answer:Problem: Find a four-digit number that satisfies the following conditions: 1. Adding 1 to the number results in a multiple of 15; 2. Subtracting 3 from the number results in a multiple of 38; 3. Reversing the digits of the number and adding it to the original number results in a multiple of 10. Solution: # Step 1: Define the number Let the four-digit number be ( x ), which can be expressed as ( overline{abcd} ) where ( a, b, c, d ) are its digits. # Step 2: Analyze the conditions - From Condition (1), ( 15 mid x + 1 ), meaning ( x + 1 ) is a multiple of 15. Thus, the last digit ( d ) must be either 4 or 9 (since adding 1 to a number ending in 4 or 9 gives a number ending in 5 or 0 respectively, which are multiples of 15). - From Condition (2), ( 38 mid x - 3 ), meaning ( x - 3 ) is a multiple of 38. Therefore, ( x ) must be an odd number (since ( x - 3 ) must be even). Combining these two results, since ( x ) must be odd, ( d ) must be 9 (since 4 is even). # Step 3: Determine the digit ( a ) - From Condition (3): ( 10 mid overline{abcd} + overline{dcba} ), meaning the sum of the number and its reverse is a multiple of 10. The smallest four-digit number that meets this criterion and has ( d = 9 ) is considered, and we note that ( a ) must be non-zero. By reversing ( overline{abcd} = overline{dcba} ): [ overline{abcd} + overline{dcba} = a cdot 1000 + b cdot 100 + c cdot 10 + d + d cdot 1000 + c cdot 100 + b cdot 10 + a = 1001a + 100b + 10c + d + 1000d + 100c + 10b + a = 1001(a+d) + 110(b+c) ] Given that ( d = 9 ), let's test ( a = 1 ): [ a + d = 1 + 9 = 10 ] Thus, ( x = overline{1bcd} ). # Step 4: Using Condition (2) Given that ( 38 mid x - 3 ): [ overline{1bc9} - 3 = 1000 + 100b + 10c + 9 - 3 = 1006 + 100b + 10c ] Let ( k ) be an integer such that: [ 1006 + 100b + 10c = 38k ] Solving for ( 100c + 10b ), we consider the digits must align to form a valid four-digit number: [ 26 < k < 53 text{ (since the resultant can only assume specific values between 1006 and 1996)} ] # Step 5: Simplification and Verification Given: [ l = 26,27,32,37,42,47,52 text{ are valid as } l ] Checking for these values: Using ( l = 27,37,42,52 ): [ 38 cdot 27 + 3 = 1026 + 3 = 1023 text{ (invalid for a four-digit number)} ] Repeating for other values we check: [ text{ only valid for } 1409 , text{ Is valid four-digit number: }text{ which satisfies all criterion. }] So, we have: [ x = 1409 text{ and } 1979 ] # Conclusion: [boxed{1409, 1979} ]

answer:To determine the possible values of x, we examine the total amounts spent by the 7th and 8th graders: - 7th graders: 36 - 8th graders: 90 Next, we find the greatest common divisor (GCD) of the two amounts, as x must be a common divisor: - Using the Euclidean algorithm: - 90 = 36 times 2 + 18 - 36 = 18 times 2 + 0 Thus, the GCD of 36 and 90 is 18. Now, we determine the divisors of 18 (possible values for x). The divisors of 18 are 1, 2, 3, 6, 9, 18. - **Count the number of divisors**: - There are 6 divisors of 18. Thus, there are 6 possible values for the ticket price x. textbf{6} Conclusion: The solution is valid and confirms that there are 6 ways the price per ticket could be set such that the total amounts paid by both groups are integer multiples of x. The final answer is boxed{textbf{(C)} 6}

answer:To solve the quadratic equation x^2 - 9x - 22 = 0, we start by attempting to factor it. We look for two numbers that multiply to -22 and add to -9. The numbers -11 and 2 satisfy these conditions. Therefore, we can factor the quadratic equation as: x^2 - 9x - 22 = (x - 11)(x + 2) = 0. Setting each factor equal to zero gives the solutions: x - 11 = 0 quad Rightarrow quad x = 11, x + 2 = 0 quad Rightarrow quad x = -2. The distinct solutions to this equation are 11 and -2. The larger of these two solutions is boxed{11}.

answer:1. Let the initial length of the rectangular plot be ( a ) and the initial width be ( b ). 2. The initial area of the plot is given by: [ A_{text{initial}} = a cdot b ] 3. The length is increased by ( 35% ). Therefore, the new length is: [ a_{text{new}} = a times (1 + 0.35) = a times 1.35 ] 4. The width is decreased by ( 14% ). Therefore, the new width is: [ b_{text{new}} = b times (1 - 0.14) = b times 0.86 ] 5. The new area of the plot is: [ A_{text{new}} = a_{text{new}} cdot b_{text{new}} = (a times 1.35) times (b times 0.86) = 1.35a times 0.86b ] 6. Simplifying, we get: [ A_{text{new}} = 1.35 times 0.86 times a times b = 1.161 times a times b ] 7. Comparing the new area with the initial area: [ frac{A_{text{new}}}{A_{text{initial}}} = frac{1.161 times a times b}{a times b} = 1.161 ] 8. This means the area has increased by ( 16.1% ), since ( 1.161 ) in percentage form is ( 116.1% ). Conclusion: The area of the plot has changed by ( 16.1% ). [ boxed{16.1%} ]