Thursday**: Extra days are Thursday and Friday. Sundays occur 4 times; Fridays occur 5 times. - **Friday**: Extra days are Friday and Saturday. Fridays occur 5 times; Sundays occur 4 times. - **Saturday**: Extra days are Saturday and Sunday. Fridays occur 4 times; Sundays occur 5 times. - **Sunday**: Extra days are Sunday and Monday. Fridays occur 4 times; Sundays occur 5 times. 4. **Equal Fridays and Sundays**: Analyzing the results, the number of Fridays equals the number of Sundays only when the month starts on: - **Monday** - **Tuesday** - **Wednesday** 5. **Conclusion**: The month can start on Monday, Tuesday, or Wednesday to ensure equal numbers of Fridays and Sundays. Therefore, the answer is 3. The final answer is boxed{C}.

answer:Joan needs to add 4 more cups of flour to the cake mixture. (7 cups required - 3 cups already added = boxed{4} cups needed)